An Extension of a Theorem of Marczewski

In [3], Marczewski proved that convergence almost surely (a.s.) is equivalent to convergence in probability precisely when {\Omega} is the (at most) countable union of disjoint atoms (i.e. purely atomic). In particular, convergence a.s. is defined by a topology when {\Omega} is purely atomic. It is well known that convergence a.s. need not be defined by a topology when {\Omega} is not purely atomic. Try to construct a counter-example on the unit interval with Borel {\sigma}-algebra and Lebesgue measure.

Today, I want to give an extension of Marczewski’s result which characterizes when convergence a.s. is defined by a topology. Using arguments similar to those in my last post, we will prove the following theorem.

Theorem 1 Let {(X,\mathcal{A},\mu)} be a finite measure space. There exists a topology {\tau} on the space of measurable functions {L^{0}(\mu)} coinciding with convergence a.s. if and only if {X} is purely atomic.

Proof: Marczewski’s result proves sufficiency, so we prove necessity. If {X} is not purely atomic, then there exists a measurable subset {A} which is not an atom and on which {\mu} satisfies the intermediate value property. In particular, for each {n\in\mathbb{N}}, there exists a partition of {A} into {m_{n}} measurable sets {A_{1}^{n},\ldots,A_{m_{n}}^{n}} with {0<\mu(A_{i}^{n})\leq n^{-1}}. Consider the sequence {f_{k}} formed by indicator functions of the {A_{i}^{n}}:

\displaystyle 1_{A_{1}^{1}},\ldots,1_{A_{m_{1}}^{1}},1_{A_{1}^{2}},\ldots,1_{A_{m_{2}}^{2}},1_{A_{1}^{3}},\ldots,

I claim that {f_{k}} converges to zero in measure. Indeed, {f_{k}} is supported on {A_{i}^{n}}, for some {n} and {1\leq i\leq m_{n}}. Note tht {\mu(A_{i}^{n})\rightarrow 0} as {n\rightarrow\infty} and {k\rightarrow\infty} implies {n\rightarrow\infty}. For any {x\in X}, {f_{k}(x)=1} for infinitely many {k}, whence {f_{k}} does not converge to zero a.s. Recall that if every subsequence of {f_{k}} has a further subsequence which converges to some {f} in a topology {\tau}, then {f_{k}\rightarrow f} in {\tau}. But convergence in measure implies convergence of a subsequence a.s. to the same limit, which is a contradiction. \Box

  1. S.K. Berberian, Lectures in Functional Analysis and Operator Theory, Spring Verlag 1974.
    1. V.I. Bogachev, Measure Theory Volume I, Springer 2007.
  2. E. Marczewski, “Remarks on the Convergence of Measurable Sets and Measurable Functions”, Colloquium Math. (1955), 118-124.
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On the Normability of the Space of Measurable Functions

Suppose {(X,\mathcal{A},\mu)} is a finite measure space, and let {L^{0}(X)} denote the space of real- or complex-valued measurable functions on {X}. We can topologize {L^{0}(X)} by saying a sequence {f_{n}\rightarrow f} in {L^{0}(X)} if and only if

\displaystyle \lim_{n\rightarrow\infty}\mu\left\{\left|f_{n}-f\right|\geq\varepsilon\right\}=0,\indent\forall\varepsilon>0

We say that {f_{n}} converges in measure to {f}. It is an easy exercise to verify that convergence in measure can be realized by the metric topologies

\displaystyle \rho_{1}(f,g):=\int_{X}\dfrac{\left|f-g\right|}{1+\left|f-g\right|}d\mu,\indent\rho_{2}(f,g):=\int_{X}\min\left(\left|f-g\right|,1\right)d\mu

and in fact, these metrics are complete.

For {p\geq 1}, the {L^{p}} spaces have a metric topology which is actually given by the {L^{p}} norm. The integrands in the expressions for {\rho_{1}} and {\rho_{2}} are not homogeneous in their arguments, so we should not expect these metrics to be norms. However, there could still be a norm lurking out there on {L^{0}(X)} with an equivalent norm topology. This past summer, I became interested in the following question after reading a comment on Math.SE: What are necessary and sufficient conditions for {L^{0}(X)} to be normable (i.e., there exists a norm {\left\|\cdot\right\|} which induces the same topology on {L^{0}(X)} as convergence in measure)?

For “typical” spaces such as the measurable functions on {[0,1]}, it is easy to disprove normability with simple counterexamples. Suppose there is a norm {\left\|\cdot\right\|} on {L^{0}[0,1]}. Consider the measurable functions defined by

\displaystyle f_{n}:=\left\|1_{A_{n}}\right\|^{-1}1_{A_{n}},\indent A_{n}:=\left\{\frac{n-1}{n}\leq x<\frac{n}{n+1}\right\}

It is evident that {f_{n}} converges to zero in measure since {\left|A_{n}\right|\rightarrow 0}, as {n\rightarrow\infty}. However, {\left\|f_{n}\right\|=1} for all {n}. Normability fails because we able to find measurable subsets of {[0,1]} of arbitrarily small measure. This is was possible because the unit interval with Lebesgue measure has no atoms.

It turns out that my question was answered by A.J. Thomasian in 1957, who showed that the space of random variables on some probability space {(\Omega,\mathcal{F},P)} is normable if and only if {\Omega} is the finite union of disjoint atoms. It follows from normalizing a finite measure that Thomasian’s result holds for finite measure spaces.

The other day, I was thinking about Thomasian’s result in terms of the Kolmogorov normability criterion, which says that a separated topological vector space (tvs) is normable if and only if it has a bounded, convex neighborhood of zero. And it seemed to be that this was the right context for his result, rather than the specific language of probability theory. So today I’m going to relate the two.

First, we prove a technically convenient characterization of boundedness for tvs.

Lemma 1 A subset {A} of a tvs {(X,\tau)} is bounded if and only if for any sequence of elements {x_{n}\in A} and scalars {\lambda_{n}\rightarrow 0}, the sequence {\lambda_{n}x_{n}\rightarrow 0}.

Proof: Suppose {A} is bounded. Let {U} be an open neighborhood of zero, and let {\lambda_{n}} a sequence of scalars tending to {0}. Without loss of generaliy, we may assume that {U} is balanced, so that {A\subset\lambda_{n}^{-1}U} for almost all {n}. Given a sequence {x_{n}\in A}, it follows that {\lambda_{n}x_{n}\in U} for almost all {n}.

If {A} is unbounded, then there exists an open neighborhood {U} of zero such that for all {n\in\mathbb{N}}, there is an element {x_{n}\in A\setminus (n\cdot U)\Leftrightarrow n^{-1}x_{n}\notin U}. \Box

Suppose {X} is not the finite union of disjoint atoms. Recall that we can write the space {X} as {X=A\cup\bigcup_{n=1}^{\infty}A_{n}}, where {A} is either of positive measure and not an atom or of measure zero and {\left\{A_{n}\right\}} are pairwise disjoint atoms. There are two cases to consider.

  1. {\mu(A)>0}: By the intermediate value property for measures, for any {\varepsilon>0} there exists a measurable set {B\subset A} with {\mu(B)=\varepsilon}. Define a sequence of measurable functions by {f_{n}:=n^{2}\cdot1_{B}}. Given any open neighborhood {U} of zero in {L^{0}(X)}, we can take {\varepsilon} sufficiently small so that {f_{n}\in U}. But {n^{-1}\cdot f_{n}} does not converge to zero in measure, so by Lemma 1, {U} is not bounded.
  2. {\mu(A_{n})>0} for all {n\in\mathbb{N}}: It follows from {\sigma}-additivity that {\mu(A_{n})\rightarrow 0} as {n\rightarrow\infty}. Given {\varepsilon>0}, we can find {B:=A_{n}} with {\mu(A_{n})<\varepsilon}. The argument in the first case shows any open neighborhood {U} of zero is not bounded.

Now suppose {X} is the finite union of disjoint atoms {A_{1},\ldots,A_{n}}. Any measurable function {f} is a.s. constant on each {A_{j}}, so we can write {f=\sum_{j=1}^{n}a_{j}1_{A_{j}}}, where {a_{1},\ldots,a_{n}} are constants. Set {\varepsilon:=\min_{j}\mu(A_{j})/2}. With {\rho_{1}} as above, we have

\displaystyle \varepsilon>\rho_{1}(f,0)=\sum_{j=1}^{n}\frac{a_{j}}{1+a_{j}}\mu(A_{j})\geq\sum_{j\atop{a_{j}\neq0}}\mu(A_{j})\geq\sum_{j\atop{a_{j}\neq 0}}\varepsilon,

which implies that {\left\{j:a_{j}\neq 0\right\}=\emptyset}. So {\left\{0\right\}=\left\{f\in L^{0}(X):\rho_{1}(f,0)<\varepsilon\right\}} is an open neighborhood of zero which is trivially bounded, whence by Kolmogorov’s criterion, {L^{0}(X)} is normable.

For completeness, it’s worth mentioning that Thomasian extended an earlier result of Marczewski which showed that convergence in probability implies convergence a.s. if and only if {\Omega} is the (at most) countable union of disjoint atoms. So in particular, if {L^{0}(X)} is normable, then convergence in measure implies converge a.s.

1. A.J. Thomasian, “Metrics and Norms on Spaces of Random Variables”, The Annals of Mathematical Statistics 28 (1957), 512-514.

2. E. Marczewski, “Remarks on the Convergence of Measurable Sets and Measurable Functions”, Colloquium Math. (1955), 118-124.

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Harmonic Functions: Weyl Lemma

In this post, we prove a generalization of the result that any (clasically) harmonic function {u} defined on a bounded, open set {\Omega\subset\mathbb{R}^{n}} is analytic. If {u\in L_{loc}^{1}(\Omega)} is weakly harmonic, i.e. {\int(\Delta u)\varphi=\int u\Delta\varphi=0} for any compactly supported {C^{\infty}} function {\varphi} (the space of such functions is denoted {C_{0}^{\infty}(\mathbb{R}^{n})}, then {u} is strongly or classically harmonic and therefore equal a.e. to an analytic function. This result is due to Hermann Weyl and appears to be known in the literature as the Weyl Lemma.

The proof I will give is based on an outline in a problem set I found online.

Let {\eta:\mathbb{R}^{n}\rightarrow\mathbb{R}} be the radial mollifier defined by

\displaystyle \eta(x):=\begin{cases} C\exp\left(\dfrac{1}{\left|x\right|^{2}-1}\right) & {\left|x\right|<1}\\ 0 & {\left|x\right|\geq 1} \end{cases}

where {C} is the normalizing constant. Define an approximate identity {\eta_{\varepsilon}=\varepsilon^{-n}\eta(-/\varepsilon)}. I leave it to the reader to verify that {\eta} is indeed {C^{\infty}} (this is a standard calculus problem) and that {\left\{\eta_{\varepsilon}\right\}_{\varepsilon>0}} is an approximate identity.

For {\varepsilon>0}, define an open subset of {\Omega} by {U_{\varepsilon}:=\left\{x\in\Omega:d(x,\partial\Omega)>\varepsilon\right\}} (i.e. the set of points in {\Omega} which are at least {\varepsilon} distance from the boundary of {\Omega}). As {\eta_{\varepsilon}} is supported on {B(0,\varepsilon)}, the convolution {u_{\varepsilon}:=u\ast\eta_{\varepsilon}} is well-defined. Furthermore, {u_{\varepsilon}} is {C^{\infty}}. Iif {x\in U_{\varepsilon}}, then {B(x,\varepsilon)\subset\Omega}, so {u_{\varepsilon}} is integrable on {U_{\varepsilon}}. By Young’s inequality, we have the estimate

\displaystyle \left\|u_{\varepsilon}\right\|_{L^{1}(U_{\varepsilon})}\leq\left\|u\right\|_{L^{1}(\Omega)}\left\|\eta_{\varepsilon}\right\|_{L^{1}(\Omega)}=\left\|u\right\|_{L^{1}(\Omega)}

I claim that {u_{\varepsilon}} is (classicaly) harmonic on {U_{\varepsilon}}. Indeed, as {\eta_{\varepsilon}\in C_{0}^{\infty}}, we can use the Lebesgue dominated convergence theorem to differentiate inside the integral in the definition of {u_{\varepsilon}} (look at the Taylor expansion of {\eta_{\varepsilon}}) to obtain

\displaystyle \Delta u_{\varepsilon}(x)=\int_{\Omega}[\Delta\eta_{\varepsilon}](x-y)u(y)dy=0,

since {\eta_{\varepsilon}(x-\cdot)\in C_{0}^{\infty}} and {u} is weakly harmonic. Consequently, {u_{\varepsilon}} has the mean value property.

We now want to extract a sequence of the {u_{\varepsilon}} which converges to some continuous function {v} and then argue that this function {v} must be harmonic. To accomplish the first task, we use the Arzelà-Ascoli theorem. First, we have that {\left\{u_{\varepsilon}\right\}} is equicontinuous and uniformly bounded. Fix an R>0, and set V:=U_{R}.

Lemma 1 The family {\left\{u_{\varepsilon}\right\}_{\varepsilon>0}} is uniformly bounded on {\overline{V}} for {0<\varepsilon<R/2}.

Proof: For {\varepsilon>0}, {u_{\varepsilon}} satisfies the mean value property. For {B(x,r)\subset\Omega},

\displaystyle \left|u_{\varepsilon}(x)\right|=\dfrac{1}{\left|B(x,r)\right|}\left|\int_{B(x,r)}u_{\varepsilon}(y)dy\right|\leq\dfrac{1}{\left|B(x,r)\right|}\int_{B(x,r)}\left|u_{\varepsilon}(y)\right|dy

If {x\in\overline{V}}, then {B(x,R/2)\subset\Omega}, so we obtain the estimate

\displaystyle \left|u_{\varepsilon}(x)\right|\leq\dfrac{1}{\left|B(x,R/2)\right|}\int_{B(x,R/2)}\left|u_{\varepsilon}(y)\right|dy=\dfrac{2^{n}}{\alpha(n)R^{n}}\leq\dfrac{2^{n}}{\alpha(n)R^{n}}\left\|u\right\|_{L^{1}(\Omega)}

for all {x\in\overline{V}}. \Box

Lemma 2 The family {\left\{u_{\varepsilon}\right\}_{\varepsilon>0}} is equicontinuous on {\overline{V}} for {0<\varepsilon<R/2}.

Proof: For any {x,y\in\overline{V}} and {0<\varepsilon<R/2}.

\displaystyle \begin{array}{lcl} \displaystyle\left|u_{\varepsilon}(x)-u_{\varepsilon}(y)\right|&=&\displaystyle\left|\dfrac{1}{\left|B(x,s)\right|}\int_{B(x,s)}u_{\varepsilon}(z)dz-\dfrac{1}{\left|B(y,s)\right|}\int_{B(y,s)}u_{\varepsilon}(z)dz\right|\\ [2 em]&=&\displaystyle\dfrac{1}{\alpha(n)s^{n}}\int_{B(x,s)}u_{\varepsilon}(z)dz-\int_{B(y,s)}\left|u_{\varepsilon}(z)dz\right|\\ [2 em]&\leq&\displaystyle\dfrac{1}{\alpha(n)s^{n}}\left[\int_{B(x,s)\setminus B(y,s)}\left|u_{\varepsilon}(z)\right|dz+\int_{B(y,s)\setminus B(x,s)}\left|u_{\varepsilon}(z)\right|dz\right] \end{array}

As {u_{\varepsilon}} is uniformly bounded on {\overline{V}} for all {0<\varepsilon<R/2} (with constant independent of {\varepsilon}) and the Lebesgue measure of the two domains depends only on {\left|x-y\right|} and goes to zero as {\left|x-y\right|\rightarrow 0}, we obtain equicontinuity. \Box

By the Arzelà-Ascoli theorem, there exists a sequence {u_{\varepsilon_{k}}} which converges uniformly to a continuous function {v\in C(\overline{V})}. Replacing {\left\{\eta_{\varepsilon}\right\}_{\varepsilon}} with a countable subset with indices tending to zero, we may assume that {\varepsilon_{k}\rightarrow 0}.

To prove that {v} is harmonic, we need the following lemma.

Lemma 3 If {v\in L_{loc}^{1}(\Omega)} has the mean value property, then {v} is (classically) harmonic.

Proof: Set {\chi_{r}(x):=\left|B(0,r)\right|^{-1}\chi_{B(0,r)}(x)}. Observe that by the Lebesgue dominated convergence theorem {v\ast\chi_{r}} is continuous. By the mean value property,

\displaystyle v\ast\chi_{r}(x)=\dfrac{1}{\left|B(0,r)\right|}\int_{\Omega}v(x-y)\chi_{B(0,r)}(y)dy=\dfrac{1}{\left|B(x,r)\right|}\int_{B(x,r)}v(z)dz=v(x),

where we use the translation invariance of the Lebesgue measure to obtain the penultimate equality. Hence, {v\in C(\Omega)}.

We now show that {v} is equal to a smooth function a.e. on {\Omega}. Set {v_{\varepsilon}:=v\ast\eta_{\varepsilon}}. Then

\displaystyle \begin{array}{lcl} v_{\varepsilon}(x)&=&\displaystyle\int_{B(0,\varepsilon)}v(x-y)\eta_{\varepsilon}(y)dy\\ [2 em]&=&\displaystyle\varepsilon^{-n}\int_{B(0,\varepsilon)}v(x-y)\eta(y/\varepsilon)dy\\ [2 em]&=&\displaystyle\int_{B(0,1)}v(x-\varepsilon y)\eta(y)dy\\ [2 em]&=&\displaystyle\int_{0}^{1}\int_{\partial B(0,s)}v(x-\varepsilon z)\eta(z)dS(z)ds\\ [2 em]&=&\displaystyle\int_{0}^{1}v(x)n\alpha(n)s^{n-1}\eta(s)ds\\ [2 em]&=&\displaystyle v(x)\cdot\int_{B(0,\varepsilon)}\eta_{\varepsilon}(y)dy=v(x) \end{array}

In particular, {v\in C^{2}(\Omega)}, so by the converse to the mean value property (see [Evans] Theorem 3, pp. 26), {v} is harmonic. \Box

As {\left\{\eta_{\varepsilon}\right\}_{\varepsilon>0}} is an approximate identity, {\left\|u_{\varepsilon}-u\right\|_{L^{1}}\rightarrow 0} as {\varepsilon\rightarrow 0}. Hence, {v=u} a.e. As we proved last time, u is therefore equal to an analytic function a.e.

Next time, we will extend the Weyl lemma to distributions that satisfy the Laplace equation.

  1. L.C. Evans, Partial Differential Equations (Second Edition), AMS, 2010.
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Harmonic Functions: Regularity

I have a confession: I have never actually taken a PDE course. I took an ODE and PDE class offered by the Applied Math department my freshman year, but it wasn’t really a mathematics class–most of the students were physics or engineering sciences concentrators. I have learned a fair amount about PDE in various analysis classes and a mini-course at Princeton’s Analysis and Geometry summer session, but with the exception of the latter, these were not devoted to PDE. So while I impose myself on Georgia Tech’s math department this Fall as a non-degree student, I thought would take their graduate PDE course.

Functions {u:\Omega\rightarrow\mathbb{R}} which arise as solutions of Laplalce’s equation {\Delta u=0} on some open set {\Omega\subset\mathbb{R}^{n}}, called harmonic functions have a lot of neat properties, many of them analogous to holomorphic functions on {\mathbb{C}}. And indeed, harmonic functions on {\mathbb{R}^{2}} are the real parts of holomorphic functions.

Today, I want to talk about the regularity of harmonic functions that comes as a consequence of their satisfying the mean value property:

\displaystyle u(x_{0})=\dfrac{1}{S(\partial B(x_{0},r))}\int_{\partial B(x_{0},r)}u(y)dS(y)=\dfrac{1}{\left|B(x_{0},r)\right|}\int_{B(x_{0},r)}u(y)dy

for all open balls with closure {\overline{B}(x_{0},r)\subset\Omega}. (Here, {S(\cdot)} denotes the surface measure and {\left|\cdot\right|} denotes the {n}-dimensional Lebesgue measure.)

Some intuition for why this averaging property might give us regularity can be gained from thinking about time series, say for a stock price. The path of stock price over some time period looks differentiable almost nowhere. It may not even be continuous, as stock prices can gap up or down between market closures and opens. However, if we consider a moving average over some window (e.g. 30 days), we get a path that looks `smoother’.

With little effort, we get that a harmonic function, which is only a priori {C^{2}(\Omega)}, is in fact smooth (i.e. {C^{\infty}}). But as you learned in calculus, smoothness does not imply (real) analyticity (Try to come up with a counterexample). Nevertheless, harmonic functions are analytic. To prove this we will only need three things: any derivative of a harmonic function is harmonic, a local estimate for the {k^{th}} order derivative of a harmonic function in terms of its {L^{1}} norm, and Taylor’s theorem.

Our plan of attack is to prove smoothness is as follows: 1) prove smoothness, 2) prove the local estimate, and 3) prove analyticity. But first, a brief digression on notation. I will use the multi-index notation {\alpha=(\alpha_{1},\ldots,\alpha_{n})}, where {\alpha_{i}\geq 0} is integral. We say that {\alpha} is of order {k} if {\left|\alpha\right|:=\sum_{i}\alpha_{i}=k}. We define

\displaystyle \alpha!=\alpha_{1}!\cdots\alpha_{n}!,\indent {\left|\alpha\right|\choose\alpha}=\dfrac{\left|\alpha\right|!}{\alpha!},\indent x^{\alpha}=x_{1}^{\alpha_{1}}\cdots x_{n}^{\alpha_{n}}

and for any sufficiently regular function {u}, we define {D^{\alpha}u=\partial_{1}^{\alpha_{1}}\cdots\partial_{n}^{\alpha_{n}}u}.

In what follows, {\Omega\subset\mathbb{R}^{n}} will denote an open set.

Recall that if {u:\Omega\rightarrow\mathbb{R}} is harmonic, then it satisfies the mean value property on {\Omega}. To prove {u} is smooth, it suffices to prove a more general result: if a locally integrable function {u} satisfies the mean value property for each ball {B(x,r)\subset\Omega}, then {u\in C^{\infty}}

Proposition 1 If {u:\Omega\rightarrow\mathbb{R}} in {L_{loc}^{1}(\Omega)} satisfies the mean value property, then {u} is {C^{\infty}}.

Proof: We will use the mean value property to show that {u} is equal to smooth function on an open neighborhood of every point {x_{0}\in\Omega}. Let {\left\{\eta_{\varepsilon}\right\}_{\varepsilon>0}} be a {C^{\infty}} compactly supported, radial approximate identity (my survey on Fourier analysis has some notes on these; also, Appendix C.5 [Evans]). Define a {C^{\infty}} function {u_{\varepsilon}:=\eta_{\varepsilon}\ast u} on {\Omega_{\varepsilon}:=\left\{x\in U|d(x_{0},\partial \Omega)>\varepsilon\right\}}. For {x\in\Omega_{\varepsilon}},

\displaystyle \begin{array}{lcl} \displaystyle u_{\varepsilon}(x_{0})&=&\displaystyle\int_{U}\eta_{\varepsilon}(x-y)u(y)dy\\ [1.5 em]&=&\displaystyle\dfrac{1}{\varepsilon^{n}}\int_{B(x,\varepsilon)}\eta\left(\dfrac{\left|x-y\right|}{\varepsilon}\right)u(y)dy\\ [1.5 em]&=&\displaystyle\dfrac{1}{\varepsilon^{n}}\int_{0}^{\varepsilon}\eta\left(\dfrac{r}{\varepsilon}\right)\left(\int_{\partial B(x,r)}udS\right)dr \end{array}

where the second line is just the definition of {\eta_{\varepsilon}} and the third line is Fubini’s theorem. Observe that for {0<r<\varepsilon} and {x\in \Omega_{\varepsilon}}, {B(x,\varepsilon)\subset\Omega}. Using the mean value property, we obtain

\displaystyle u_{\varepsilon}(x)=\dfrac{u(x)}{\varepsilon^{n}}\int_{0}^{\varepsilon}\eta\left(\dfrac{r}{\varepsilon}\right)n\alpha(n)r^{n-1}dr=u(x)\underbrace{\int_{B(0,\varepsilon)}\eta_{\varepsilon}(y)dy}_{=1}=u(x)

\Box

Lemma 2 For any multi-index {\alpha=(\alpha_{1},\ldots,\alpha_{n})} and harmonic function {u}, {D^{\alpha}u} is harmonic (note this differentiation is well-defned since {u} is necessarily smooth).

Proof: Fix a multi-index {\alpha} and set {v:=D^{\alpha}u}. Using the equality of mixed partials, we see that

\displaystyle\partial_{x_{i}}^{2}v=D^{\alpha}\partial_{x_{i}}^{2}u,

whence by linearity,

\displaystyle\Delta v=\sum_{i=1}^{n}\partial_{x_{i}}^{2}v=D^{\alpha}\left[\sum_{i=1}^{n}\partial_{x_{i}}^{2}u\right]=D^{\alpha}\left[\Delta u\right]=0

\Box

Proposition 3 If {u:\Omega\rightarrow\mathbb{R}} is harmonic, then for any multi-index {\alpha}, with {\left|\alpha\right|=k},

\displaystyle \left|D^{\alpha}u(x_{0})\right|\leq\dfrac{n^{k}e^{k-1}k!}{r^{k}}\left\|u\right\|_{L^{\infty}(B(x_{0},r))}

for all balls {B(x_{0},r)\subset\Omega}.

Note that without the result that {u} is {C^{\infty}}, it would not make sense to write {D^{\alpha}u} for an arbitrary multi-index.

Proof: The proof will be by induction on the order {\left|\alpha\right|}. It is clear from the mean value property that {C_{0}} satisfies Suppose we have a constant {C_{k}} such that

\displaystyle \left|D^{\alpha}u(x)\right|\leq\dfrac{C_{k}}{r^{k}}\sup_{\overline{B}_{r}(x)}\left|u\right|

for all balls with {\overline{B}(x,r)\subset\Omega}, for all multi-indices {\alpha} of order {k}. Suppose {\left|\alpha\right|=k}. Then for some {i}, we can write {D^{\alpha}=\partial_{i}D^{\beta}}, where {\left|\beta\right|=k}. As {D^{\beta}u} is harmonic, the mean value property gives

\displaystyle \begin{array}{lcl}\left|D^{\alpha}u(x)\right|&=&\displaystyle\dfrac{1}{\left|B(x,s)\right|}\left|\int_{B(x,s)}\partial_{i}D^{\beta}u(y)dy\right|\\ &=&\displaystyle\dfrac{1}{\left|B(x,s)\right|}\left|\int_{\partial B(x,s)}D^{\beta}u(y)\cdot\nu_{i}dS(y)\right| \end{array}

for {0<s<r}. If {y\in \partial B(x,s)}, then {B(y,r-s)\subset B(x,r)}. So by the induction hypothesis,

\displaystyle \left|D^{\alpha}u(x)\right|\leq\dfrac{1}{\left|B(x,s)\right|}\dfrac{C_{k}}{\left|r-s\right|^{k}}\int_{\partial B(x,s)}\left\|D^{\beta}u\right\|_{L^{\infty}(B(x,r))}dS(y)

Writing {s=\theta r}, where {0<\theta<1}, the RHS simplifies to

\displaystyle \left|D^{\alpha}u(x)\right|\leq\dfrac{nC_{k}}{\theta (1-\theta)^{k}r^{k+1}}\left\|D^{\beta}u\right\|_{L^{\infty}(B(x,r))}

I leave it to the reader to verify that {\theta=1/k} minimizes the RHS, whence

\displaystyle \left|D^{\alpha}u(x)\right|\leq n(k+1)\left(\dfrac{k+1}{k}\right)^{k}\dfrac{C_{k}}{r^{k+1}}\leq en(k+1)\dfrac{C_{k}}{r^{k+1}}

Set {C_{k+1}:=en(k+1)C_{k}}. \Box

Theorem 4 If {u:\Omega\rightarrow\mathbb{R}} is harmonic, then {u} is real analytic.

Proof: As {u} is smooth, it has a Taylor expansion about {x\in\Omega}

\displaystyle u(x+h)=\sum_{\left|\alpha\right|\leq k-1}\dfrac{D^{\alpha}u(x)}{\alpha!}h^{\alpha}+R_{k}(x,h)

where {R_{k}(x,h)=O(\left\|h\right\|^{k})}. Actually, one form of Taylor’s theorem gives the remainder

\displaystyle R_{k}(x,h)=\sum_{\left|\alpha\right|=k}\dfrac{D^{\alpha}u(x+th)}{\alpha!}h^{\alpha}

for {0<t<1}. Let {r>0} be such that {\overline{B}(x,2r)\subset\Omega}. From our work above, we have the estimate

\displaystyle \left|D^{\alpha}u(x+th)\right|\leq\dfrac{n^{k}e^{k-1}k!}{r^{k}}\sup_{\overline{B}(x+th,r)}\left|u(x)\right|\leq\dfrac{n^{k}e^{k-1}k!}{r^{k}}\underbrace{\sup_{\overline{B}(x,2r)}\left|u(x)\right|}_{:=M}

Substituting this result into the expression for {R_{k}(x,h)}, we obtain the estimate

\displaystyle \begin{array}{lcl}\displaystyle\left|R_{k}(x,h)\right|&\leq&\displaystyle\dfrac{Mn^{k}e^{k-1}\left\|h\right\|^{k}k!}{r^{k}}\left(\sum_{\left|\alpha\right|=k}\dfrac{1}{\alpha!}\right)\\ [1.5 em]&=&\displaystyle\dfrac{Mn^{k}e^{k-1}\left\|h\right\|^{k}k!}{r^{k}}\dfrac{n^{k}}{k!}\\ [1.5 em]&=&\displaystyle\dfrac{M}{e}\left(\dfrac{n^{2}\left\|h\right\|e}{r}\right)^{k}\\ \end{array}

If {\left\|h\right\|^{-1}<n^{2}e/r}, then {R_{k}(x,h)\rightarrow 0} as {k\rightarrow\infty}. \Box

In the next installment on harmonic functions, we will prove the Weyl Lemma which states that any weak solution {u} of the Laplace equation is equal a.e. to a harmonic function. We will also extend the Weyl Lemma to distributions which satisfy the Laplace equation.

L.C. Evans, Partial Differential Equations (Second Edition), AMS, 2010.

G.B. Folland, Introduction to Partial Differential Equations (Second Edition), Princeton UP, 1995.

J.K. Hunter, PDE Notes Chapter 2, https://www.math.ucdavis.edu/~hunter/pdes/ch2.pdf, Accessed 9/5/14.

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Kolmogorov Normability Criterion

This past week I have been refreshing my knowledge of topological vector spaces (tvs) while reading some papers on generalizations of the Mazur-Ulam theorem to metrizable tvs. I intend to upload a typed set of notes on the subject which I have been putting together, but they are still mostly handwritten. For now, I want to present a theorem of A.N. Kolmogorov, the Kolmogorov normability criterion, that gives necessary and sufficient conditions for a tvs {(X,\tau)} to be normable (i.e. there exists a norm {x\mapsto\left\|x\right\|} such that the norm topology coincides with {\tau}). Two reasons motivate me to share it: 1) the proof is elegant; 2) I have spent a fair amount of time this summer studying when topological spaces, in particular function spaces, are normable.

To minimize any confusion over terminology, I include the following definitions. A notational note: we denote the line segment between two points {x,y} by {[x,y]:=\left\{tx+(1-t)y:0\leq t\leq 1\right\}}.

Let {X} be a vector space. A subset {A} is…

  • convex if {[x,y]\subset A} for any {x,y\in A}.
  • absorbing (or absorbent or radial) if for all {x\in X}, there exists a scalar {\alpha_{0}>0} such that {\left|\alpha\right|\leq\alpha_{0}\Rightarrow \alpha x\in A}.
  • circled (or balanced if {[-x,x]\subset A} for any {x\in A}.
  • bounded if for every neighborhood U there exists a scalar \alpha\in\mathbb{K} such that A\subset\alpha U.

It is a well-known that a Hausdorff tvs is metrizable if and only if it has a countable base of neighborhoods at zero (Theorem 5.12 [1]). However, there are plenty of familiar spaces which are metrizable, but normable. Consider the vector space X of real sequences a=(\alpha_{k})_{k=1}^{\infty} with addition and scalar multiplication defined pointwise. Define a metric d by

\displaystyle d(a,b):=\sum_{k=1}^{\infty}2^{-k}\dfrac{\left|\alpha_{k}-\beta_{k}\right|}{1+\left|\alpha_{k}-\beta_{k}\right|},\ \forall a=(\alpha_{k}), b=(\beta_{k})

The reader may verify that that (X,d) is a complete, metrizable tvs. Let a\mapsto\left\|a\right\|. Observe that \left|c\alpha_{k}\right|/(1+\left|c\alpha_{k}\right|)\rightarrow 1 as c\rightarrow\infty. Since a norm is positively homogeneous, \left\|c\cdot a\right\|=\left|c\right|\left\|a\right\|\rightarrow\infty as \left|c\right|\rightarrow\infty. But then for any r>0, there does not exists an \varepsilon>0 such that

\displaystyle B_{\varepsilon}:=\left\{a\in X:d(0,a)<\varepsilon\right\}\subset\left\{a\in X:\left\|a\right\|<r\right\}=:B'_{r}

Indeed, if e_{m} be a standard basis vector such that 2^{-m}<\varepsilon, then ne_{m}\in B_{\varepsilon} for all n\geq 1, but the norm of the norm tends to \infty as n\rightarrow\infty.

A more interesting example is the space {L^{p}[0,1]} of Lebesgue measurable functions with {\int_{0}^{1}\left|f\right|^{p}<\infty}, where {0<p<1}, lacks a norm equivalent to its natural topology. It turns out that the non-normability is intimately connected to the lack of nontrivial convex open sets in this space, a result known as Day’s theorem. The above example might suggest to us that convexity plays a role in any criterion for normability of a tvs.

Recall that a tvs is said to be separated if for any {x,y\in X}, there exists a nbhd of {x} which does contain {y}. Equivalently, {X} is separated if there exists a nbhd of {0} which does not contain {y}. The following theorem of A.N. Kolmogorov characterizes the normability of a separated (not a priori Hausdorff) tvs in terms of the existence of a bounded convex neighborhood of zero. The proof presented here is that of [2].

Theorem (A.N. Kolmogorov) A tvs {(X,\tau)} is normable if and only if it is separated and contains a bounded convex nbhd of zero.

Proof: The {\Rightarrow} direction is obvious. For {\Leftarrow}, suppose that {X} is separated and {U} is a bounded convex nbhd of zero. Taking a balanced nbhd of zero {V\subset U} and noting that {\text{conv}(V)\subset\text{conv}(U)=U}, we may assume that {U} is balanced, bounded, and convex.

For {x\neq 0}, define a set of scalars {A(x):=\left\{\lambda:x\notin\lambda U\right\}}; define {A(0):=\left\{0\right\}}. I claim that {A(x)} contains nonzero scalars when {x\neq 0}. Indeed, for {x\neq 0}, there exists an open nbhd {V} of zero such that {x\notin V}. Since {U} is bounded, there exists {\alpha\in\mathbb{K}} such that {U\subset\alpha V\Leftrightarrow \alpha^{-1}U\subset V}. As {x\notin V}, {x\notin\alpha^{-1}U\Leftrightarrow\alpha^{-1}\in A(x)}. Although we know that {A(x)} contains nontrivial scalars, it is not obvious that {A(x)} contains an interval or disk when {x\neq 0}. We will establish this result below in order to define a norm structure on {X}.

Define a nonnegative (possibly) extended-valued function {x\mapsto\left\|x\right\|:=\sup\left\{\left|\lambda\right|:\lambda\in A(x)\right\}}. We will show that {\left\|\cdot\right\|} is a norm. It is clear that {\left\|0\right\|=0} and the preceding claim shows that {x\neq 0\Rightarrow\left\|x\right\|>0}. I claim that

\displaystyle {\left\{\lambda:\left|\lambda\right|<\left\|x\right\|\right\}\subset A(x)} \ \forall x\in X.     (1)

If {x=0}, then the LHS is empty. If x\neq 0, then there exists a sequence of scalars \lambda_{n}\in A(x) increasing to \left\|x\right\|. For \left|\lambda\right|<\left|\lambda_{n}\right|, we have \lambda/\lambda_{n}\cdot U\subset U, as U is balanced. Hence if \lambda_{n}^{-1}x\notin U, then \lambda_{n}^{-1}x\notin(\lambda/\lambda_{n})\cdot U\Leftrightarrow\lambda x\notin U. As a consequence, we obtain {\left\|x\right\|<\infty} for all {x}. Indeed, note that {U} is a nbhd of zero, therefore absorbing. So given {x}, there exists {\alpha_{0}>0} such that {0\leq\left|\alpha\right|\leq\alpha_{0}\Rightarrow \alpha x\in U}. It follows that {\left\|x\right\|\leq\alpha_{0}^{-1}}.

For homogeneity, let \alpha be a nonzero scalar. Observe that by (1), x\notin\lambda U\Leftrightarrow\alpha x\notin\alpha\cdot\lambda U if and only if \left|\lambda\right|<\left\|x\right\|. It follows that \left|\alpha\right|\left\|x\right\|\leq\left\|\alpha x\right\|. If \left|\lambda\right|>\left\|x\right\|, then (1) implies \left|\alpha\lambda\right|>\left\|\alpha x\right\|. Letting \left|\lambda\right|\downarrow\left\|x\right\|, we obtain the reverse inequality.

So far we have not used the hypothesis that {U} is convex, but we will need it to show that {x\mapsto\left\|x\right\|} satisfies the triangle inequality. Fix {x,y\in X}. If either {x=0} or {y=0}, then the triangle inequality is trivial, so assume otherwise. Take {\varepsilon>0} sufficiently small so that {\max\left\{\left\|x\right\|,\left\|y\right\|\right\}+\varepsilon<\left\|x\right\|+\left\|y\right\|}. Then {x\in(\left\|x\right\|+\varepsilon)\cdot U} and {y\in(\left\|y\right\|+\varepsilon)\cdot U}. Since {U} is convex,

\displaystyle\dfrac{\left\|x\right\|+\varepsilon}{\left\|x\right\|+\left\|y\right\|}\cdot\dfrac{1}{\left\|x\right\|+\varepsilon}x+\dfrac{\left\|y\right\|+\varepsilon}{\left\|x\right\|+\left\|y\right\|}\cdot\dfrac{1}{\left\|y\right\|+\varepsilon}y\in U,

whence {x+y\in (\left\|x\right\|+\left\|y\right\|)U}. As {\left\{\left|\lambda\right|<\left\|x+y\right\|\right\}\subset A(x+y)}, it follows that {\left\|x+y\right\|\leq\left\|x\right\|+\left\|y\right\|}.

Lastly, we need to show that norm topology coincides with {\tau}. Let {B_{\varepsilon}:=\left\{x\in X:\left\|x\right\|<\varepsilon\right\}} be a ball of radius {\varepsilon>0} centered at zero. If {x\in (\varepsilon/2)\cdot U}, then {\left\|x\right\|\leq\varepsilon/2}, whence {B_{\varepsilon}} is {\tau}-open. Conversely, suppose {V} is an open nbhd of zero. Since {U} is bounded, there is {\alpha\in\mathbb{K}} such that {U\subset\alpha V\Leftrightarrow \alpha^{-1}U\subset V}. If {\left\|x\right\|<\alpha^{-1}}, then {x\in\alpha^{-1}U}, whence {B_{\alpha^{-1}}\subset V}. \Box

1. S.K. Berberian, Lectures in Functional Analysis and Operator Theory, Spring Verlag 1974.

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Banach Measures

If you have never read one of Elias Stein’s books, you are missing out. “Harmonic Analysis: Real-Variables, Orthogonality, and Oscillatory” and “Introduction to Fourier Analysis on Euclidean Spaces” (co-authored by Guido Weiss) are not light bedtime reading, but they make the reader–or at least, me–get excited about Analysis. Moreover, I increasingly find the benefits of reading the masters when trying to learn something. Today, I want to share a result from Book 4 of the four-volume “Princeton Lectures in Analysis” geared towards undergraduates and beginning graduate students.

We will show that there exists a finitely-additive measure {\hat{m}} defined on all subsets of {\mathbb{R}^{d}} that is translation-invariant and agrees with the Lebesgue measure on the Lebesgue {\sigma}-algebra. Of course, such a measure has no hope of being countably additive ({\sigma}-additive) as we know from the existence of Vitali sets. Such a measure {\hat{m}} is called a Banach measure.

Theorem 1 There exists an extended-valued nonnegative function {\hat{m}:\mathcal{P}(\mathbb{R}^{d})\rightarrow[0,\infty]} satisfying

(Finitely Additive) {\hat{m}(E_{1}\cup E_{2})=\hat{m}(E_{1})+\hat{m}(E_{2})}, if {E_{1},E_{2}\subset\mathbb{R}^{d}} are disjoint;

(Agreement) {\hat{m}(E)=m(E)} if {E\subset\mathbb{R}^{d}} is Lebesgue measurable;

(Translation Invariance) {\hat{m}(E+h)=\hat{m}(E)} for all {E\subset\mathbb{R}^{d}} and {h\in\mathbb{R}^{d}}.

To prove Theorem 1, we will follow the functional analytic approach to measure and integration by first defining an integral as a linear functional on a suitable space of functions and then defining a measure by evaluating the integral of characteristic functions. Specifically, we use the Hahn-Banach theorem to extend the Lebesgue integral to the space of bounded (possibly nonmeasurable) functions on the {d}-torus {\mathbb{R}^{d}/\mathbb{Z}^{d}}.

A quick remark on notation: we denote the power set and characteristic function of a set {X} by {\mathcal{P}(X)} and {\chi_{E}}, respectively.

Theorem 2 There exists a linear functional {f\mapsto I(f)} defined on all bounded functions on {\mathbb{R^{d}}/\mathbb{Z^{d}}} such that(Positive Semidefinite) {I(f)\geq 0}, if {f\geq 0};

(Linearity) {I(\alpha f_{1}+\beta f_{2})=\alpha I(f_{1})+\beta I(f_{2})} for all {f_{1}, f_{2}} and {\alpha,\beta\in\mathbb{R}};

(Agreement) {I(f)=\int_{[0,1]^{d}}f(x)dx}, if {f} is Lebesgue measurable;

(Translation Invariance) {I(f_{h})=I(f)}, where {f_{h}(x):=f(x-h)}, for all {h\in\mathbb{R}^{d}}.

Proof: Let {V} denote the real vector space of bounded functions {f:\mathbb{R}^{d}/\mathbb{Z}^{d}}. Let {V_{0}} denote the subspace of Lebesgue measurable functions. Denote the functional on {V_{0}} defined by Lebesgue integration by {I_{0}}. To use the Hahn-Banach theorem, we need to find a suitable seminorm {p} such that {I_{0}(f)\leq p(f)} for all {f\in V_{0}}. The clever construction of {p} is due to Polish mathematician Stefan Banach.

Let {A=\left\{a_{1},\ldots,a_{N}\right\}} be a sequence in {\mathbb{R}^{d}} of length {N}. Given {A}, define

\displaystyle M_{A}(f):=\sup_{x\in\mathbb{R}^{d}}\left(\dfrac{1}{N}\sum_{j=1}^{N}f(x+a_{j})\right)\in\mathbb{R}.

We define {p(f):=\inf_{A}\left\{M_{A}(f)\right\}}, where the infimum is taken over all finite sequencess {A} of arbitrary length. The definition of {p(f)} should remind you of upper Darboux sums used in Riemann integration. It is clear that {p} is well-defined, as {f} is bounded.

I claim that {p} is a sublinear functional on {V}. Indeed, it is evident that {p} is positively homgeneous. Given {\varepsilon>0}, we can find find sequences {A} and {B} of length {N_{1}} and {N_{2}}, respectively, such that {M_{A}(f_{1})\leq p(f_{1})+\varepsilon} and {M_{A}(f_{2})\leq p(f_{2})+\varepsilon}. Define a sequence {C:=\left\{a_{i}+b_{j}\right\}_{1\leq i\leq N_{1},1\leq j\leq N_{2}}} of length {N_{1}\cdot N_{2}}. It is evident that {M_{C}(f_{1}+f_{2})\leq M_{C}(f_{1})+M_{C}(f_{2})}. Since {M_{E}(f)=M_{E'}(f_{h})}, where any {h\in\mathbb{R}^{d}} and {E'} is a translate of {E}, we see that

\displaystyle\dfrac{1}{N_{2}}\sum_{j=1}^{N_{2}}\dfrac{1}{N_{1}}\sum_{i=1}^{N_{1}}f_{1}(x+a_{i}+b_{j})\leq\dfrac{1}{N_{2}}\sum_{j=1}^{N_{2}}M_{A-b_{j}}(f_{1})=M_{A}(f_{1}).

Taking the supremum of the LHS over all {x\in\mathbb{R}^{d}}, we see that {M_{C}(f_{1})\leq M_{A}(f_{1})}. A completely analogous argument shows that {M_{C}(f_{2})\leq M_{C}(f_{2})}. Putting these inequalities together, we conclude that

\begin{array}{lcl}\displaystyle p(f_{1}+f_{2})&\leq&\displaystyle M_{C}(f_{1}+f_{2})\\[1.5 em]\displaystyle&=&\displaystyle\leq M_{C}(f_{1})+M_{C}(f_{2})\\[1.5 em]&\leq&\displaystyle M_{A}(f_{1})+M_{B}(f_{2})\\[1.5 em]&\leq&\displaystyle p(f_{1})+p(f_{2})+2\varepsilon.\end{array}

Letting {\varepsilon\downarrow 0} completes the proof of sublinearity.
I claim that {I_{0}(f)\leq p(f)} for all {f\in V_{0}}. Indeed, by translation invariance and linearity,

\displaystyle I_{0}(f)=\dfrac{1}{N}\sum_{i=1}^{N}\int_{[0,1]^{d}}f(x+a_{i})dx=\int_{[0,1]^{d}}\left(\dfrac{1}{N}\sum_{i=1}^{N}f(x+a_{i})\right)dx\leq M_{A}(f)

for all finite sequences {A\subset\mathbb{R}^{d}}. Taking the infimum of the RHS over all such {A}, we conclude that {I_{0}(f)\leq p(f)}.

We apply the Hahn-Banach theorem to {I_{0}} to obtain a linear functional {I:V\rightarrow\mathbb{R}} bounded by {p}. Properties (ii) and (iii) have already been proven. For (i), observe that if {f\leq 0}, then {M_{A}(f)\leq 0} for all sequencess {A}, whence {p(f)\leq 0}. If {f\leq 0}, then applying this observation to {-f} shows that {I(f)\geq 0}. To establish (iv), it suffices by symmetry to show that {p(f-f_{h})\leq 0}, where {h\in\mathbb{R}^{d}} is fixed. Consider the set {A:=\left\{h,2h,\ldots,Nh\right\}}, for some integer {N\geq 1}. Then

\displaystyle \begin{array}{lcl}\displaystyle\dfrac{1}{N}\sum_{j=1}^{N}(f-f_{h})(x+jh)&=&\displaystyle\dfrac{1}{N}\sum_{j=1}^{N}\left[f(x+jh)-f(x+(j-1)h)\right]\\[2 em]&=&\displaystyle\dfrac{1}{N}\left[f(x+Nh)-f(x)\right]\\[2 em]&=&\displaystyle\dfrac{2\left\|f\right\|_{\infty}}{N}\end{array}

Letting {N\uparrow\infty}, we obtain that {p(f-f_{h})\leq M_{A}(f-f_{h})\leq 0}. \Box

By defining {\hat{m}:\mathcal{P}(\mathbb{R}^{d}/\mathbb{Z}^{d})\rightarrow[0,\infty]} by {\hat{m}(E):=I(\chi_{E})}, we obtain the following corollary.

Corollary 3 There exists a nonnegative set function {\mathcal{P}(\mathbb{R}^{d}/\mathbb{Z}^{d})\rightarrow[0,\infty]} such that

(Finite Additivity) {\hat{m}(E_{1}\cup E_{2})=\hat{m}(E_{1})+\hat{m}(E_{2})}, if {E_{1},E_{2}\subset\mathbb{R}^{d}/\mathbb{Z}^{d}} are disjoint;

(Agreement) {\hat{m}(E)=m(E)}, if {E} is Lebesgue measurable;

(Translation Invariance) {\hat{m}(E+h)=\hat{m}(E)} for all {h\in\mathbb{R}^{d}}.

We are now ready to prove Theorem 1. The plan is to partition {\mathbb{R}^{d}} into countably many disjoint cubes that can be translated to the unit cube {Q_{0}:=[0,1)^{d}} by an element of {\mathbb{Z}^{d}}. We first define {\hat{m}} locally on disjoint cubes by computing {\hat{m}_{0}} on the translate to {Q_{0}}. For an arbitrary set {E}, we define {\hat{m}(E)} as the “infinite sum” of these local quantities.

Proof of Theorem 1: Let {Q_{n}=Q_{0}+n} denote the denote the translate of the unit cube by {n=(n_{1},\ldots,n_{d})\in\mathbb{Z}^{d}}. The collection {\left\{Q_{n}:n\in\mathbb{Z}^{d}\right\}} forms a countable partition of {\mathbb{R}^{d}/\mathbb{Z}^{d}}. For a subset {E\subset\mathbb{R}^{d}}, define

\displaystyle \hat{m}(E):=\sum_{n\in\mathbb{Z}^{d}}\hat{m}_{0}(E\cap Q_{n}-n)

Observe that if {E\subset Q_{n}}, for some {n\in\mathbb{Z}^{d}}, then {\hat{m}(E)=\hat{m}_{0}(E-n)}. Properties (i) and (ii) are evident from the corresponding properties of {\hat{m}_{0}} and the definition of {\hat{m}}.

To prove (iii), we will first show translation invariance for {h\in\mathbb{Z}^{d}}. Then by considering {h\pmod{1}}, it suffices to show translation invariance for {h\in Q_{0}}.
For {h=m\in\mathbb{Z}^{d}}, observe that {(E\cap Q_{n})-n=(E\cap Q_{n-m})-(n-m)}, whence

\begin{array}{lcl}\displaystyle \hat{m}(E_{n}+m)&=&\displaystyle\sum_{n\in\mathbb{Z}^{d}}\hat{m}_{0}((E+m)\cap Q_{n}-n)\\[1.5 em]&=&\displaystyle\sum_{n\in\mathbb{Z}^{d}}\hat{m}_{0}(E\cap Q_{n-m}-(n-m))\\[1.5 em]&=&\displaystyle\hat{m}_{0}(E)\end{array}

since {n\mapsto n-m} is a bijection of {\mathbb{Z}^{d}}. Now suppose that {h=(h_{1},\ldots,h_{d})\in Q_{0}}. For {n\in\mathbb{Z}^{d}}, set {E_{n}:=E\cap Q_{n}} and define sets

\displaystyle \begin{array}{lcl} \displaystyle E_{n}'&:=&\displaystyle E_{n}\cap (n_{1},n_{1}+1-h_{1}]\times\cdots\times (n_{d},n_{d}+1-h_{d}]\\[1.5 em]\displaystyle E_{n}''&:=&\displaystyle E_{n}\cap (n_{1}+1-h_{1},n_{1}+1]\times\cdots\times (n_{d}+1-h_{d},n_{d}+1]\end{array}

We can write {E} as the disjoint union {\bigcup E_{n}'\cup\bigcup E_{n}''}. Observe that {(E_{n}+h)\cap Q_{n}\subset Q_{n}} while {E_{n}''+h\subset Q_{\tilde{n}}}, where {\tilde{n}=(n_{1}+1,\ldots,n_{d}+1)}. Using the finite additivity of {\hat{m}_{0}} and the {\mathbb{Z}^{d}}-translation invariance of {\hat{m}}, we obtain

\displaystyle \begin{array}{lcl} \displaystyle\hat{m}(E+h)&=&\displaystyle\hat{m}\left(\bigcup_{n\in\mathbb{Z}^{d}}E_{n}'+h\right)+\hat{m}\left(\bigcup_{n\in\mathbb{Z}^{d}}E_{n}''+h\right)\\[2 em]&=&\displaystyle\sum_{n\in\mathbb{Z}^{d}}\hat{m}_{0}((E_{n}'+h)-n)+\sum_{n\in\mathbb{Z}^{d}}\hat{m}_{0}(E_{n}''+h-\tilde{n})\\[2 em]&=&\displaystyle\sum_{n\in\mathbb{Z}^{d}}\left[\hat{m}_{0}(E_{n}'-n)+\hat{m}_{0}(E_{n}''+2h-\tilde{n})\right]\\[2 em]&=&\displaystyle\sum_{n\in\mathbb{Z}^{d}}\left[\hat{m}_{0}(E_{n}'-n)+\hat{m}_{0}(E_{n}''-n)\right]\\[2 em]&=&\displaystyle\sum_{n\in\mathbb{Z}^{d}}\hat{m}_{0}(E_{n}-n)=\hat{m}_{0}(E)\end{array}

\Box

For {d=1}, the existence of {\hat{m}} shows that there is no Banach-Tarski paradox in the real line, as the only nontrivial rigid motions are translations. It turns out that the Banach-Tarski paradox is also false in the plane. The interested reader in this result and more may consult Banach’s 1923 article “Sur le probleme de la mesure” and Banach’s 1924 article co-authored with Alfred Tarski. A more recent reference is Stan Wagon’s “The Banach-Tarski Paradox”.

[1] S. Banach, A. Tarski, Sur la decomposition des ensembles de points en parties respectivement congruentes, Fund. math 6 (1924), 118-148.

[2] S. Banach, Sur le probleme de la mesure, Fund. math 4 (1923), 7-33.

[3] E.M. Stein and R. Shakarchi, Functional Analysis: Introduction to Further Topics in Analysis (Princeton Lectures in Analysis) (Bk. 4), Princeton UP, 2011.

[4] S. Wagon, The Banach-Tarski Paradox, Cambridge UP, 1993.

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Intermediate Value Property of Measures

It is well known that the Lebesgue measure {\lambda} has an intermediate value property like that of real-valued functions defined on some bounded interval. For any Lebesgue measurable set {A}, the function {t\mapsto \lambda(A\cap[0,t])} is continuous (continuity of measure) and bounded from above by {\lambda(A)}. By the ordinary intermediate value theorem, for any {\alpha\in[0,\lambda(A)]}, we can find {t\in [0,\infty]} such that {\lambda(A\cap [0,t])=\alpha}. We conclude that there exists a measurable set {B} with {\lambda(B)=\alpha}. Replacing intervals with balls, we see that a completely analogous result holds for the {n}-dimensional Lebesgue measure.

The Lebesgue measure is a very `nice’ measure, so even without considering counterexamples, it would be premature to conclude that every measure space {(X,\mathcal{A},\mu)} has the intermediate value property. A simple counterexample would be a set {X} with the trivial {\sigma}-algebra {\mathcal{A}=\left\{\emptyset,X\right\}} and the measure {\mu(\emptyset):=0,\mu(X):=\infty}. A slightly less trivial counterexample is the counting measure on an infinite set. In both cases, we see that there are measurable sets {A} whose only subsets have either measure {\mu(A)} or measure zero. Recall that such sets are called atoms.

We will prove that every bounded atomless measure space {(X,\mathcal{A},\mu)} has the intermediate value property in three steps. First, we will show that a finite measure space has at most countably many pairwise disjoint atoms. Second, we will show that we can partition {X} into finitely many sets in such a way that we separate the atoms from the non-atoms with the hyperplane {\left\{\mu(A)=\varepsilon\right\}}. Third, we will use Step 2 together with an induction argument to construct the desired set.

A few quick remarks about terminology. A measure {\mu} is respectively atomic and nonatomic or atomless if its {\sigma}-algebra has and does not have {\mu}-atoms. Two measurable sets {A,B} are equivalent if {\mu(A\triangle B)=0}, where {\triangle} denotes the symmetric difference. In particular, if two atoms {A,B} are nonequivalent (i.e. {\mu(A\triangle B)>0}), then {\mu(A\cap B)=0}. So without loss of generality, we may assume that {A,B} are disjoint.

Lemma 1 A finite measure space {(X,\mathcal{A},\mu)} has at most countably many nonequivalent atoms.

To prove this lemma, we use the same argument used to show that at most countably many Fourier coefficients of a Hilbert space element are nonzero.

Proof: Let {\left\{U_{\alpha}\right\}_{\alpha\in I}} be a collection of nonequivalent atoms. For each {n\geq 1}, define {I_{n}:=\left\{\alpha\in I:\mu(U_{\alpha})\geq n^{-1}\right\}}. Since {\mu(X)<\infty}, each {I_{n}} is finite, whence {I=\bigcup_{n=1}^{\infty}I_{n}} is at most countable. \Box

A useful consequence of the lemma is that we can always decompose a measure space as {X=A\cup\bigcup_{n=1}^{\infty}A_{n}}, where {A} is not an atom and each {A_{n}} is either {\emptyset} or a disjoint atom.

Theorem 2 Let {(X,\mathcal{A},\mu)} be a finite measure space. For all {\varepsilon>0}, there exists a partition of {X} into pairwise disjoint sets {X_{1},\ldots,X_{n}\in\mathcal{A}} with either {\mu(X_{i})\leq\varepsilon} or {X_{i}} is an atom with {\mu(X_{i})>\varepsilon}.

Proof: Let {\varepsilon>0} be given. Since {\mu} is bounded, {\mathcal{A}} contains finitely many disjoint atoms {A_{1},\ldots,A_{p}} with measure {\geq\varepsilon}. Note that no atom of measure {>\varepsilon} is contained in {Y:=\bigcup_{i=1}^{n}X_{i}}. I claim that every set {B\in\mathcal{A}} of positive measure contained in {Y} itself contains a measurable set {C} with {0<\mu(C)\leq\varepsilon}. Suppose there exists {B\in\mathcal{A}} for which the claim is false. If {\mu(B)\leq\varepsilon}, then we cane take {C:=B}; so, {\mu(B)>\varepsilon}. Since {B} is not an atom, there exists a measurable set {B_{1}} with {\varepsilon<\mu(B_{1})<\mu(B)} and {\mu(C_{1}:=B\setminus B_{1})>\varepsilon} by our hypothesis. By the same argument, there exists a measurable set {B_{2}\subset C_{1}} with {\mu(C_{1})>\mu(B_{2})>\varepsilon}. Set {C_{2}:=C_{1}\setminus B_{2}}. By induction, we obtain a sequence of pairwise disjoint measurable sets {\left\{B_{n}\right\}_{n=1}^{\infty}} with {\mu(B_{n})>\varepsilon} for all {n}. At each step {n}, we are doubling our lower bound for {\mu(B)<\infty}, which gives a contradiction.

For {A\in\mathcal{A}}, define

\displaystyle \eta(A):=\sup\left\{\mu(B):B\subset A, B\in\mathcal{A},\mu(B)\leq\varepsilon\right\}

Our first result shows that {0<\eta(A)\leq\varepsilon}. If {\mu(Y)\leq\varepsilon}, then we are done. Otherwise, there exists a measurable set {B_{1}\subset Y} with {0<\mu(B_{1})\leq\eta(Y)}. If {\mu(Y\setminus B_{1})\leq\varepsilon}, then we are done. Otherwise, there exists a measurable set {B_{2}\subset Y\setminus B_{1}} with {0<\eta(Y\setminus B_{1})/2\leq\mu(B_{2})\leq\varepsilon}. Continuing by induction, we obtain an at most countable sequence {\left\{B_{n}\right\}} of pairwise disjoint measurable sets {B_{n}} with {B_{n}\subset Y} and {0<\eta(Y\setminus\bigcup_{i=1}^{n}B_{i})/2\leq\mu(B_{n+1})\leq\varepsilon}.

If the collection is infinite, then for {B_{0}:=Y\setminus\bigcup_{i=1}^{\infty}},

\displaystyle \eta(B_{0})\leq\eta\left(Y\setminus\bigcup_{i=1}^{n}B_{i}\right)\leq 2\mu(B_{n+1}),\ \forall n

As {\sum_{n=1}^{\infty}\mu(B_{n})\leq\mu(Y)<\infty}, we obtain {\lim\mu(B_{n})=0}, whence {\eta(B_{0})=\mu(B_{0})=0}. Take {m} sufficiently large so that {\sum_{n=m}^{\infty}\mu(B_{n})<\varepsilon}. Then a {A_{1},\ldots,A_{p},B_{1},\ldots,B_{m},B_{0}\cup\bigcup_{n=m+1}^{\infty}B_{n}} forms a partition of {X} satisfying the hypotheses of the theorem. \Box

For given {\alpha\in[0,\mu(X)]}, we want to construct an increasing sequence of measurable sets {\left\{A_{n}\right\}_{n=1}^{\infty}} where {\mu(A_{n})\rightarrow\alpha}, as {n\rightarrow\infty}. We ca then conclude from the continuity of measure that {\mu(\bigcup_{n}A_{n})=\alpha}. With an atomless measure, the preceding theorem allows us control the measure of {A_{n+1}\setminus A_{n}} at each step. By `chipping away’ at the quantity {\alpha-\mu(\bigcup_{i=1}^{n}A_{n})>0}, we choose {A_{n+1}}.

Corollary 3 Let {(x,\mathcal{A},\mu)} be a measure space where {\mu} is atomless. For all {\alpha\in[0,\mu(X)]}, there exists a set {A\in\mathcal{A}} such that {\mu(A)=\alpha}.

Proof: By the previous theorem, we can partition {X} into finitely many measurable sets {\left\{X_{1},\ldots,X_{n_{1}}\right\}} with {\mu(X_{i})<1/2} for {1\leq i\leq n_{1}}. Let {1\leq m\leq n_{1}} be the maximal integer such that {\mu(\bigcup_{i=1}^{m}X_{i})\leq\alpha}. Set {A_{1}:=\bigcup_{i=1}^{m}X_{i}} and note that {0\leq\alpha-\mu(A_{1})<1/2}. Partition {X\setminus A_{1}} into sets of measure less than {1/2^{2}}. Let {B_{1}} be the largest union of these sets with total measure less than {\alpha-\mu(A_{1})}. Observe that {(\alpha-\mu(A_{1}))-\mu(B_{1})<1/4}. Set {A_{2}:=A_{1}\cup B_{1}}, and note that {0\leq\alpha-\mu(A_{2})<1/4}. By induction, we obtain a sequence of increasing sets {\left\{A_{n}\right\}_{n=1}^{\infty}} and decreasing sets {\left\{B_{n}\right\}_{n=1}^{\infty}} where {A_{n+1}=A_{n}\cup B_{n}} and {0\leq (\alpha-\mu(A_{n}))-\mu(B_{n})<1/2^{n+1}}. Rewriting our upper bound, we see that {\mu(A_{n})\rightarrow\alpha}. \Box

The converse to the preceding result is false: there exists an atomic measure {\mu} with the intermediate value property. For a counterexample, consider the natural numbers equipped with the probability measure {\mu} induced by the function

\displaystyle f(x):=\sum_{n=1}^{\infty}{2^{-n}}\chi_{\left\{n\right\}}(x) \ \forall x\in\mathbb{R}

For any {\alpha\in[0,1]}, we can find a sequence {\left\{n_{i}\right\}_{i=1}^{\infty}\subset\mathbb{N}} such that {\alpha=\sum_{i}2^{-n_{i}}}, whence {\alpha=\mu(\bigcup_{i}\left\{n_{i}\right\})}.

  1. V.I. Bogachev, Measure Theory Volume I, Springer, 2007.
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