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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One Response to On the Normability of the Space of Measurable Functions

  1. Pingback: An Extension of a Theorem of Marczewski | Math by Matt

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