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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One Response to Intermediate Value Property of Measures

  1. Pingback: On the Normability of the Space of Measurable Functions | Math by Matt

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