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.