It is well known that the Lebesgue measure has an intermediate value property like that of real-valued functions defined on some bounded interval. For any Lebesgue measurable set , the function is continuous (continuity of measure) and bounded from above by . By the ordinary intermediate value theorem, for any , we can find such that . We conclude that there exists a measurable set with . Replacing intervals with balls, we see that a completely analogous result holds for the -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 has the intermediate value property. A simple counterexample would be a set with the trivial -algebra and the measure . A slightly less trivial counterexample is the counting measure on an infinite set. In both cases, we see that there are measurable sets whose only subsets have either measure or measure zero. Recall that such sets are called *atoms*.

We will prove that every bounded atomless measure space 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 into finitely many sets in such a way that we separate the atoms from the non-atoms with the hyperplane . Third, we will use Step 2 together with an induction argument to construct the desired set.

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

Lemma 1A finite measure space 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 be a collection of nonequivalent atoms. For each , define . Since , each is finite, whence is at most countable.

A useful consequence of the lemma is that we can always decompose a measure space as , where is not an atom and each is either or a disjoint atom.

Theorem 2Let be a finite measure space. For all , there exists a partition of into pairwise disjoint sets with either or is an atom with .

*Proof:* Let be given. Since is bounded, contains finitely many disjoint atoms with measure . Note that no atom of measure is contained in . I claim that every set of positive measure contained in itself contains a measurable set with . Suppose there exists for which the claim is false. If , then we cane take ; so, . Since is not an atom, there exists a measurable set with and by our hypothesis. By the same argument, there exists a measurable set with . Set . By induction, we obtain a sequence of pairwise disjoint measurable sets with for all . At each step , we are doubling our lower bound for , which gives a contradiction.

For , define

Our first result shows that . If , then we are done. Otherwise, there exists a measurable set with . If , then we are done. Otherwise, there exists a measurable set with . Continuing by induction, we obtain an at most countable sequence of pairwise disjoint measurable sets with and .

If the collection is infinite, then for ,

As , we obtain , whence . Take sufficiently large so that . Then a forms a partition of satisfying the hypotheses of the theorem.

For given , we want to construct an increasing sequence of measurable sets where , as . We ca then conclude from the continuity of measure that . With an atomless measure, the preceding theorem allows us control the measure of at each step. By `chipping away’ at the quantity , we choose .

Corollary 3Let be a measure space where is atomless. For all , there exists a set such that .

*Proof:* By the previous theorem, we can partition into finitely many measurable sets with for . Let be the maximal integer such that . Set and note that . Partition into sets of measure less than . Let be the largest union of these sets with total measure less than . Observe that . Set , and note that . By induction, we obtain a sequence of increasing sets and decreasing sets where and . Rewriting our upper bound, we see that .

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

For any , we can find a sequence such that , whence .

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

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