In this post, we are going to discuss mutually singular and absolutely continuous measures with the goal of proving the Lebesgue-Radon-Nikod\'{y}m theorem.

\bigskip In what follows, will denote a measurable space. We begin with a review of signed measures.

Definition 1Asigned measureon a -algebra is a set function such that

- is extended-valued in the sense that for all
- If are disjoint subsets of , then

Note that in order for condition (2) to hold, the sum must be independent of any rearrangement. If is finite, then it follows from a theorem of Riemann that the sum converges absolutely. I have included a proof of this result for the interested reader.

Lemma 2(Riemann Rearrangement Lemma) Let be a series of real numbers which converges, but not absolutely. Suppose . Then there exists a rearrangement with partial sums such that

*Proof:* For , set

Then . I claim that the series diverge. If either or converges, then since converges, it follows that converges, which contradicts our hypothesis that does not converge absolutely.

Let denote the nonnegative terms of , in the order in which they occur, and let denote the absolute values of the negative terms of also in their original order. Since and for all , both diverge.

We will construct sequences , such that

satisfies the conclusion of the lemma. Choose real sequences such that . Let be the minimal positive integers such that

and

Let be the minimal positive integers such that

and

We continue in this fashion. This selection process is possible since diverge. Let denote the partial sums whose last terms are and , respectively. Then

Since is convergent, as . We conclude that . Since the subsequential limits of the rearrangement series are bounded from above by and bounded from below by (this is evident from the squeeze theorem), it follows immediately that

Corollary 3If every rearrangement of real-valued series converges, then .

Given a signed measure on a measure space , one might ask if it is always possible to find a positive measure which dominates in the following sense:

Moreover, one might ask if there is a smallest such in the sense that if is any other positive measure which dominates , then for all . It turns out the answer is yes. First, we need to define the notion of the total variation of a measure, analogous to the variation of a measurable function.

Definition 4Define a set function , called thetotal variationof , by

where the supremum is taken over all countable partitions of .

Lemma 5The total variation of a signed measure is a positive measure which satisfies .

*Proof:* The only axiom of measures which does not obviously satisfy is -additivity. Let be a countable collection of disjoint sets in , and set . For each , let such that . It follows from the definition of supremum and of that, for each , there exists a countable collection of disjoint sets such that and

Since is a partition of , we have

Taking the supremum over all which satisfy yields the inequality

For the reverse inequality, let be any other partition of . For fixed, is a partition of . By the -additivity of ,

If , then . It follows from the definition of total variation that , and the inequality

holds trivially. If , then we can interchange the order of summation to obtaion

since is a partition of for each fixed . Since was an arbitrary partition of , we obtain the reverse inequality

which completes the proof.

Analogous to the decomposition of a function into a difference of two nonnegative functions, we can write a signed measure as the difference of two positive measures.

Definition 6For a signed measure , we define thepositive variationandnegative variationof byIf is such that , then . It follows from the preceding proposition that are both (positive) measures, and moreover,

We say that the signed measure is-finiteif the measure is -finite.

Definition 7Two signed measures and on a measure space aremutually singularif there are disjoint subsets so thatIf are mutually singular, we write .

If is a signed measure and is a positive measure on , we say that is

absolutely continuouswith respect to if

If is absolutely continuous with respect to , we write .

Lemma 8If and are mutually singular, and is also absolutely continuous with respect to , then vanishes identically.

*Proof:* Let be disjoint subsets such that

For any , . By absolute continuity, . Since was arbitrary, we conclude that .

Lemma 9Let be a positive measure and be a signed measure.

- If for every , there exists such that , then .
- If is a finite measure, then the converse holds.

*Proof:* We first prove (1). Let be such that . It follows from our hypothesis that for every , from which we conclude that .

We prove (2) by contradiction. Suppose there exists such that for all , there exists with and . For each choose with and . Then

for all . We conclude that . Since by hypothesis, we have . But this is a contradiction since

There are several proofs of the (Lebesgue-)Radon-Nikodym theorem, but I am fond of the one given below because it uses the theory of Hilbert spaces, which is quite elegant. The exposition closely follows that of Stein and Shakarchi in *Real Analysis: Measure, Integration, and Hilbert Spaces*.

Theorem 10Suppose is a -finite positive measure on the measure space and is a -finite signed measure on . Then there exist unique signed meaures and on such that , and . Furthermore, the measure is given by

for some extended -integrable function .

*Proof:* We first consider the case where both and are positive and finite. Set . We define a functional by

Clearly, is linear. I claim that is bounded. Indeed, since are both positive,

where the last inequality follow from the Cauchy-Schwarz inequality. Since is a Hilbert space, the Riesz representation theorem tells us that there exists a unique (up to a.e. equivalence) such that

If with , when we set and recall that , we obtain

, I claim that -a.e. Indeed, for all sets implies that

Taking the intersection yields . By the same argument, for all implies that -a.e. Thus, we may assume that for all , and we have

Consider the two sets

and define two measures and on by

The diligent inclined reader can verify that are indeed measures. I claim that . It is tautological that are disjoint and for all measurable subsets , so we need only show that for all measurable subsets . Indeed, taking in the identity yields

I now claim that . Let be such that . Then

Since , we conclude that a.e., which imples that . I now claim that . Let , and set . Then

If , then , and if , then . In other words, . Since and our measure space is finite, the dominated convergence theorem implies that

Observe that

Set . From the monotone convergence theorem we conclude that

Furthermore, since .

We now consider the case where are -finite positive measures. It follows from the definition of -finite that we can find pairwise disjoint sets such that and for all . We define positive, finite measures on by

For each , we write , where and . Defining

completes the argument.

If is signed, then we apply the preceding argument separately to the positive and negative variations of .

To see the uniqueness of the decomposition, suppose we also have , where and . Then

Clearly, the LHS is absolutely continuous with respect to . I claim that the RHS is singular with respect to . Indeed, let be paritions guaranteed in the definition of singular measures. Then for all measurable subset , and for any ,

We conclude that .

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