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# 8.11: The Radon–Nikodym Theorem. Lebesgue Decomposition

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I. As you know, the indefinite integral

$\int f dm$

is a generalized measure. We now seek conditions under which a given generalized measure $$\mu$$ can be represented as

$\mu=\int f dm$

for some $$f$$ (to be found). We start with two lemmas.

## Lemma $$\PageIndex{1}$$

Let $$m, \mu : \mathcal{M} \rightarrow[0, \infty)$$ be finite measures in $$S.$$ Suppose $$S \in \mathcal{M}, \mu S>0$$ (i.e., $$\mu \not \equiv 0$$) and $$\mu$$ is $$m$$-continuous (Chapter 7, §11).

Then there is $$\delta>0$$ and $$a$$ set $$P \in \mathcal{M}$$ such that $$m P>0$$ and

$(\forall X \in \mathcal{M}) \quad \mu X \geq \delta \cdot m(X \cap P).$

Proof

As $$m<\infty$$ and $$\mu S>0,$$ there is $$\delta>0$$ such that

$\mu S-\delta \cdot m S>0.$

Fix such a $$\delta$$ and define a signed measure (Lemma 2 of Chapter 7, §11)

$\Phi=\mu-\delta m,$

so that

$(\forall Y \in \mathcal{M}) \quad \Phi Y=\mu Y-\delta \cdot m Y;$

hence

$\Phi S=\mu S-\delta \cdot m S>0.$

By Theorem 3 in Chapter 7, §11 (Hahn decomposition), there is a $$\Phi$$-positive set $$P \in \mathcal{M}$$ with a $$\Phi$$-negative complement $$-P=S-P \in \mathcal{M}.$$

Clearly, $$m P>0;$$ for if $$m P=0,$$ the $$m$$-continuity of $$\mu$$ would imply $$\mu P=0$$, hence

$\Phi P=\mu P-\delta \cdot m P=0,$

contrary to $$\Phi P \geq \Phi S>0$$.

Also, $$P \supseteq Y$$ and $$Y \in \mathcal{M}$$ implies $$\Phi Y \geq 0;$$ so by (1),

$0 \leq \mu Y-\delta \cdot m Y.$

Taking $$Y=X \cap P,$$ we get

$\delta \cdot m(X \cap P) \leq \mu(X \cap P) \leq \mu X,$

as required.$$\quad \square$$

## Lemma $$\PageIndex{2}$$

With $$m, \mu,$$ and $$S$$ as in Lemma 1, let $$\mathcal{H}$$ be the set of all maps $$g : S \rightarrow E^{*}, \mathcal{M}$$-measurable and nonnegative on $$S,$$ such that

$\int_{X} g dm \leq \mu X$

for every set $$X$$ from $$\mathcal{M}$$.

Then there is $$f \in \mathcal{H}$$ with

$\int_{S} f dm=\max _{g \in \mathcal{H}} \int_{S} g dm.$

Proof

$$\mathcal{H}$$ is not empty; e.g., $$g=0$$ is in $$\mathcal{H}.$$ We now show that

$(\forall g, h \in \mathcal{H}) \quad g \vee h=\max (g, h) \in \mathcal{H}.$

Indeed, $$g \vee h$$ is $$\geq 0$$ and $$\mathcal{M}$$-measurable on $$S,$$ as $$g$$ and $$h$$ are.

Now, given $$X \in \mathcal{M},$$ let $$Y=X(g>h)$$ and $$Z=X(g \leq h).$$ Dropping "$$dm$$" for brevity, we have

$\int_{X}(g \vee h)=\int_{Y}(g \vee h)+\int_{Z}(g \vee h)=\int_{Y} g+\int_{Z} h \leq \mu Y+\mu Z=\mu X,$

proving (2).

Let

$k=\sup _{g \in \mathcal{H}} \int_{S} g d m \in E^{*}.$

Proceeding as in Problem 13 of Chapter 7, §6, and using (2), one easily finds a sequence $$\left\{g_{n}\right\} \uparrow, g_{n} \in \mathcal{H},$$ such that

$\lim _{n \rightarrow \infty} \int_{S} g_{n} dm=k.$

(Verify!) Set

$f=\lim _{n \rightarrow \infty} g_{n}.$

(It exists since $$\left\{g_{n}\right\} \uparrow.$$) By Theorem 4 in §6,

$k=\lim _{n \rightarrow \infty} \int_{S} g_{n}=\int_{S} f.$

Also, $$f$$ is $$\mathcal{M}$$-measurable and $$\geq 0$$ on $$S,$$ as all $$g_{n}$$ are; and if $$X \in \mathcal{M},$$ then

$(\forall n) \quad \int_{X} g_{n} \leq \mu X;$

hence

$\int_{X} f=\lim _{n \rightarrow \infty} \int_{X} g_{n} \leq \mu X.$

Thus $$f \in \mathcal{H}$$ and

$\int_{S} f=k=\sup _{g \in H} \int_{S} g,$

i.e.,

$\int_{S} f=\max _{g \in \mathcal{H}} \int_{S} g \leq \mu S<\infty.$

This completes the proof.$$\quad \square$$

Note 1. As $$\mu<\infty$$ and $$f \geq 0,$$ Corollary 1 in §5 shows that $$f$$ can be made finite on all of $$S.$$ Also, $$f$$ is $$m$$-integrable on $$S.$$

## Theorem $$\PageIndex{1}$$ (Radon-Nikodym)

If $$(S, \mathcal{M}, m)$$ is a $$\sigma$$-finite measure space, if $$S \in \mathcal{M},$$ and if

$\mu : \mathcal{M} \rightarrow E^{n}\left(C^{n}\right)$

is a generalized $$m$$-continuous measure, then

$\mu=\int f dm \text { on } \mathcal{M}$

for at least one map

$f : S \rightarrow E^{n}\left(C^{n}\right),$

$$\mathcal{M}$$-measurable on $$S$$.

Moreover, if $$h$$ is another such map, then $$m S$$ $$(f \neq h)=0$$

The last part of Theorem 1 means that $$f$$ is "essentially unique." We call $$f$$ the Radon-Nikodym $$(RN)$$ derivative of $$\mu,$$ with respect to $$m.$$

Proof

Via components (Theorem 5 in Chapter 7, §11), all reduces to the case

$\mu : \mathcal{M} \rightarrow E^{1}.$

Then Theorem 4 (Jordan decomposition) in Chapter 7, §11, yields

$\mu=\mu^{+}-\mu^{-},$

where $$\mu^{+}$$ and $$\mu^{-}$$ are finite measures $$(\geq 0),$$ both $$m$$-continuous (Corollary 3 from Chapter 7, §11). Therefore, all reduces to the case $$0 \leq \mu<\infty.$$

Suppose first that $$m,$$ too, is finite. Then if $$\mu=0,$$ just take $$f=0$$.

If, however, $$\mu S>0,$$ take $$f \in \mathcal{H}$$ as in Lemma 2 and Note 1; $$f$$ is nonnegative, bounded, and $$\mathcal{M}$$-measurable on $$S$$,

$\int f \leq \mu<\infty,$

and

$\int_{S} f dm=k=\sup _{g \in \mathcal{H}} \int_{S} g dm.$

We claim that $$f$$ is the required map.

Indeed, let

$\nu=\mu-\int f dm;$

so $$\nu$$ is a finite $$m$$-continuous measure $$(\geq 0)$$ on $$\mathcal{M}.$$ (Why?) We must show that $$\nu=0$$.

Seeking a contradiction, suppose $$\nu S>0.$$ Then by Lemma 1, there are $$P \in \mathcal{M}$$ and $$\delta>0$$ such that $$m P>0$$ and

$(\forall X \in \mathcal{M}) \quad \nu X \geq \delta \cdot m(X \cap P).$

Now let

$g=f+\delta \cdot C_{P};$

so $$g$$ is $$\mathcal{M}$$-measurable and $$\geq 0.$$ Also,

\begin{aligned}(\forall X \in \mathcal{M}) \quad \int_{X} g=\int_{X} f+\delta \int_{X} C_{P} &=\int_{X} f+\delta \cdot m(X \cap P) \\ & \leq \int_{X} f+\nu(X \cap P) \\ & \leq \int_{X} f+\nu X=\mu X \end{aligned}

by our choice of $$\delta$$ and $$\nu.$$ Thus $$g \in \mathcal{H}.$$ On the other hand,

$\int_{S} g=\int_{S} f+\delta \int_{S} C_{P}=k+\delta m P>k,$

contrary to

$k=\sup _{g \in \mathcal{H}} \int_{S} g.$

This proves that $$\int f=\mu,$$ indeed.

Now suppose there is another map $$h \in \mathcal{H}$$ with

$\mu=\int h d m=\int f d m \neq \infty;$

so

$\int(f-h) dm=0.$

(Why?) Let

$Y=S(f \geq h) \text { and } Z=S(f<h);$

so $$Y, Z \in \mathcal{M}$$ (Theorem 3 of §2) and $$f-h$$ is sign-constant on $$Y$$ and $$Z.$$ Also, by construction,

$\int_{Y}(f-h) dm=0=\int_{Z}(f-h) dm.$

Thus by Theorem 1(h) in §5, $$f-h=0$$ a.e. on $$Y,$$ on $$Z,$$ and hence on $$S=Y \cup Z$$ that is,

$mS(f \neq h)=0.$

Thus all is proved for the case $$mS<\infty$$.

Next, let $$m$$ be $$\sigma$$-finite:

$S=\bigcup_{k=1}^{\infty} S_{k} \text { (disjoint)}$

for some sets $$S_{k} \in \mathcal{M}$$ with $$m S_{k}<\infty$$.

By what was shown above, on each $$S_{k}$$ there is an $$\mathcal{M}$$-measurable map $$f_{k} \geq 0$$ such that

$\int_{X} f_{k} dm=\mu X$

for all $$\mathcal{M}$$-sets $$X \subseteq S_{k}.$$ Fixing such an $$f_{k}$$ for each $$k,$$ define $$f : S \rightarrow E^{1}$$ by

$f=f_{k} \quad \text { on } S_{k}, \quad k=1,2, \ldots.$

Then (Corollary 3 in §1) $$f$$ is $$\mathcal{M}$$-measurable and $$\geq 0$$ on $$S$$.

Taking any $$X \in \mathcal{M},$$ set $$X_{k}=X \cap S_{k}.$$ Then

$X=\bigcup_{k=1}^{\infty} X_{k} \text { (disjoint)}$

and $$X_{k} \in \mathcal{M}.$$ Also,

$(\forall k) \quad \int_{X_{k}} f d m=\int_{X_{k}} f_{k} d m=\mu X_{k}.$

Thus by $$\sigma$$-additivity (Theorem 2 in §5),

$\int_{X} f d m=\sum_{k=1}^{\infty} \int_{X_{k}} f d m=\sum_{k} \mu X_{k}=\mu X<\infty \quad (\mu \text { is finite!}).$

Thus $$f$$ is as required, and its "uniqueness" follows as before.$$\quad \square$$

Note 2. By Definition 3 in §10, we may write

$"d \mu=f dm"$

for

$"\int f dm=\mu."$

Note 3. Using Definition 2 in §10 and an easy "componentwise" proof, one shows that Theorem 1 holds also with $$m$$ replaced by a generalized measure $$s$$. The formulas

$\mu=\int f dm \text { and } mS(f \neq h)=0$

then are replaced by

$\mu=\int f ds \text { and } v_{s}S(f \neq h)=0.$

II. Theorem 1 requires $$\mu$$ to be $$m$$-continuous $$(\mu \ll m).$$ We want to generalize Theorem 1 so as to lift this restriction. First, we introduce a new concept.

## Definition

Given two set functions $$s, t : \mathcal{M} \rightarrow E\left(\mathcal{M} \subseteq 2^{S}\right),$$ we say that $$s$$ is $$t$$-singular $$(s \perp t)$$ iff there is a set $$P \in \mathcal{M}$$ such that $$v_{t} P=0$$ and

$(\forall X \in \mathcal{M} | X \subseteq-P) \quad s X=0.$

(We then briefly say "s resides in $$P.$$")

For generalized measures, this means that

$(\forall X \in \mathcal{M}) \quad s X=s(X \cap P).$

Why?

## Corollary $$\PageIndex{1}$$

If the generalized measures $$s, u : \mathcal{M} \rightarrow E$$ are $$t$$-singular, so is $$k s$$ for any scalar $$k$$ (if $$s$$ is scalar valued, $$k$$ may be a vector).

So also are $$s \pm u,$$ provided $$t$$ is additive.

Proof

(Exercise! See Problem 3 below.)

## Corollary $$\PageIndex{2}$$

If a generalized measure $$s : \mathcal{M} \rightarrow E$$ is $$t$$-continuous $$(s \ll t)$$ and also $$t$$-singular $$(s \perp t),$$ then $$s=0$$ on $$\mathcal{M}.$$

Proof

As $$s \perp t,$$ formula (3) holds for some $$P \in \mathcal{M}, v_{t} P=0.$$ Hence for all $$X \in \mathcal{M},$$

$s(X-P)=0 \text { (for } X-P \subseteq-P \text{)}$

and

$v_{t}(X \cap P)=0 \text { (for } X \cap P \subseteq P \text{).}$

As $$s \ll t,$$ we also have $$s(X \cap P)=0$$ by Definition 3(i) in Chapter 7, §11. Thus by additivity,

$sX=s(X \cap P)+s(X-P)=0,$

as claimed.$$\quad \square$$

## Theorem $$\PageIndex{2}$$ (Lebesgue decomposition)

Let $$s, t : \mathcal{M} \rightarrow E$$ be generalized measures.

If $$v_{s}$$ is $$t$$-finite (Definition 3(iii) in Chapter 7, §11), there are generalized measures $$s^{\prime}, s^{\prime \prime} : \mathcal{M} \rightarrow E$$ such that

$s^{\prime} \ll t \text { and } s^{\prime \prime} \perp t$

and

$s=s^{\prime}+s^{\prime \prime}.$

Proof

Let $$v_{0}$$ be the restriction of $$v_{s}$$ to

$\mathcal{M}_{o}=\left\{X \in \mathcal{M} | v_{t} X=0\right\}.$

As $$v_{s}$$ is a measure (Theorem 1 of Chapter 7, §11), so is $$v_{0}$$ (for $$\mathcal{M}_{0}$$ is a $$\sigma$$-ring; verify!).

Thus by Problem 13 in Chapter 7, §6, we fix $$P \in \mathcal{M}_{0},$$ with

$v_{s} P=v_{0} P=\max \left\{v_{s} X | X \in \mathcal{M}_{0}\right\}.$

As $$P \in \mathcal{M}_{0},$$ we have $$v_{t} P=0;$$ hence

$|sP| \leq v_{s} P<\infty$

(for $$v_{s}$$ is $$t$$-finite).

Now define $$s^{\prime}, s^{\prime \prime}, v^{\prime},$$ and $$v^{\prime \prime}$$ by setting, for each $$X \in \mathcal{M}$$,

\begin{aligned} s^{\prime} X &=s(X-P); \\ s^{\prime \prime} X &=s(X \cap P); \\ v^{\prime} X &=v_{s}(X-P); \\ v^{\prime \prime} X &=v_{s}(X \cap P). \end{aligned}

As $$s$$ and $$v_{s}$$ are $$\sigma$$-additive, so are $$s^{\prime}, s^{\prime \prime}, v^{\prime},$$ and $$v^{\prime \prime}$$. (Verify!) Thus $$s^{\prime}, s^{\prime \prime} : \mathcal{M} \rightarrow E$$ are generalized measures, while $$v^{\prime}$$ and $$v^{\prime \prime}$$ are measures $$(\geq 0)$$.

We have

$(\forall X \in \mathcal{M}) \quad s X=s(X-P)+s(X \cap P)=s^{\prime} X+s^{\prime \prime} X;$

i.e.,

$s=s^{\prime}+s^{\prime \prime}.$

Similarly one obtains $$v_{s}=v^{\prime}+v^{\prime \prime}$$.

Also, by (5), since $$X \cap P=\emptyset$$,

$-P \supseteq X \text { and } X \in \mathcal{M} \Longrightarrow s^{\prime \prime} X=0,$

while $$v_{t} P=0$$ (see above). Thus $$s^{\prime \prime}$$ is $$t$$-singular, residing in $$P$$.

To prove $$s^{\prime} \ll t,$$ it suffices to show that $$v^{\prime} \ll t$$ (for by (4) and (6), $$v^{\prime} X=0$$ implies $$\left|s^{\prime} X\right|=0$$).

Assume the opposite. Then

$(\exists Y \in \mathcal{M}) \quad v_{t} Y=0$

(i.e., $$Y \in \mathcal{M}_{0}$$), but

$0<v^{\prime} Y=v_{s}(Y-P).$

$v_{s}(Y \cup P)=v_{s} P+v_{s}(Y-P)>v_{s} P,$

with $$Y \cup P \in \mathcal{M}_{0},$$ contrary to

$v_{s} P=\max \left\{v_{s} X | X \in \mathcal{M}_{0}\right\}.$

This contradiction completes the proof.$$\quad \square$$

Note 4. The set function $$s^{\prime \prime}$$ in Theorem 2 is bounded on $$\mathcal{M}.$$ Indeed, $$s^{\prime \prime} \perp t$$ yields a set $$P \in \mathcal{M}$$ such that

$(\forall X \in \mathcal{M}) \quad s^{\prime \prime}(X-P)=0;$

and $$v_{t} P=0$$ implies $$v_{s} P<\infty.$$ (Why?) Hence

$s^{\prime \prime} X=s^{\prime \prime}(X \cap P)+s^{\prime \prime}(X-P)=s^{\prime \prime}(X \cap P).$

As $$s=s^{\prime}+s^{\prime \prime},$$ we have

$\left|s^{\prime \prime}\right| \leq|s|+\left|s^{\prime}\right| \leq v_{s}+v_{s^{\prime}};$

so

$\left|s^{\prime \prime} X\right|=\left|s^{\prime \prime}(X \cap P)\right| \leq v_{s} P+v_{s^{\prime}} P.$

But $$v_{s^{\prime}} P=0$$ by $$t$$-continuity (Theorem 2 of Chapter 7, §11). Thus $$\left|s^{\prime \prime}\right| \leq v_{s} P<\infty$$ on $$\mathcal{M}.$$

Note 5. The Lebesgue decomposition $$s=s^{\prime}+s^{\prime \prime}$$ in Theorem 2 is unique. For if also

$u^{\prime} \ll t \text { and } u^{\prime \prime} \perp t$

and

$u^{\prime}+u^{\prime \prime}=s=s^{\prime}+s^{\prime \prime},$

then with $$P$$ as in Problem 3, $$(\forall X \in \mathcal{M})$$

$s^{\prime}(X \cap P)+s^{\prime \prime}(X \cap P)=u^{\prime}(X \cap P)+u^{\prime \prime}(X \cap P)$

and $$v_{t}(X \cap P)=0.$$ But

$s^{\prime}(X \cap P)=0=u^{\prime}(X \cap P)$

by $$t$$-continuity; so (8) reduces to

$s^{\prime \prime}(X \cap P)=u^{\prime \prime}(X \cap P),$

or $$s^{\prime \prime} X=u^{\prime \prime} X$$ (for $$s^{\prime \prime}$$ and $$u^{\prime \prime}$$ reside in $$P$$). Thus $$s^{\prime \prime}=u^{\prime \prime}$$ on $$\mathcal{M}$$.

By Note 4, we may cancel $$s^{\prime \prime}$$ and $$u^{\prime \prime}$$ in

$s^{\prime}+s^{\prime \prime}=u^{\prime}+u^{\prime \prime}$

to obtain $$s^{\prime}=u^{\prime}$$ also.

Note 6. If $$E=E^{n}\left(C^{n}\right),$$ the $$t$$-finiteness of $$v_{s}$$ in Theorem 2 is redundant, for $$v_{s}$$ is even bounded (Theorem 6 in Chapter 7, §11).

We now obtain the desired generalization of Theorem 1.

## Corollary $$\PageIndex{3}$$

If $$(S, \mathcal{M}, m)$$ is a $$\sigma$$-finite measure space $$(S \in \mathcal{M}),$$ then for any generalized measure

$\mu : \mathcal{M} \rightarrow E^{n}\left(C^{n}\right),$

there is a unique $$m$$-singular generalized measure

$s^{\prime \prime} : \mathcal{M} \rightarrow E^{n}\left(C^{n}\right)$

and a ("essentially" unique) map

$f : S \rightarrow E^{n}\left(C^{n}\right),$

$$\mathcal{M}$$-measurable and $$m$$-integrable on $$S,$$ with

$\mu=\int f dm+s^{\prime \prime}.$

(Note 3 applies here.)

Proof

By Theorem 2 and Note 5, $$\mu=s^{\prime}+s^{\prime \prime}$$ for some (unique) generalized measures $$s^{\prime}, s^{\prime \prime} : \mathcal{M} \rightarrow E^{n}\left(C^{n}\right),$$ with $$s^{\prime} \ll m$$ and $$s^{\prime \prime} \perp m.$$

Now use Theorem 1 to represent $$s^{\prime}$$ as $$\int f dm,$$ with $$f$$ as stated. This yields the result.$$\quad \square$$