8.11: The Radon–Nikodym Theorem. Lebesgue Decomposition
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.
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.\(\quad \square\)
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 \(S.\)
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.
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?
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.)
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.\(\quad \square\)
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
\[0<v^{\prime} Y=v_{s}(Y-P).\]
So by additivity,
\[v_{s}(Y \cup P)=v_{s} P+v_{s}(Y-P)>v_{s} P,\]
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.
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.\(\quad \square\)