**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} ightarrow[0, infty)) be finite measures in (S.) Suppose (S in mathcal{M}, mu S>0) (i.e., (mu ot 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

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 ightarrow 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} ight} uparrow, g_{n} in mathcal{H},) such that

[lim _{n ightarrow infty} int_{S} g_{n} dm=k.]

(Verify!) Set

[f=lim _{n ightarrow infty} g_{n}.]

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

[k=lim _{n ightarrow 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 ightarrow 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

This completes the proof.(quad square)

**Note 1.** As (mu

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} ightarrow E^{n}left(C^{n} ight)]

is a generalized (m)-continuous measure, then

[mu=int f dm ext { on } mathcal{M}]

for at least one map

[f : S ightarrow E^{n}left(C^{n} ight),]

(mathcal{M})-measurable on (S).

Moreover, if (h) is another such map, then (m S) ((f eq 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} ightarrow 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

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

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

[ u=mu-int f dm;]

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

Seeking a contradiction, suppose ( u 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 u 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,

[egin{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+ u(X cap P) & leq int_{X} f+ u X=mu X end{aligned}]

by our choice of (delta) and ( u.) 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 eq infty;]

so

[int(f-h) dm=0.]

(Why?) Let

[Y=S(f geq h) ext { and } Z=S(f

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 eq h)=0.]

Thus all is proved for the case (mS

Next, let (m) be (sigma)-finite:

[S=igcup_{k=1}^{infty} S_{k} ext { (disjoint)}]

for some sets (S_{k} in mathcal{M}) with (m S_{k}

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 ightarrow E^{1}) by

[f=f_{k} quad ext { 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=igcup_{k=1}^{infty} X_{k} ext { (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

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 ext { and } mS(f eq h)=0]

then are replaced by

[mu=int f ds ext { and } v_{s}S(f eq 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} ightarrow Eleft(mathcal{M} subseteq 2^{S} ight),) 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} ightarrow 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} ightarrow 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 ext { (for } X-P subseteq-P ext{)}]

and

[v_{t}(X cap P)=0 ext { (for } X cap P subseteq P ext{).}]

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} ightarrow 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} ightarrow E) such that

[s^{prime} ll t ext { 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 ight}.]

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} ight}.]

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

[|sP| leq v_{s} P

(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}),

[egin{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} ightarrow 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 ext { 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 ight|=0)).

Assume the opposite. Then

[(exists Y in mathcal{M}) quad v_{t} Y=0]

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

[0

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} ight}.]

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

[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} ight| leq|s|+left|s^{prime} ight| leq v_{s}+v_{s^{prime}};]

so

[left|s^{prime prime} X ight|=left|s^{prime prime}(X cap P) ight| 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}
ight| leq v_{s} P

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

[u^{prime} ll t ext { 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}
ight),) 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} ightarrow E^{n}left(C^{n} ight),]

there is a unique (m)-singular generalized measure

[s^{prime prime} : mathcal{M} ightarrow E^{n}left(C^{n} ight)]

and a ("essentially" unique) map

[f : S ightarrow E^{n}left(C^{n} ight),]

(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} ightarrow E^{n}left(C^{n} ight),) 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)