Proof of Radon-Nikodym theorem

Proof of Radon-Nikodym theorem

Theorem (Radon-Nikodym). Suppose Ω is a nonempty set andA aσ-field on it. Supposeµand ν are σ-finite measures on (Ω,A) such that for all A∈A, µ(A) = 0 impliesν(A) = 0. Then

• There exists z : Ω→[0,∞) measurable such that for all A∈A, ν(A) =R

Az dµ.

• Such a z is unique upto a.e. equality (w.r.t.µ).

• z is integrable w.r.t. µif and only if ν is a finite measure.

Proof. Assuming we have obtained the z, it is easy to see that z is integrable if and only ifν is a finite measure, since ν(Ω) =R

z dµ.

Uniqueness

We show uniqueness first. Suppose there are two z1, z2 : Ω → [0,∞) such that for all A ∈ A, R

Az₁ dµ=ν(A) =R

Az₂ dµ. We would like to say R

Az₁ dµ−R

Az₂ dµ= 0, but both the integrals might be ∞. Both µ and ν are σ-finite. Let C be a countable partition of Ω such that for all C ∈ C, µ(C) < ∞ and ν(C) <∞ (to obtain this we can take such a partition for µ, and such a partition for ν, and take pairwise intersections). Fix a C ∈ C. Let

B =C∩ {ω :z₁(ω)−z₂(ω)>0}

Note that 1_B(z₁−z₂) = 1_C(z₁−z₂)⁺(evaluate both sides atω: forω /∈C, both are 0; forω∈C\B, LHS is 0 and since (z₁ −z₂)(ω) ≤ 0, RHS is 0; for ω ∈ B both side are z₁(ω)−z₂(ω)). ν(B) = R 1Bz1dµ=R

1Bz2dµ. Sinceν(B)≤ν(C)<∞, it is meaningful to considerR

1Bz1dµ−R

1Bz2dµ.

0 = Z

1_Bz₁ dµ− Z

1_Bz₂ dµ

= Z

1_B(z₁−z₂) dµ

= Z

1C(z1−z2)⁺ dµ

If the integral of a nonnegative function is 0, it must be 0 almost everywhere. Applying to the above:

µ({ω :ω ∈C and z₁(ω)−z₂(ω)>0}) = 0

We can interchange the roles of z₁ and z₂ in the above argument. Combining the two, we get µ({ω :ω ∈C and z₁(ω)−z₂(ω)6= 0}) = 0

This holds for all C∈ C. SinceC is countable, and a countable union of null sets is a null set, µ({ω :z₁(ω)6=z₂(ω)}) = 0

This proves that the z promised in the theorem (if it exists) is unique upto a.e. (w.r.t.µ) equality.

Existence

We now show the existence of such a z. We reduce the general case to the case when both µand ν are finite, and then show existence for the finite case.

Reduction to the finite measure case

Suppose we know the Radon-Nikodym theorem holds for the case when the measures involved are finite. In general, assume µ and ν are σ-finite, and let C be a countable partition of Ω such that for all C ∈ C,µ(C) and ν(C) are finite. Define finite measures µ_C and ν_C for all C ∈ C,

µ_C(A) = µ(C∩A), ν_C(A) =ν(C∩A)

If µ_C(A) = 0, then ν_C(A) = 0. Using the finite-measure case of the theorem, let z_C : Ω →[0,∞) be such that for all A∈A and C∈ C,

ν_C(A) = Z

A

z_C dµ_C

z_C and z_C1_C differ at most on Ω\C, which has 0 µ_C measure. So we can replace z_C by z_C1_C, and so assume that for ω /∈ C, z_C(ω) = 0. Define z : Ω → [0,∞), by requiring that z|_C = (z_C)|_C (that is, for ω ∈ C, z(ω) = z_C(ω)). z can also be described as P

C∈Cz_C - this shows that z is measurable.

If g : Ω→[0,∞) is 0 outside C, then R

g dµ=R

g dµ_C. This can be proven as usual by proving it for simple functions and then for all nonnegative measurable functions. This fact is applied to z_C in what follows. For anyA ∈A,

ν(A) =X

C∈C

ν(A∩C)

=X

C∈C

νC(A)

=X

C∈C

Z

A

zC dµC

=X

C∈C

Z

A

z_C dµ

= Z

A

X

C∈C

z_C dµ (by taking partial sums; MCT)

= Z

A

z dµ

It remains to prove the theorem for the finite measure case:

Finite measures - getting hold of the z:

Now assume µand ν are finite measures. Let C =

f : (f : Ω→[0,∞]R¯),∀A∈A(R

Af dµ≤ν(A))

We want to identify f such that equality holds, that is, for all A∈A, R

Af dµ=ν(A).

C is nonempty, because the zero function belongs toC. Intuitively, we want to choose the “largest”

function in C as a candidate for our z. Does such a “largest” function exist? We observe some properties of C first:

• Iff, g ∈C, thenf∨g (which is the pointwise maximum off and g) belongs to C. Take any A ∈ A. We apply the condition defining C to the sets A₁ := A∩ {ω : f(ω) ≥ g(ω)} and A₂ :=A∩ {ω :f(ω)< g(ω)}:

Z

A1

f ≤ν(A1), Z

A2

g ≤ν(A2)

Note thatf1_A1 = (f∨g)1_A1 andg1_A2 = (f∨g)1_A2. Substituting in the above equations and adding them, we get

Z

A

(f ∨g)≤ν(A)

• Suppose {f_n}^∞_n=1 is an increasing sequence in C. Let f be their supremum (equivalently limit). Then f ∈C: take any A ∈A. {f_n1_A}^∞_n=1 increases to f1_A, and so by the monotone convergence theorem the corresponding limit with integrals holds. Since eachR

f_n1_A≤ν(A), we have R

f1A≤ν(A).

We identify the “largest” function in C by considering the function with the largest integral. For all f ∈C, R

f dµ≤ν(Ω)<∞. Let

α= supn

R f dµ :f ∈Co

0 ≤α ≤ν(Ω). In particular, α 6=∞. Let {f_n}^∞_n=1 be a sequence from C such that R

f_n dµ ^∞

n=1

converges to α (this can be done because of the supremum definition of α). Consider the partial maxima of {f_n}^∞_n=1, that is, define {g_n}^∞_n=1 by

g_n = max{f₁, f₂, . . . , f_n} We know that g_n ∈ C, so R

g_n dµ ≤ α. Also, f_n ≤ g_n, so R

f_n dµ ≤ R

g_n dµ ≤ α. Thus limn→∞

R gn dµ = α. {gn}^∞_n=1 is an increasing sequence; let z = limn→∞gn. By the monotone convergence theorem, R

z dµ=α. Since z is integrable, z is finite almost everywhere. We modify z on a set of measure 0 suitably so thatz never takes the value∞. Thisz is our “largest” function in C, as we show now:

Let g ∈ C. We will show that g ≤ z a.e. (w.r.t µ). Let A = {ω : g(ω) > z(ω)}. Let z∨g = h.

h∈C. A={ω:h(ω)> z(ω)}={ω :h(ω)−z(ω)>0}. h ≥z;R

h dµ≥R

z dµ=α, so R

h=α.

h−z is a nonnegative function whose µintegral is 0, so h−z is 0 almost everywhere. µ(A) = 0.

g ≤z almost everywhere (w.r.t µ).

Finite measures - showing that the z works Define λ:A →[0,∞),

λ(A) = ν(A)− Z

A

z dµ

Note that this is well-defined, that is, the expression on the RHS indeed belongs to [0,∞). λ is a measure: clearly λ(∅) = 0. LetC ⊆A be a countable collection of disjoint sets.

X

C∈C

ν(C) +X

C∈C

Z

C

z dµ =ν[ C

+ Z

SC

z dµ <∞

Since the relevant series is absolutely convergent, we can rearrange terms, and λ[

C

=ν[

C

− Z

SC

z dµ =X

C∈C

ν(C)−X

C∈C

Z

C

z dµ=X

C∈C

ν(C)− Z

C

z dµ

=X

C∈C

λ(C)

λ is a measure. We want to show that λ is the zero measure, and then we will be done.

If for all A ∈ A and all k ∈ N, λ(A) ≤ ¹_kµ(A) then since µ(A) is finite, λ(A) = 0, and we are done. So there exists A ∈A and k ∈ N such that λ(A)− ¹_kµ(A)> 0. Fix such an A and k. We will derive a contradiction.

Call a set Z ∈A to be good if the following two conditions hold:

• λ(Z)− ¹_kµ(Z)>0.

• For all B ⊆Z with B ∈A, λ(B)− ¹_kµ(B)≥0.

The first condition implies that a good set Z must haveµ(Z)>0: otherwise ifµ(Z) = 0, then by the hypothesis, ν(Z) = 0, and soλ(Z) = 0. A satisfies the first condition, but need not satisfy the second condition.

We will identify a good set - A itself may not be a good set, but we can “scrape off” some parts of it to get a good set. Once we identify a good set Z, we will use the function z +_k¹1_Z to get a contradiction.

Let A₀ =A. Let β₀ = inf{λ(B)− ¹_kµ(B) : B ⊆A, B ∈A}. If β₀ ≥0, then A itself is a good set.

So assume β₀ <0. β₀ cannot be −∞, since it is at least −(ν(Ω) +R

z dµ+¹_kµ(Ω)). Pick B₀ ⊆A₀ such that

λ(B₀)− 1

kµ(B₀)< 1 2β₀ Put A₁ =A\B₀.

λ(A₁)− ¹_kµ(A₁) = λ(A)−λ(B₀)

− _k¹µ(A)− ¹_kµ(B₀)

≥λ(A)−_k¹µ(A)>0 (because λ(B₀)− ¹_kµ(B₀)< ¹₂β₀ <0.)

An important observation is this: if B ⊆A₁, thenλ(B)−_k¹µ(B)≥ ¹₂β₀. Otherwise, we could take a B which violates this, and B∪B₀ would violate the infimum definition of β₀.

If A₁ is a good set, we have found a good set. Otherwise, repeat with A₁ what we did with A₀. The above observation says that theβ₁so obtained will satisfyβ₁ ≥ ¹₂β₀. Inductively, we construct a sequences of sets {An}^∞_n=1 and {Bn}^∞_n=1, and a sequence of real numbers {βn}^∞_n=1, such that the following holds: some A_n is good, or for all n∈N,

• B_n⊆A_n.

• A_n+1 =A_n\B_n.

• β_n = inf

λ(B)− ¹_kµ(B) :B ⊆A_n, B ∈A ,β_n <0.

• λ(A_n+1)− ¹_kµ(A_n+1)≥λ(A_n)− ¹_kµ(A_n).

• β_n+1 ≥ ¹₂β_n.

This immediately implies that for all n,β_n ≥ ₂¹nβ₀ and λ(A_n)− ¹_kµ(A_n)≥λ(A)− ¹_kµ(A)>0.

If no A_n was good, define A_∞ =T

n∈NA_n. Since λ and µare finite measures, λ(A∞) = lim

n→∞λ(A_n) and µ(A∞) = lim

n→∞µ(A_n)

So λ(A∞) − ¹_kµ(A∞) ≥ λ(A)− ¹_kµ(A) > 0. If B ⊆ A∞ and B ∈ A, then since B ⊆ A_n, λ(B)−_k¹µ(B)≥ ₂¹nβ0. Since this holds for alln,λ(B)−¹_kµ(B)≥0. A∞ is a good set! If someAn

was good, let A∞ be that good set.

We will show that z+ ¹_k1A∞ ∈C. For any S ∈A, if S ⊆Ω\A∞, then Z

S

z+ 1

k1A∞

dµ=

Z

S

z dµ ≤ν(S)

If S ⊆A_∞, then R

S z+¹_k1_A∞

dµ=R

Sz dµ+ ¹_kµ(S)

≤R

Sz dµ+λ(S) (since A_∞ is a good set)

=ν(S) (by definition of λ)

For general S, we just combine the part ofS in A and the part in Ω\A:

Z

S

z+1

k1_A∞

dµ= Z

S∩A

z+1

k1_A∞

dµ+

Z

S\A

z+ 1

k1_A∞

dµ≤ν(S∩A)+ν(S\A) =ν(S)

We have shown that z+ ¹_k1_A∞ ∈ C. Since A∞ is a good set, µ(A∞) >0. It is not the case that z+ ¹_k1_A∞ ≤z a.e. (w.r.t. µ), which contradicts what we have shown earlier.

λ must be the zero measure, and so for all D∈A, ν(D) =

Z

D

z dµ