Chapter 8 Radon-Nikodym Theorem

8.1 Signed measures

We introduce the notion of signed measures which will be useful in the proof of the Radon-Nikodym theorem.

Definition 8.1 (Finite signed measure) A function μ from a σ-algebra E to R is a finite signed measure if

  • μ()=0,
  • If (An)n1 is a sequence of disjoint sets then μ(nAn)=nμ(An)

Example 8.1 If (E,E,μ) is a measure space and fL1(E) then ν defined by ν(A)=μ(f1A) is a signed measure.

We want to show two decomposition theorems which basically allow us to reduce the situation back to measures. First we need some more definitions and a useful Lemma.

Definition 8.2 If (E,E) is a measurable space and ν is a finite signed measure then we call A a positive set if for every BE with BA then ν(B)0. The negative sets are defined analogously.

Lemma 8.1 Suppose that ν is a finite signed measure on (E,E) and suppose AE with ν(A)<0 then there exists a negative set B with BA and ν(B)ν(A).

Proof. We will produce this set A by an itterative process, define

δ1=sup

then since \emptyset \subseteq A we have that \delta_1 \geq 0. If \delta_1 = 0 then we have a negative set so are done. If not we can find a set C_1 \subseteq A with

\nu(C_1) \geq \min\{\delta_1/2 ,1\}.

(We take the minimum here because we don’t know that \delta_1 is finite.) Now we will define a sequence of \delta_n and C_n by setting

\delta_n = \sup\{ \nu(C) \,:\, C \subseteq (A \setminus \bigcup_{i=1}^{n-1}C_i)\}

and C_n \subseteq A a set so that C_n \cap \left( \bigcup_{i=1}^{n-1}C)i\right) = \emptyset and

\nu(C_n) \geq \min\{ \delta_n /2 ,1\}.

Now let C_\infty = \bigcup_n C_n and B = A \setminus C_\infty. We now need to check that B has the required properties,

\nu(A) = \nu(C_\infty) + \nu(B) \geq \nu(B),

as \nu(C_\infty) \geq 0 by construction. As \nu is a finite measure we must have \nu(C_\infty) < \infty and as the C_n are constructed to be disjoint this means we must have \lim_n \nu(C_n) = 0. Therefore \lim_n \delta_n = 0. If D \subseteq B then we must have that \nu(D) \leq \delta_n for every n, therefore \nu(D) \leq 0.

Now we are able to state and prove our two decomposition theorems.

Theorem 8.1 (Hahn Decomposition theorem) Let (E, \mathcal{E}) be a measure space and \nu a finite signed measure. Then there exists a positive set P and a negative set N for \nu such that E = P \cup N.

Proof. Let L = \inf\{ \nu(A)\,:\, \mbox{$A$ is a negative set for $\nu$}\} then L is finite as otherwise we could construct a set with measure -\infty. Then let A_n be a negative set with \nu(A_n) \leq L+1/n then let N = \bigcup_n A_n.

We can check that N is a negative set and that \nu(N) = L. If A \subseteq N then let B_n = A_n \setminus \bigcup_{k=1}^{n-1}A_k then A = \bigcup_n (A \cap B_n) and A \cap B_n \subseteq A_n so \nu(A \cap B_n) \leq 0 and \nu(A) = \sum_n \nu(A \cap B_n) \leq 0. Now since N is a negative set \nu(N \setminus A_n) \leq 0, therefore \nu(N) = \nu(N \setminus A_n) + \nu(A_n) \leq \nu(A_n) \leq L +1/n. This is true for any n so \nu(N) \leq L and since L is defined to be the infinmum over \nu(A) for all negative sets A, we will have \nu(N) \geq L, therefore \nu(N) = L.

Let P = N^c we want to check that P is a positive set. Suppose there exists a set A \subseteq P with \nu(A) < 0, then by our lemma there exists a negative set B \subseteq A with \nu(B) \leq \nu(A)<0. Then N \cup B is a negative set and N and B are disjoint so \nu(N \cup B) = \nu(N) + \nu(B) < \nu(N) which contradicts the fact that \nu(N) = L = \inf\{ \nu(A)\,:\, \mbox{$A$ is a negative set for $\nu$}\} so we are done.

Theorem 8.2 (Jordan decompostion theorem) Every finite signed measure is the difference of two positive measures. Precisely, if (E, \mathcal{E}) is a measure space and \nu is a signed measure then there exist measures \nu_+ and \nu_- such that for every A \in \mathcal{E} we have \nu(A) = \nu_+(A)- \nu_-(A).

Proof. Take some Hahn decomposition (P, N) then let \nu_+(A) = \nu(A \cap P), as A \cap P \subseteq P then \nu(A \cap P) \geq 0. Similarly let \nu_-(A) = -\nu(A \cap N). By additivity of \nu we have that \nu(A) = \nu_+(A) - \nu_-(A). Countable additivity of \nu_+ and \nu_- follow immediately from countable additivity of \nu.

Now we notice that if B \subseteq A then

\nu(B) = \nu_+(B) - \nu_-(B) \leq \nu_+(B) \leq \nu_+(A)

and \nu_+(A) = \nu(A \cap P) therefore we have that

\nu_+(A) = \sup\{\nu(B) \,:\, B \subseteq A, B \in \mathcal{E}\}

in the same way

\nu_-(A) = \sup\{ - \nu(B) \,:\, B \subseteq A, B \in \mathcal{E}\}.

This shows that the values of \nu_+, \nu_- do not depend on the particular choice of Hahn decomposition.

8.2 Absolute Continuity

We now move on to the main focus of this section, the Radon-Nikodym theorem. In order to understand the theorem we need a definition.

Definition 8.3 Let (E, \mathcal{E}) be a measurable space and \mu and \nu be two measures then we say that \nu is absolutely continuous with respect to \mu of \nu \ll \mu if for every A \in \mathcal{E} with \mu(A) = 0 we also have \nu(A) = 0.

We can characterise absolute continuity

Lemma 8.2 Suppose that (E, \mathcal{E}) is a measurable space and \mu a measure, \nu a finite measure then \nu \ll \mu if and only if for earch \epsilon >0 there exists a \delta>0 such that \mu(A) < \delta implies that \nu(A) < \epsilon.

Proof. First let us suppose there exists such at \delta for each \epsilon, then if \mu(A) = 0 we have that \mu(A)< \delta for every \delta so we must have \nu(A) < \epsilon for every \epsilon so \nu(A) =0.

Now let us suppose that \nu \ll \mu. We prove the result by contradiction. Suppose there exists an \epsilon such that for every \delta there exists a set A with \mu(A)< \delta but \nu(A) >\epsilon. Then we can find a sequence of sets A_k such that \mu(A_k) < 2^{-k} but \nu(A_k) \geq \epsilon. By the first Borel-Cantelli lemma we have that

\mu \left( \bigcap_n \bigcup_{m \geq n} A_m \right) = 0.

We also have that \nu(\bigcup_{m \geq n} A_m) \geq \nu(A_n) \geq \epsilon and

\nu \left( \bigcap_n \bigcup_{m \geq n} A_m\right) = \lim_n \nu \left( \bigcup_{m \geq n} A_m \right) \geq \epsilon.

This show gives us a set with \mu(B) = 0 but \nu(B) > 0 which contradicts \nu \ll \mu.

Now we can prove the main theorem for this section.

Theorem 8.3 (Radon-Nikodym Theorem) Let (E, \mathcal{E}) be a measure space and let \mu, \nu be two finite measures with \nu \ll \mu. Then there exists a measurable function g: E \rightarrow [0, \infty) such that \nu(A) = \mu(g1_A). The function g is unique up to identifying almost everywhere equal functions. We write g = \mathrm{d}\nu/\mathrm{d}\mu and call it the Radon-Nikodym derivative of \nu with respect to \mu.

Proof. Let us define the set \mathcal{F} which is the set of all measurable functions, f, with \mu(f1_A) \leq \nu(A) for every A \in \mathcal{E}. The idea is that \mathcal{F} contains a function g which achieves \mu(g) = \sup_{f \in \mathcal{F}} \mu(f).

As a first step we show that f_1 \vee f_2 = \max\{ f_1, f_2\} \in \mathcal{F} when f_1, f_2 \in \mathcal{F}. Let us take any A \in \mathcal{E} then let A_1 = A \cap \{ f_1 \geq f_2\} and A_2 = A \cap \{ f_1 < f_2\}. Then

\mu(f_1 \vee f_2 1_A) = \mu(f_1 \vee f_2 1_{A_1}) + \mu(f_1 \vee f_2 1_{A_2}) = \mu(f_1 1_{A_1}) + \mu(f_2 1_{A_2}) \leq \nu(A_1) + \nu(A_2) = \nu(A).

Therefore f_1 \vee f_2 \in \mathcal{F}.

Now take a sequence f_n such that \mu(f_n) \geq \sup_{f \in \mathcal{F}} \mu(f) - 1/n. Then let g_n = f_1 \vee f_2 \vee \dots \vee f_n, so that the sequence of function g_n is increasing and \mu(g_n) \geq \sup_{f \in \mathcal{F}} \mu(f) - 1/n. Then as g_n is increasing it has a limit g and the monotone convergence theorem shows that

\mu(g1_A) = \lim_n \mu(g_n1_A) \leq \nu(A).

So g \in \mathcal{F}.

Now we can define another positive measure \nu_0(A) = \nu(A) - \mu(g1_A). We want to show that \nu_0 =0 and will do this by contradiction. Suppose that there exists A \in \mathcal{E} such that \nu_0(A)>0 then by monotonicity we will have \nu_0(E) >0 and since \mu is a finite measure there exists a number \epsilon >0 such that \nu_0(E) > \epsilon \mu(E). Now \nu_0 - \epsilon \mu is a finite signed measure. Let (P, N) be a Hahn decomposition for this signed measure. Then (\nu_0 - \epsilon \mu)(A \cap P) \geq 0 so \nu_0(A\cap P) \geq \epsilon \mu(A \cap P). Hence we have

\begin{align*} \nu(A) &= \mu(g1_A) + \nu_0(A) \geq \mu(g1_A) + \nu_0(A \cap P) \\ & \geq \mu(g1_A) + \epsilon \mu(A \cap P) = \mu(1_A(g +\epsilon 1_{P})). \end{align*}

We also have that \mu(P) >0 as if \mu(P) = 0 then we would have \nu_0(P)=0 as \nu_0 \ll \nu \ll \mu, and this would mean

(\nu_0 - \epsilon \mu)(E) = (\nu_0 - \epsilon \mu)(N) \leq 0,

which would contradict \nu_0(E) > \epsilon \mu(E). Therefore, g+ \epsilon 1_P belongs to \mathcal{F} but \mu(g+1_P) > \mu(g) which contradicts the fact that g achieves \mu(g) = \sup_{f \in \mathcal{F}} \mu(f). Hence \nu(A) = \mu(g1_A).

Now we turn to uniqueness suppose that we have two positive functions g,h such that \nu(A) = \mu(g1_A) = \mu(h1_A) for every A, then as \nu is finite g and h are integrable so g-h is integrable and \mu((g-h)1_A) = 0 for every A. As g-h is measurable then \{x \in E\,:\, g-h \geq 0\} is a measurable set so \mu((g-h)1_{\{x \in E\,:\, g-h \geq 0\}}) = 0. This shows that (g-h)1_{\{x \in E\,:\, g-h \geq 0\}} = 0 almost everywhere. In the same way (g-h)1_{\{ x \in E\,:\, g-h \leq 0\}} =0 almost everywhere. Therefore g=h \mu-almost everywhere.

8.3 Duality in L^p spaces - NONEXAMINABLE

The goal of this section is to prove that if 1/p+1/q =1 then the dual space of L^p(E) is isomorphic to the space L^q(E). First let us define a dual space.

Definition 8.4 Let \mathcal{V} be a Banach space (a complete, normed vector space) then the dual space of \mathcal{V} is written \mathcal{V}' and is the space of all bounded linear operators from \mathcal{V} to \mathbb{R}. We recall that we call an operator K on \mathcal{V} bounded if |K(v)| \leq C\|v\| for all v \in \mathcal{V}. We can define a norm on \mathcal{V}' by \|K\| = sup_{\|v\| =1}|K(v)|.

The first thing to note is that if g \in L^q(E) then we can define a bounded linear operator on L^p(E) by K_g(f) = \mu(fg). This is bounded by Hölder’s inequality |\mu(fg)| \leq \mu(|fg|) = \|fg\|_1 \leq \|f\|_p \|g\|_p. It is also linear thanks to the linearity of the integral. Therfore we can produce a map from L^q(E) \rightarrow (L^p(E))' by g \mapsto K_g.

Theorem 8.4 Let (E, \mathcal{E}, \mu) be a finite measure space and p \in (1, \infty). The dual space of L^p(E) is L^q(E) where 1/p +1/q = 1. Furthermore the map defined by g \mapsto K_g is an isometry.

Remark. This result also holds for arbitrary measure spaces (without the finite assumption). Extending to \sigma-finite measure spaces is relatively straightforward and then to any measure space is more complicated.

Proof. Remark: This result is similar in spirit to the Riesz representation result that was a non-examinable topic in week 6.

First we note that the map g \mapsto K_g is linear and \|K_g\|_{(L^p)'} \leq \|g\|_q. Therefore the map is injective we want to show that \|K_g\| = \|g\| and that it is surjective.

First for the fact that \|K_g\| = \|g\| we look at the function f(x) = sgn(g)|g(x)|^{q-1} then \mu(|f|^p) = \mu(|g|^q) < \infty. Therefore we can look at the action of K_g on f and we have K_g(f) = \mu(|g|^q) so we know that \|K_g\| \geq K_g(f)/\|f\|_p = \mu(|g|^q)/\mu(|g|^q)^{1/p} = \mu(|g|^q)^{1-1/p} = \|g\|_q. Therefore g \mapsto K_g preserves norms.

Now we want to show that this map is surjective. Let us take K an arbitrary element of (L^p(E))'. Since \mu(E) < \infty, 1_A \in L^p(E) for every A \in \mathcal{E} so we can define a function on \mathcal{E} by k(A)=K(1_A). We want to show that k is a signed measure. k(\emptyset)=K(0)=0 and let A_1,A_2,\dots be a sequence of disjoint measurable sets. Then 1_{\bigcup_{j=1}^n A_j} = \sum_{j=1}^n 1_{A_j} then k(\bigcup_{j=1}^n A_j) = K(1_{\bigcup_{j=1}^n A_j}) = K( \sum_{j=1}^n 1_{A_j}) = \sum_{j=1}^n K(1_{A_j}) = \sum_{j=1}^n k(A_j). We also have that \| 1_{\bigcup_j A_j} - 1_{\bigcup_{j=1}^n A_j} \|_p \rightarrow 0 as n \rightarrow 0. Therefore, as K is a continuous map on L^p we have K(1_{\bigcup_n A_n}) = \sum_n K(1_{A_n}), so k(\bigcup_n A_n) = \sum_n k(A_n) Therefore k is indeed a signed measure. By the Hahn decomposition and the Jordan decomposition we can write k = k_+ - k_- and there exists P \cup N a Hahn decomposition with k being positive on P and negative on N.

Next we want to show that k_+ \ll \mu and k_- \ll \mu. If A \in \mathcal{E} is such that \mu(A) = 0 then \mu(A \cap P)=0 and \mu(A \cap N) = 0 and as 1_{A \cap P}= 0 \mu-a.e. we have K(1_{A \cap P}) = K(0) = 0 and K(1_{A \cap N}) = K(0) = 0. Therefore k_+(A) = 0 and k_-(A) = 0.

Then by the Radon-Nikodym theorem there exists functions g_+ and g_- such that k_+(A) = \mu(g_+1_A) and k_-(A) = - \mu(g_- 1_A). Now let g = g_+ - g_- we want to show that g \in L^q and that K = K_g. This is complicated!

Let us define E_n by E_n = \{ x \,:\, |g(x)| \leq n\} then g1_{E_n} is bounded and so in L^q as \mu is finite. Then define a linear functional on L^p by K_n(f) = \mu(fg1_{E_n}) and another by \tilde{K}_n (f) = K(f1_{E_n}). Then if A is a measurable set we have K_n (1_A) = \tilde{K}_n (1_A), by linearity if h is a simple function then K_n(h) = \tilde{K}_n(h).

We showed in Assignment 4 that given a function f \in L^p, \epsilon >0 there exists a simple function h with \|f-h\|_p \leq \epsilon. Then we have that |K_n(f) - \tilde{K}_n(f)| \leq |K_n(f) - K_n(h)| + |\tilde{K}_n(f) - \tilde{K}_n (h)| \leq \|K\| \|f-h\|_p + \|g1_{E_n}\|_q \|f-h\|_p \leq (\|K\| + \|g\|_q) \epsilon. Since \epsilon is arbitrary this shows that K_n(f) = \tilde{K}_n(f). \tilde{K}_n(f) = K_{g1_{E_n}}(f) so by our isometry we have \|\tilde{K}_n\| = \|g1_{E_n}\|_q. We also have that \|\tilde{K}_n\| \leq \|K\| as \| \tilde{K}_n\| = \sup_{\|f\|_p =1} K_n(f) = \sup_{\|f\|_p = 1} K(f1_{E_n}) \leq \sup_{\|f\|_p=1} K(f) = \|K\|. Therefore \|g1_{E_n}\|_q \leq \|K\| therefore \|g1_{E_n}\|_p^p = \int |g|^p 1_{E_n} \mu(\mathrm{d}x) and then by monotone convergence we get that \|g\|_q = \lim_n \|g 1_{E_n}\|_q \leq \|K\|. Therefore, g \in L^q. Then by exactly the same argument with which we showed K_n = \tilde{K}_n we have that K = K_g. This concludes the proof in the finite case.

Extending to the sigma finite case works very similarly to previous results.