Prove Lebesgue-Radon-Nikodym (LRN) Theorem in Three Ways
“Most mathematicians prove what they can, von Neumann proves what he wants.”
Theorem 1 (LRN): Let $\nu,\mu$ be finite positive measures on a measure space $(X,\mathcal{M})$. There exists positive measures $\lambda,\rho$ on $(X,\mathcal{M})$ and an integrable function $f:X\rightarrow [0,\infty]$ such that
\[\lambda \perp \mu, \rho \ll \mu, \nu=\lambda+\rho, d \rho=f d \mu\]This is actually a simplified version. However, the uniqueness of this decomposition and generalization to the case of finite signed measures or complex measures are easy. Therefore, I’ll mainly focus on Theorem 1 and present three different mathematical ideas related to the existence of this decomposition in this article.
Self-duailty of $L^2$
This argument is due to von Neumann. It also appears in one of the exercises on Folland’s book and is presented in Rudin’s book. Let $\lambda=\mu+\nu$. Consider the map $f \in L^2(\lambda) \rightarrow \int_X f d \nu$. This is a bounded linear functional on $L^2(\lambda)$ because by the Cauchy-Schwartz inequality, we have
\[\left|\int_X f d \nu\right| \leq(\nu(X))^{1 / 2}\|f\|_{L^2(\nu)} \leq(\nu(X))^{1 / 2}\|f\|_{L^2(\lambda)}\]We have the last inequality because
\[\begin{aligned} \|f\|_{L^2(\nu)} & =\left(\int_X f^2 d \nu\right)^{1 / 2} \\ & \leq\left(\int_X f^2 d \mu+\int_X f^2 d \nu\right)^{1 / 2} \\ & =\left(\int_X f^2 d \lambda\right)^{1 / 2} \\ & =\|f\|_{L^2(\lambda)} \end{aligned}\]Since $L^2(\lambda)$ is a Hilbert space, it is self-dual by the Riesz representation theorem, which suggests that there exists some $g\in L^2(\lambda)$ with $|g|_{L^2(\lambda)} \leq(\nu(X))^{1 / 2}$ such that for all $f \in L^2(\lambda)$,
\[\begin{aligned} \int_X f d \nu & =\int_X f g d \lambda\hspace{1mm}(1) \\ \int_X f(1-g) d \nu & =\int_X f g d \mu\hspace{1mm}(2) \end{aligned}\]Now suppose that $\lambda(\{g<0\})>0$.
\[\{g<0\}=\bigcup_{n \in \mathbb{N}}\left\{g<-\frac{1}{n}\right\}\]There exists some $n$ such that $\lambda(\{g<-\frac{1}{n}\}) \neq 0$. Replace $f$ with $1_{\{g<-\frac{1}{n}\}}$ in equality (1), we get
\[\begin{aligned} 0 & \leq \int_{\left\{g<-\frac{1}{n}\right\}} 1 d \nu=\int_{\left\{g<-\frac{1}{n}\right\}} g d \lambda \\ & \leq-\frac{1}{n} \lambda\left(\left\{g<-\frac{1}{n}\right\}\right)<0 \end{aligned}\]We have a strict inequality, which is a contradiction and suggests that $g\geq 0$ $\lambda-$a.e. Similarly, if $\lambda(\{g>1\})>0$, then
${g>1}=\bigcup_{n \in \mathbb{N}}{g>1+\frac{1}{n}}$
Choose an $n$ such that $\lambda\left(\{g>1+\frac{1}{n}\}\right) \neq 0$. Replace $f$ with $1_{\{g>1+\frac{1}{n}\}}$ in equality (1) to get,
\[\begin{aligned} \lambda\left(\left\{g>1+\frac{1}{n}\right\}\right) & \geq \nu\left(\left\{g>1+\frac{1}{n}\right\}\right)=\int_{\left\{g>1+\frac{1}{n}\right\}} 1 d \nu \\ & =\int_{\left\{g>1+\frac{1}{n}\right\}} g d \lambda \geq\left(1+\frac{1}{n}\right) \lambda\left(\left\{g>1+\frac{1}{n}\right\}\right) \\ & >\lambda\left(\left\{g>1+\frac{1}{n}\right\}\right) \end{aligned}\]Again, we obtain a contradiction because of the strict inequality. This means that $g\leq 1$ $\lambda-$a.e. We may thus assume that $0\leq g\leq 1$ everywhere. Consider the following two set and two measures induced by them:
\[\begin{aligned} & A=\{x: g(x)<1\}, \forall E \in \mathcal{M}, \nu_A(E)=\nu(A \cap E) \\ & B=\{x: g(x)=1\}, \forall E \in \mathcal{M}, \nu_B(E)=\nu(B \cap E) \end{aligned}\]Clearly $\nu=\nu_A+\nu_B$. Our goal now is to show that $\nu_B\perp \mu$ and $d \nu_A=1_A(1-g)^{-1} g d \mu$.
-
Since $\nu_B(A)=0$, we only need to show $\mu(B)=0$ to conclude that $\nu_B\perp \mu$, which follows from replacing $f$ with $1_B$ in equality (2):
\[\begin{aligned} 0 & =\int_B 0 d \nu=\int_B(1-g) d \nu \\ & =\int_B g d \mu=\int_B 1 d \mu=\mu(B) \end{aligned}\] -
$d \nu_A=1_A(1-g)^{-1} g d \mu$ follows from replacing $f$ with $1_E1_A(1+g+g^2+\cdots+g^n)$ in equality (2) for any measurable set $E$:
\[\begin{aligned} \nu_A(E) & =\nu(A \cap E)=\int_E 1_A d \nu \\ & =\int_E 1_A\left(1+g+g^2+\cdots+g^n\right)(1-g) d \nu \\ & =\int_E 1_A\left(1+g+g^2+\cdots+g^n\right) g d \mu \end{aligned}\]
Since $0\leq g< 1$ on $A$, $\sum_ng^n$ is an increasing, convergent geometric series. Thus, we can use MCT to change the order of limit and integral:
\[\nu_A(E)=\lim \int_E 1_A\left(1+g+g^2+\cdots+g^n\right) g d \mu=\int_E 1_A(1-g)^{-1} g d \mu\]A final remark: In the last step, we can not directly replace $f$ with $\chi_A(1-g)^{-1}$ in equality (2) because $\chi_A(1-g)^{-1}$ may not be in $L^2(\lambda)$.
Measure Theoretic Viewpoint
This argument is taken from Folland’s Real Analysis. Let’s reexamine the theorem and notice that the RN derivative $f$ satisfies the following property:
\[\forall E \in \mathcal{M}, \int_E f d \mu=\rho(E) \leq \nu(E)\]This naturally leads us to consider the set of all functions that satisfy the property above.
\[\mathcal{F}=\left\{f: X \rightarrow[0, \infty]: \forall E \in \mathcal{M}, \int_E f d \mu \leq \nu(E)\right\}\]Let $\alpha=\sup \{\int_X f d \mu: f \in \mathcal{F}\}$. If we can prove that a maximizer $g$ (i.e. $\int_X g d \mu=\alpha$ ) satisfies $d \nu-g d \mu$ is singular with respect to $d \mu$ (which is a consequence of Lemma 3.7 on Folland’s book), then we are done. How should we find this maximizer then? Well, a maximizing sequence $\{f_n\}$ clearly exists (i.e. $\int_X f_n d \mu \rightarrow \alpha$ ). Naively we want to take the pointwise limit of $\{f_n\}$, but this does not guarantee the convergence of the integral. The trick is to use the property that $\mathcal{F}$ is closed under $\max$ to show that $\{\max \{f_i: 1 \leq i \leq n\}\}_n$ is still a maximizing sequence that allows us to use MCT to change the order of limit and integral.
RN Derivative as a Martingale
This argument is taken from Durrett’s Probability. We’ll actually prove a different version of Theorem 1.
Theorem 2: Let $\mathcal{F}$ be a countably generated $\sigma$-algebra. If $\mu$ is a finite measure and $\nu$ is a probability measure such that $\mu \ll \nu$, then there is an integrable function $g: X \rightarrow[0, \infty]$ so that $d \mu=g d \nu$.
The idea is to approximate the RN derivative $g$ by a sequence of random variables that is measurable with respect to a $\sigma$-algebra generated by finitely many sets. More specifically, let $\{A_i\}_{i=1}^{\infty}$ be the sequence of sets that generates $\mathcal{F}$. Let $\mathcal{F}_n=\sigma(\{A_i\}_{i=1}^n)$. For each $n$, we can find a collections of set $\{I_{k, n}\}_{k=1}^{K_n} \subset \mathcal{F}_n$ that forms a partition of $X$ such that every set $E \in \mathcal{F}_n$ is disjoint union of some $I_{k, n}$. Moreover, $\{I_{k, n+1}\}_{k=1}^{K_{n+1}}$ is always a refinement of $\{I_{k, n}\}_{k=1}^{K_n}$. We now define the random variables $X_n \in \mathcal{F}_n$ :
\[X_n=\sum_{k=1}^{K_n} \frac{\mu\left(I_{k, n}\right)}{\nu\left(I_{k, n}\right)} 1_{I_{k, n}}\]The most amazing part about this proof is that $\{X_n\}$ is a martingale. Let’s first show that
\[E\left[1_{I_{k, n+1}} \mid \mathcal{F}_n\right]=\frac{\nu(I_{k, n+1})}{\nu(I_{j, n})} 1_{I_{j, n}}\]where $I_{k, n+1} \subset I_{j, n}$. Our proposed random variable is clearly measurable with respect to $\mathcal{F}_n$. In addition,
\[\begin{aligned} \int_{I_{j, n}} \frac{\nu\left(I_{k, n+1}\right)}{\nu\left(I_{j, n}\right)} 1_{I_{j, n}} d \nu & =\nu\left(I_{k, n+1}\right) \\ & =\int_{I_{j, n}} 1_{I_{k, n+1}} d \nu \end{aligned}\]which means that our proposed random variable is correct. Now we use linearity of conditional expectation to see that
\[\begin{aligned} E\left[X_{n+1} \mid \mathcal{F}_n\right] & =\sum_{k=1}^{K_{n+1}} \frac{\mu\left(I_{k, n+1}\right)}{\nu\left(I_{k, n+1}\right)} E\left[1_{I_{k, n+1}} \mid \mathcal{F}_n\right] \\ & =\sum_{j=1}^{K_n} \sum_{I_{k, n+1} \subset I_{j, n}} \frac{\mu\left(I_{k, n+1}\right)}{\nu\left(I_{k, n+1}\right)} \frac{\nu\left(I_{k, n+1}\right)}{\nu\left(I_{j, n}\right)} 1_{I_{j, n}} \\ & =\sum_{j=1}^{K_n} \sum_{I_{k, n+1} \subset I_{j, n}} \frac{\mu\left(I_{k, n+1}\right)}{\nu\left(I_{j, n}\right)} 1_{I_{j, n}} \\ & =\sum_{j=1}^{K_n} \frac{\mu\left(I_{j, n}\right)}{\nu\left(I_{j, n}\right)} 1_{I_{j, n}} \\ & =X_n \end{aligned}\]Since we just show that $\{X_n\}$ is a (nonnegative) martingale, it converges a.e. to some $L^1$ random variable $g$. For any $A \in \mathcal{F}_m \subset \mathcal{F}_n$, where $m \leq n, A$ is some finite union of $I_{k, n}$.
\[\begin{aligned} \mu(A) & =\sum_{I_{k, n} \in A} \mu\left(I_{k, n}\right) \\ & =\sum_{I_{k, n} \in A} \frac{\mu\left(I_{k, n}\right)}{\nu\left(I_{k, n}\right)} \int_A 1_{I_{k, n}} d \nu \\ & =\int_A X_n d \nu \end{aligned}\]Thus, $\lim \int_A X_n d \nu=\mu(A)$ for all $A \in \mathcal{G}:=\bigcup_n \mathcal{F}_n$. If we can change the order of limit and integral, then $d \mu=g d \nu$ on $\mathcal{F}$ by the $\pi-\lambda$ theorem. In order to justify this, one need to prove that $X_n$ is actually uniformly integrable (then we are done because uniform integrability implies $L^1$ convergence), which follows from the absolute continuity of $\mu$ with respect to $\nu$. We omit the details here.