📂Measure Theory

Radon-Nikodym Theorem Proof

Theorem ¹

If two sigma-finite measure $\nu$, $\mu$ in a measure space $( \Omega , \mathcal{F} )$ satisfy $\nu \ll \mu$, then for all $A \in \mathcal{F}$, there exists a unique $\mathcal{F}$-measurable function $h$ satisfying $h \ge 0$ almost everywhere according to $\mu$ and $$\nu (A) = \int_{A} h d \mu$$.

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• That $h$ is almost everywhere according to $\mu$ means $\mu \left( h^{-1} ( -\infty , 0 ) \right) = 0$, similar to almost everywhere. $\nu (A) \ll \mu (A)$ signifies that $\nu$ is absolutely continuous with respect to $\mu$, and for all $A \in \mathcal{F}$, the following holds: $$\mu (A) =0 \implies \nu (A) =0$$

Statement

According to the properties of the Lebesgue integral, $\displaystyle \nu (A) = \int_{A} \mathbb{1}_{A} d \nu$, but bringing another measure $\mu$ also satisfies $\nu (A) = \int_{A} h d \mu$ through a uniquely existing intermediary $h$, and it is also looked for what exactly it is. This theorem’s $h$ is called the Radon-Nikodym derivative. The Radon-Nikodym theorem immediately assures the existence of conditional expectations in probability theory, and its importance is immense.

Proof

Assume first that $\mu ( \Omega ) = \nu ( \Omega ) < \infty$, meaning $\nu$ and $\mu$ are finite measures.

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Part 1. $\displaystyle \int_{\Omega} g d \mu = \int_{\Omega} g h_{\mu} d \varphi$

Suppose the two finite measures $\mu$, $\varphi$ in $( \Omega , \mathcal{F} )$ satisfy $0 \le \mu \le \varphi$. To compute $\displaystyle \int_{\Omega} g d \mu$ for any $\mathcal{F}$-measurable function $g \ge 0$, define the simple function $g_{n}$ with a finite number of values $n$ and the smallest metric with $g$ as follows. $$g_{n} := \sum_{k=1}^{n} a_{k} \mathbb{1}_{A_{k}} = \argmin | g - G |$$ Here, let $\mathcal{Q}_{n} := \left\{ A_{k} \right\}_{k=1}^{n}$ be the partition for all $n$, and say $\mathcal{Q}_{n+1}$ is the refinement of $\mathcal{Q}_{n}$. Then, by definition, if $g_{n} \nearrow g$, then $$\begin{align*} \int_{\Omega} g_{n} d \mu =& \int_{\Omega} \sum_{k=1}^{n} a_{k} \mathbb{1}_{A_{k}} d \mu \\ =& \sum_{k=1}^{n} a_{k} \int_{A_{k}} \mathbb{1}_{A_{k}} d \mu \\ =& \sum_{k=1}^{n} a_{k} \mu (A_{k}) \\ =& \sum_{k=1}^{n} a_{k} {{ \mu (A_{k}) } \over { \varphi (A_{k}) }} \varphi (A_{k}) \\ =& \sum_{k=1}^{n} a_{k} {{ \mu (A_{k}) } \over { \varphi (A_{k}) }} \int_{A_{k}} \mathbb{1}_{A_{k}} d \varphi \\ =& \sum_{k=1}^{n} \int_{A_{k}} a_{k} \mathbb{1}_{A_{k}} {{ \mu (A_{k}) } \over { \varphi (A_{k}) }} d \varphi \\ =& \int_{\Omega} g_{n} {{ \mu } \over { \varphi }} d \varphi \end{align*}$$

Monotone convergence theorem: If a sequence of non-negative measurable functions $\left\{ f_{n} \right\}$ satisfies $f_{n} \nearrow f$, then $$\lim_{n \to \infty} \int_{E} f_{n} dm = \int_{E} f dm$$

Radon-Nikodym derivative: If for all $n \in \mathbb{N}$, $\mathcal{Q}_{n+1}$ is a refinement of $\mathcal{Q}_{n}$, then $$\lim_{n \to \infty} h_{\mathcal{Q}_{n}} = \lim_{n \to \infty} {{\nu} \over {\mu}} = {{d \nu } \over {d \mu }}$$

Then, by the properties of the monotone convergence theorem and the Radon-Nikodym derivative, $$\begin{align*} \int_{\Omega} g d \mu =& \lim_{n \to \infty} \int_{\Omega} g_{n} d \mu \\ =& \lim_{n \to \infty} \int_{\Omega} g_{n} {{ \mu } \over { \varphi }} d \varphi \\ =& \int_{\Omega} g {{d \mu } \over {d \varphi }} d \varphi \end{align*}$$

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Part 2. Existence of $h$

If it is $\varphi = \nu + \mu$, obviously $0 \le \nu \le \varphi$ and therefore $0 \le \mu \le \varphi$, the condition of Part 1 that $\varphi$ is greater than both $\nu$ and $\mu$ is met, the Radon-Nikodym derivative $\displaystyle h_{\mu} = {{ d \mu } \over { d \varphi }}$, $\displaystyle h_{\nu} = {{ d \nu } \over { d \varphi }}$ can be well defined. Consider two sets from $\mathcal{F}$ $$\begin{align*} F &:= \left\{ \omega \in \Omega : h_{\mu} (\omega) > 0 \right\} \\ G &:= \left\{ \omega \in \Omega : h_{\mu} (\omega) = 0 \right\} \end{align*}$$ For the subset $F$ of $A \subset F$, if defined as $\displaystyle h := \mathbb{1}_{A} {{ h_{\nu} } \over { h_{\mu} }}$, then by Part 1, $$\begin{align*} \nu (A) =& \int_{A} \mathbb{1}_{A} d \nu \\ =& \int_{A} \mathbb{1}_{A} {{ d \nu } \over { d \varphi }} d \varphi \\ =& \int_{A} \mathbb{1}_{A} h_{\nu} {{ h_{\mu} } \over { h_{\mu} }} d \varphi \\ =& \int_{A} \mathbb{1}_{A} {{ h_{\nu} } \over { h_{\mu} }} h_{\mu} d \varphi \\ =& \int_{A} h h_{\mu} d \varphi \\ =& \int_{A} h d \mu \end{align*}$$ and by the definition of $G$, if $\displaystyle \mu (G) = \int_{G} h_{\mu} d \varphi = 0$, then $\nu \ll \mu$ from the premise, so $\mu (G) = 0 \implies \nu (G) = 0$. Therefore, $h$ satisfies $\nu (A) = \int_{A} h d \mu$.

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Part 3. Uniqueness of $h$

$$\int_{A} f dm = 0 \iff f = 0 \text{ a.e.}$$

If $f = g$ and $f = h$ exist satisfying $\nu (A) = \int_{A} f d \mu$ for $A \in \mathcal{F}$, $$\begin{align*} 0 =& \nu (A) - \nu (A) \\ =& \int_{A} h d \mu - \int_{A} g d \mu \\ =& \int_{A} (h - g ) d \mu \end{align*}$$ then almost everywhere $h = g$.

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Part 4. Generalization to Sigma-Finite Measures

Now, assume that $\nu$ and $\mu$ are sigma-finite measures. $$A_{k} \in \mathcal{F} \\ \nu (A_{k}) < \infty \\ \mu (A_{k}) < \infty \\ i \ne j \implies A_{i} \cap A_{j} = \emptyset \\ X = \bigcup_{k \in \mathbb{N}} A_{k}$$ Fix sequences of sets $\left\{ A_{k} \right\}_{k \in \mathbb{N}}$ and $E \in \mathcal{F}$ satisfying the above conditions and redefine finite measures over $E \cap A_{k}$ accordingly. $$\nu_{k} (E) := \nu ( E \cap A_{k} ) \\ \mu_{k} (E) := \mu ( E \cap A_{k} )$$ Then, according to Parts 1-3, there exist $h_{k}$ for all $k \in \mathbb{N}$ satisfying the following. $$\nu_{k} (E) = \int_{E} h_{k} d \mu_{k}$$ By the definitions of $\nu_{k}$, $\mu_{k}$, DCP( ) is guaranteed, and for all $k \in \mathbb{N}$, $h_{k} (A_{k}^{c}) = 0$ is assured. Accordingly, if defining $\displaystyle h := \sum_{k \in \mathbb{N}} h_{k}$ then $$\begin{align*} \nu (E) =& \sum_{k \in \mathbb{N}} \nu_{k} (E) \\ =& \sum_{k \in \mathbb{N}} \int_{E} h_{k} d \mu_{k} \\ =& \int_{E} h d \mu \end{align*}$$

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