[Measure Theory] Radon-Nikodym Theorem
Let $(X, \mathcal{A}, \mu)$ be a measure space. Suppose $f$ is non-negative and integrable with respect to $\mu$. If we define
\[\begin{aligned} \nu (A) := \int_A f \; d\mu \end{aligned}\]then it is trivial that $\nu$ is a measure. In this post, we will consider the converse: if we are given two measures $\mu$ and $\nu$, when does there exist $f$ such that $\nu (A) = \int_A f \; d\mu$ for $A \in \mathcal{A}$? The Radon-Nikodym answers this question, and provides generalizations of the relationship between the two central operations of calculus—differentiation and integration. And it will be a vital cue for our discussions of differentiation in Lebesgue measure theory.
Absolute Continuity
Suppose $\nu$ is a measure and $\mu$ is a positive measure on $(X, \mathcal{A})$. A measure $\nu$ is said to be absolutely continuous with respect to $\mu$ if $\nu (A) = 0$ whenever $\mu (A) = 0$. Or, $\mu$ is said to be dominating $\nu$. We write $\nu \ll \mu$.
The term absolute continuity stems for real analysis. We will see later; absolute continuity of measures is equivalent to absolute continuity of function. And with finiteness, it is equivalent to another condition that is a form of continuity:
Note that as $| \nu | \ll \mu$ iff $\nu \ll \mu$, it suffices to assume that $\nu$ is a signed measure.
$\mathbf{Proof.}$
$\color{#bf1100}{\boldsymbol{(\Rightarrow)}}$ If $\mu(A) = 0 < \delta = \varepsilon$, then $\nu(A) < \varepsilon$ for all $\varepsilon > 0$, hence $\nu(A) = 0$, and thus $\nu \ll \mu$.
$\color{#bf1100}{\boldsymbol{(\Leftarrow)}}$ Suppose $\nu \ll \mu$. Let any $\varepsilon > 0$ be given. Suppose that for any $\delta > 0$, there exists $A \in \mathcal{A}$ such that $\mu (A) < \delta$ but $\nu (A) \geq \varepsilon$.
Then, we can choose a sequence of measurable sets $E_k$ such that
\[\begin{aligned} \mu (E_k) < \frac{1}{2^k} \quad \text{ but } \quad \nu (E_k) \geq \varepsilon \end{aligned}\]Let $F = \bigcap_{n=1}^\infty \left( \bigcup_{k=n}^\infty E_k \right)$. Then from the fact that $\mu (\bigcup_{k=1}^\infty E_k ) < \infty$ and the assumption that $\nu$ is a finite measure, we have
\[\begin{aligned} \mu \left( \bigcap_{n=1}^\infty \left( \bigcup_{k=n}^\infty E_k \right) \right) = \lim_{n \to \infty } \mu \left( \bigcup_{k=n}^\infty E_k \right) < \lim_{n \to \infty } \sum_{k=n}^\infty E_k \varepsilon^k = 0, \end{aligned}\]but
\[\begin{aligned} \nu \left( \bigcap_{n=1}^\infty \left( \bigcup_{k=n}^\infty E_k \right) \right) = \lim_{n \to \infty } \nu \left( \bigcup_{k=n}^\infty E_k \right) \geq \varepsilon, \end{aligned}\]by continuity from below.
\[\tag*{$\blacksquare$}\]Equivalently, it is easily verified that
The followings are equivalent:
- $\nu \ll \mu$;
- $| \nu | \ll \mu$;
- $\nu^+ \ll \mu$ and $\nu^- \ll \mu$.
$\mathbf{Proof.}$
Suppose $\nu \ll \mu$. Let $(E, F)$ be a Hahn-Jordan decomposition for $\nu$ and $\nu^+ (A) = \nu (A \cap F)$ and $\nu^-(A) = \nu (A \cap E)$. If $\mu(A)=0$, then $\nu (A) = 0$. Also, $\mu( A \cap F ) = 0$ because $\mu$ is positive measure, and so we have $\nu (A \cap F) = \nu^+ (A) = 0$ from $\nu \ll \mu$. Thus $\nu^+ \ll \mu$. Since $\nu (A) = 0$, so does $\nu^- (A) = 0$.
Suppose $\nu^+ \ll \mu$ and $\nu^- \ll \mu$. Then $\nu \ll \mu$ is trivial. Thus $\nu \ll \mu$ iff $\nu^+ \ll \mu$ and $\nu^- \ll \mu$. And trivially $| \nu | \ll \mu$.
Suppose $| \nu | \ll \mu$. Then trivially $\nu^+ \ll \mu$ and $\nu^- \ll \mu$.
\[\tag*{$\blacksquare$}\]Radon-Nikodym Theorem
We now come to the main theorem of the post, so-called Radon-Nikodym theorem, or Lebesgue-Radon-Nikodym theorem. But we need a technical lemma to prove the theorem:
- $\mu \perp \nu$, or
- there exists $\varepsilon > 0$ and $G \in \mathcal{A}$ such that $\mu (G) > 0$ and $G$ is a positive set for $\nu - \varepsilon \mu$.
$\mathbf{Proof.}$
Note that $\nu - \frac{1}{n} \mu$ is a signed measure for any $n \in \mathbb{N}$ as $\nu$ and $\mu$ are finite. From the Hahn decomposition theorem, there exists a negative set $E_n$ and a positive set $F_n$ for this measure such that $E_n \cap F_n = \emptyset$, $E_n \cup F_n = X$.
Let $F = \bigcup_{n=1}^\infty F_n$ and $E = \bigcap_{n=1}^\infty E_n$. Then $E^c = \bigcup_{n=1}^\infty E_n^c = \bigcup_{n=1}^\infty F_n = F$. For each $n \in \mathbb{N}$, $E \subset E_n$ hence
\[\begin{aligned} \nu (E) & \leq \nu (E_n) \quad \because \nu \text{ is a positive measure } \\ & \leq \frac{1}{n} \mu (E_n) \quad \because E_n \text{ is a negative set for } \nu - \frac{1}{n} \mu \\ & \leq \frac{1}{n} \mu (X) \quad \because \mu \text{ is a positive measure } \end{aligned}\]Therefore $n \nu (E) \leq \mu (X)$. As $\mu (X) < \infty$ and $n$ is arbitrary, $\nu (E) = 0$.
- If $\mu (E^c) = 0$, then $\nu \perp \mu$.
- If not, $\mu (E^c) > 0$. As $\mu (E^c) \leq \sum_{n=1}^\infty \mu (F_n)$, $\mu (F_{n_0}) > 0$ for some $n_0$. Let $\varepsilon = \frac{1}{n_0}$. Then $G = F_{n_0}$ is the desired set.
Here is the main result. It allows us to express a measure in terms of an integral and indicates whether and how a transition from one measure to another is possible.
Suppose $\mu$ is a $\sigma$-finite positive measure on a measurable space $(X, \mathcal{A})$ and $\nu$ is a finite positive measure on $(X, \mathcal{A})$ such that $\nu$ is absolutely continuous with respect to $\mu$. Then there exists a $\mu$-integrable non-negative function $f$ which is measurable with respect to $\mathcal{A}$ such that $$ \begin{aligned} \nu (A) = \int_A f \; d\mu \end{aligned} $$ for all $A \in \mathcal{A}$. Moreover, if $g$ is another such function, then $f = g$ almost everywhere with respect to $\mu$.
$\mathbf{Proof.}$
Step 1. Uniqueness
\[\begin{aligned} \int_A ( f - g ) \; d\mu = \nu (A) - \nu (A) = 0 \end{aligned}\]for all $A \in \mathcal{A}$. $f - g = 0$ almost everywhere.
Step 2. Existence
(1) Assume first that $\mu$ is a finite measure. The main idea of the proof is to look at the set of $f$ such that $\int_A f \; d\mu \leq \nu (A)$ for each $A \in \mathcal{A}$, and then to choose the one such that $\int_X f \; d\mu$ is largest.
Define
\[\begin{aligned} \mathcal{F} = \left\{ g \text{ measurable } \mid g \geq 0, \int_A g\; d\mu \leq \nu(A) \text{ for all } A \in \mathcal{A} \right\}. \end{aligned}\]Note that $0 \in \mathcal{F}$. Let
\[\begin{aligned} L = \text{ sup } \left\{ \int_X g \; d\mu \mid g \in \mathcal{F} \right\} \end{aligned}\]Then construct the sequence \(\{ g_n \}\) in $\mathcal{F}$ such that $\int_X g_n \; d\mu \to L$. Let
\[\begin{aligned} h_n & = \text{ max } (g_1, \cdots, g_n) \\ f & = \underset{n}{\text{ sup }} h_n \end{aligned}\]$\mathbf{Proof.}$
We prove the claim by induction.
Since $g_1, g_2 \in \mathcal{F}$, $h_2 = \text{ max } (g_1, g_2) \in \mathcal{F}$.
Indeed,
\[\begin{aligned} \int_A \text{ max } (g_1, g_2) \; d\mu & = \int_{A \cap \{ g_1 \geq g_2 \}} g_1 \; d\mu + \int_{A \cap \{ g_1 < g_2 \}} g_2 \; d\mu \\ & \leq \nu (A \cap \{ g_1 \geq g_2 \}) + \nu (A \cap \{ g_1 \geq g_2 \}^c) = \nu (A) \end{aligned}\]Hence, $h_2 \in \mathcal{F}$. By induction, $h_n \in \mathcal{F}$.
\[\tag*{$\blacksquare$}\]Since $h_n$ increases, M.C.T implies that $f \in \mathcal{F}$: $\int_A f \; d\nu \leq \nu (A)$ for all $A \in \mathcal{A}$.
And $\int_X f \; d\mu \geq \int_X h_n \; d\mu \geq \int_X g_n \; d\mu$ for all $n \in \mathbb{N}$. Hence, $\int_X f \; d\mu = L$.
$\mathbf{Proof.}$
Define a measure $\lambda$ by
\[\begin{aligned} \lambda(A) := \nu (A) - \int_A f \; d\mu \end{aligned}\]Then $\lambda$ is a positive measure as $f \in \mathcal{F}$. We will prove that $\lambda \perp \nu$.
Suppose $\lambda$ is not mutually singular to $\mu$: there exists $\varepsilon > 0$ such that $G \in \mathcal{A}$, $\mu (G) > 0$ and $G$ is positive for $\lambda - \varepsilon \mu$ by the previous lemma. (Note that $\lambda$ is finite.) For any $A \in \mathcal{A}$,
\[\begin{aligned} \nu (A) - \int_A f \; d\mu = \lambda (A) \geq \lambda (A \cap G) \geq \varepsilon \mu (A \cap G) = \int_A \varepsilon \chi_G \; d\mu. \end{aligned}\]Hence $\nu (A) \geq \int_A (f + \varepsilon \chi_G ) \; d\mu$. Thus $f + \varepsilon \chi_G \in \mathcal{F}$, but it’s contradiction as $\int_X (f + \varepsilon \chi_G ) \; d\mu = L + \varepsilon \mu (G) > L$.
Therefore, $\lambda \perp \mu$. In other words, there exists $H \in \mathcal{A}$ such that $\mu (H) = 0$ and $\lambda (H^c) = 0$. Note that we assumed $\nu \ll \mu$, it implies $\lambda \ll \mu$ since $- \int_A f \; d\mu$ is absolutely continuous with regard to $\mu$.
Hence, $\lambda (H) = 0$. It implies that $\lambda = 0$.
\[\tag*{$\blacksquare$}\](2) Now suppose that $\mu$ is $\sigma$-finite. There exist \(\{ F_i \in \mathcal{A} \}\) with $F_i \uparrow X$ and $\mu (F_i) < \infty$.
Let
\[\begin{aligned} \mu_i (A) & := \mu (A \cap F_i) \\ \nu_i (A) & := \mu (A \cap F_i) \end{aligned}\]Then we can find $f_i$ such that $\nu_i (A) = \int_A f_i \; d\mu_i$ for all $A \in \mathcal{A}$ and each $i \in \mathbb{N}$. From the uniqueness, $f_i = f_j$ on $F_i$ for $i \leq j$. Define $f$ by $f(x) = f_i (x)$ for $x \in F_i$. Then
\[\begin{aligned} \nu (A \cap F_i) = \nu_i (A) = \int_A f_i \; d\mu_i = \int_{A \cap F_i} f \; d\mu \end{aligned}\]for all $A \in \mathcal{A}$. By letting $i \to \infty$, done.
\[\tag*{$\blacksquare$}\]-
In general, the theorem still holds if $\nu$ is a finite signed measure. But in this case, $f$ will be integrable real-valued function, not non-negative.
$\mathbf{Proof.}$
Recall that $\nu \ll \mu$ if and only if $\nu^{\pm} \ll \mu$. Thus, we apply Radon-Nikodym to both $\nu^+$ and $\nu^-$ and obtain $f^+$ and $f^-$, respectively. Define $f = f^+ - f^-$. Note that $f$ is $\mu$-integrable. Then,
\[\begin{aligned} \int_A f \; d\mu = \int_A f^+ \; d\mu - \int_A f^- \; d\mu = \nu^+ (A) - \nu^-(A) = \nu (A). \end{aligned}\]Also, the uniqueness can be shown similarly with positive measure version.
\[\tag*{$\blacksquare$}\] - Furthermore, the theorem still holds even if $\nu$ is a $\sigma$-finite signed measure. For the proof, see [2].
We usually denote the result of the theorem as $d\nu = f \; d\mu$ for some $f$. Here, the function $f$ is called the Radon-Nikodym derivative of $\nu$ with respect to $\mu$ or sometimes the density of $\nu$ with respect to $\mu$. And we denote it by $d\nu / d\mu$:
\[\begin{aligned} d\nu = \frac{d\nu}{d\mu} d\mu. \end{aligned}\]The formulas suggested by the differential notation $d \nu / d\mu$ are generally correct. For example, it is obvious that
\[\begin{aligned} \frac{d (\nu_1 + \nu_2)}{d \mu} = \frac{d \nu_1}{d \mu} + \frac{d \nu_2}{d \mu} \end{aligned}\]Also we have chain rule:
Suppose that $\nu$ is a finite signed measure and $\mu, \lambda$ are $\sigma$-finite measures on $(X, \mathcal{A})$ such that $\nu \ll \mu$ and $\mu \ll \lambda$.
- Suppose $d\nu = g \; d\mu$. If a function $f$ is $\nu$-integrable, $$ \begin{aligned} \int f \; d\nu = \int f g \; d\mu. \end{aligned} $$
- We have $\nu \ll \lambda$ and $$ \begin{aligned} \frac{d\nu}{d\lambda} = \frac{d\nu}{d\mu} \frac{d\mu}{d\lambda} \quad \lambda\text{-a.e. } \end{aligned} $$
$\mathbf{Proof.}$
(1) Suppose $f = \chi_E$. Then by definition of Radon-Nikodym derivative, the statement is true.
Suppose $f$ is a non-negative simple function. Then by linearity of integrals, it is therefore true.
Then for non-negative measurable functions $f$, it is true by the simple function approximation of non-negative functions and monotone convergence theorem.
Finally for $\nu$-integrable $f = f^+ - f^-$, apply the statement to both $f^+$ and $f^-$. Done by linearity of integrals.
(2) By (1), we have
\[\begin{aligned} \int f \; d\mu = \int f \frac{d \mu}{d\lambda} \; d\lambda. \end{aligned}\]For any $A \in \mathcal{A}$, let $f = \chi_A (d \nu / d\mu)$. Then
\[\begin{aligned} \int f \; d\mu = \nu (A) = \int \chi_A \frac{d\nu}{d\mu} \frac{d \mu}{d\lambda} \; d\lambda = \int_A \frac{d\nu}{d\mu} \frac{d \mu}{d\lambda} \; d\lambda. \end{aligned}\]for all $A \in \mathcal{A}$. It implies $\frac{d\nu}{d\mu} \frac{d \mu}{d\lambda}$ is Radon-Nikodym derivative. In other words,
\[\begin{aligned} \frac{d\nu}{d\lambda} = \frac{d\nu}{d\mu} \frac{d \mu}{d\lambda}. \end{aligned}\] \[\tag*{$\blacksquare$}\]Lebesgue Decomposition Theorem
From the Radon-Nikodym theorem, we obtain one of the most principal theorems in measure theory called Lebesgue decomposition theorem:
Suppose $\mu$ is a $\sigma$-finite positive measure and $\nu$ is a finite positive measure. There exist unique positive measures $\lambda$, $\rho$ such that $$ \begin{aligned} \nu = \lambda + \rho \quad \text{ where } \quad \lambda \perp \mu, \; \rho \ll \mu \end{aligned} $$
$\mathbf{Proof.}$
Step 1. Existence
(1) Suppose $\mu$ and $\nu$ are finite positive measures. As in the proof of the Radon-Nikodym Theorem, we define for $A \in \mathcal{A}$,
\[\begin{aligned} \lambda (A) = \nu (A) - \int_A f \; d\mu \end{aligned}\]using $f$ constructed in the proof of theorem. Then, we already show that $\lambda$ is a positive measure and $\lambda \perp \mu$.
Set
\[\begin{aligned} \rho (A) := \int_A f \; d\mu \end{aligned}\]Recall that $f$ is integrable. Then we have $\rho \ll \mu$.
(2) Generalize our statement to the case that $\mu$ is $\sigma$-finite, as in the Radon-Nikodym Theorem.
Step 2. Uniqueness
Let $\mu = \lambda_1 + \rho_1 = \lambda_2 + \rho_2$. Then we have $\lambda_1 - \lambda_2 = \rho_2 - \rho_1$.
Since $\lambda_1, \lambda_2 \perp \mu$, it is trivial that $\lambda_1 - \lambda_2 \perp \mu$. Since there exist disjoint pair $(E_1, F_1)$ and $(E_2, F_2)$ of which union is $X$ such that
\[\begin{aligned} \lambda_1 (E_1) = \mu (F_1) = \lambda_2 (E_2) = \mu (F_2) = 0 \end{aligned}\]Thus $E_1 \cap E_2$ is a null set for $\lambda_1 - \lambda_2$.
Since $\rho_1, \rho_2 \ll \mu$, it is trivial that $\rho_2 - \rho_1 \ll \mu$. Since $\mu (F_1 \cup F_2) = 0$, i.e. $\mu ((E_1 \cap E_2)^c) = 0$, $(E_1 \cap E_2)^c$ is a null set for $\rho_2 - \rho_1 = \lambda_1 - \lambda_2$.
This implies that $\lambda_1 - \lambda_2 = 0$.
\[\tag*{$\blacksquare$}\]Likewise, the theorem can be generalized to signed measures.
-
In general, the theorem still holds if $\nu$ is a finite signed measure. But in this case, $\rho$ and $\lambda$ will be signed measures, not positive.
$\mathbf{Proof.}$
Let $\nu = \nu^+ - \nu^-$ be Jordan decomposition. Note that $\nu^{\pm}$ are positive and also finite. Then there exist unique $\lambda^+, \rho^+$, and $\lambda^-, \rho^-$ such that
\[\begin{aligned} \nu^+ & = \lambda^+ + \rho^+ \quad \text{ where } \quad \lambda^+ \perp \mu, \; \rho^+ \ll \mu \\ \nu^- & = \lambda^- + \rho^- \quad \text{ where } \quad \lambda^- \perp \mu, \; \rho^- \ll \mu \\ \end{aligned}\]Let $\lambda = \lambda^+ - \lambda^-$. And let $\rho = \rho^+ - \rho^-$. Then $\nu = \lambda + \rho$. Since it is trivial that $\lambda \perp \mu$ and $\rho \ll \mu$, we show that it is the Lebesgue decomposition of $\nu$.
Also, the uniqueness can be shown similarly with positive measure version.
\[\tag*{$\blacksquare$}\] - Furthermore, the theorem still holds even if $\nu$ is a $\sigma$-finite signed measure. For the proof, see [2].
Reference
[1] Richard F. Bass, Real Analysis for Graduate Students, Version 4.3
[2] Folland, Gerald B. Real analysis: modern techniques and their applications. Vol. 40. John Wiley & Sons, 1999.
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