Radon-Nikodym Theorem for (positive) measures, chain rule

Question

Let us take some time to recall the Radon-Nikodym Theorem for (positive) measures:

If $\nu$ and $\mu$ are $\sigma$-finite measures on $(X, \mathcal{M})$ then $\nu = \nu_a + \nu_s$ with $\nu_a \ll \mu$ and $\nu_s \perp \mu$, and the decomposition is unique. Moreover, $d\nu_a = f\,d\mu$ for a nonnegative measurable function $f$ (note: $d\nu_a = f\,d\mu$ means that for every measurable $E$, $\int_E d\nu_a = \int_E f\,d\mu$). If $\nu \ll \mu$ then $\nu_s = 0$ and the function $f$ is called the Radon-Nikodym derivative and is denoted ${{d\nu}\over{d\mu}}$.

Say that $\mu$, $\nu$ and $\sigma$ are finite measures on $(X, \mathcal{M})$ and suppose that $\mu \ll\nu$ and $\nu \ll \lambda$. How do I see that $\mu \ll \lambda$ and$${{d\mu}\over{d\lambda}} = {{d\mu}\over{d\nu}}{{d\nu}\over{d\lambda}}$$for $\lambda$ almost every point?

Answer 1

First, we prove a lemma:

Lemma: Let $(X, M, \mu)$ be a measure space. Let $f$ be a nonnegative measurable function on $(X, M)$. Define a measure $\nu$ on $(X, M)$ by $\nu(E) = \int_{E}fd\mu$ for $E \in M$. Then for any nonnegative, measurable function $F$ on $(X, M)$, we have

$$\int_{X}Fd\nu = \int_{X}Ffd\mu.$$

Proof of lemma:

Suppose $F(x) = \sum_{i = 1}^{n}a_{i}\chi_{A_{i}}$, where the $a_{i} > 0$ are distinct and the $A_{i} \in M$ are disjoint. Then:

$$\int_{X}Ffd\mu = \int_{X}f\sum_{i = 1}^{n}a_{i}\chi_{A_{i}}d\mu = \sum_{i = 1}^{n}a_{i}\int_{X}f\chi_{A_{i}}d\mu = \sum_{i = 1}^{n}a_{i}\int_{A_{i}}fd\mu = \sum_{i = 1}^{n}a_{i}\nu(A_{i})$$

$$ = \int_{X}\sum_{i = 1}^{n}a_{i}\chi_{A_{i}}d\nu = \int_{X}Fd\nu.$$

Next, suppose $F$ is a measurable, nonnegative function. Since $X$ is $\sigma$-finite, we may take nonnegative simple functions $\phi_{k}$ such that $supp(\phi_{k}) \subseteq supp(F)$, and $\phi_{k} \nearrow F$. Then $\phi_{k}f \nearrow Ff$. By the monotone convergence theorem (and the proof for simple functions to give the middle equivalence),

$$\int_{X}Fd\nu = \lim_{k \to \infty}\int_{X}\phi_{k}d\nu = \lim_{k \to \infty}\int_{X}\phi_{k}fd\mu = \int_{X}Ffd\mu. $$

This proves the lemma.

By the Radon-Nikodym theorem:

• $\mu = \mu_{a} + \mu_{s}$, with $\mu_{a} \ll \nu$ and $\mu_{s} \perp \nu$, and $d\mu_{a} = fd\nu$ for some nonnegative, measurable function $f$. $\mu \ll \nu$, so $\mu_{s} = 0$ $\Rightarrow$ $\mu = \mu_{a}$ and $d\mu = fd\nu$, i.e. $f = \frac{d\mu}{d\nu}$.
• $\nu = \nu_{a} + \nu_{s}$, with $\nu_{a} \ll \lambda$ and $\nu_{a} \perp \lambda$, and $d\nu_{a} = gd\lambda$ for some nonnegative, measurable function $g$. $\nu \ll \lambda$, so $\nu_{s} = 0$ $\Rightarrow$ $\nu = \nu_{a}$ and $d\nu = gd\lambda$, i.e. $g = \frac{d\nu}{d\lambda}$.
• $\mu = \mu_{a}’ + \mu_{s}’$, with $\mu_{a}’ \ll \lambda$ and $\mu_{s}’ \perp \lambda$ and $d\mu_{a} = hd\lambda$ for some nonnegative, measurable function $h$. $\mu \ll \lambda$, so $\mu_{s}’ = 0$ $\Rightarrow$ $\mu = \mu_{a}’$ and $d\mu = hd\lambda$, i.e. $h = \frac{d\mu}{d\lambda}$.

From the above discussion, we see that for every measurable set $E$,

$$\int_{E}d\mu = \int_{E}hd\lambda.$$

By the above lemma,

$$\int_{E}d\mu = \int_{E}fd\nu = \int_{E}fgd\lambda \Rightarrow \int_{E}hd\lambda = \int_{E}fgd\lambda = \mu(E).$$

Radon-Nikodym asserts the function that is integrated is unique $\mu$-a.e., so $h = fg$ $\mu$-a.e., or $\frac{d\mu}{d\lambda} = \frac{d\mu}{d\nu}\frac{d\nu}{d\lambda}$.

Answer 2

A special case. Supose $\lambda$, $\nu$ and $\mu$ Radon measures on a a locally compact Hausdorff $X$. You can look to positive measures $\mu$, $\nu$ and $\lambda$ as functional in the space of continuous functions with compact support as follows. $$ \varphi\mapsto \mu(\varphi):=\int_X \varphi \mathrm{d}\mu, \quad \psi\mapsto \nu(\psi):=\int_X \psi \mathrm{d}\nu \;\;\; \mbox{ and} \;\;\; \xi\mapsto \lambda(\xi):=\int_X \xi \mathrm{d}\nu $$ We use the:

The Riesz-Markov Theorem: Let $X$ be a locally compact Hausdorff space and $I$ a positive linear functional on $C_c (X )$ (continuous on compact support). Then there is a unique Radon measure $\hat{\mu}$ on $\mathcal{B}(X)$, the Borel $\sigma$-algebra associated with the topology on $X$, for which $$I(f)= \int_X f d\hat{\mu} \text{ for all } f \in C_c(f)$$

Then for all continuous on compact support functions $f=1_K$, $K$ compact, we have $$ \begin{array}{rll} \mu(K)= \mu(f) = & \nu\left(f\cdot \frac{\mathrm{d}\mu}{\mathrm{d}\nu}\right) & \mu >>\nu \\ = & \lambda\left(f\cdot \frac{\mathrm{d}\mu}{\mathrm{d}\nu}\cdot \frac{\mathrm{d}\nu}{\mathrm{d}\lambda}\right) & \nu >> \lambda \end{array} $$ On the other hand, we have $\mu>>\lambda$ implies $ \mu(K)=\mu(f)=\lambda\left(f\cdot \frac{\mathrm{d}\mu}{\mathrm{d}\lambda} \right) $ and $$ \lambda\left(f\cdot \frac{\mathrm{d}\mu}{\mathrm{d}\lambda} \right) = \lambda\left(f\cdot \frac{\mathrm{d}\mu}{\mathrm{d}\nu}\cdot \frac{\mathrm{d}\nu}{\mathrm{d}\lambda}\right), \quad \forall f\in C_c(X) $$