Measure Theory – Radon-Nikodym Theorem for Positive Measures and 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,

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

This proves the lemma.

By the Radon-Nikodym theorem:

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