# Radon-Nikodym derivatives. Royden, Problem 33.

Radon-Nikodym derivatives. Show that that Radon-Nikodym derivative $$[\frac{d\nu}{d\mu}]$$ has the following properties:

a. If $$\nu\ll \mu$$ and $$f$$ is a nonnegative measurable function, then $$\int fd\nu=\int f[\frac{d\nu}{d\mu}]d\mu.$$

Solution by Royden's solution: Let $$(X,\mathcal{B},\mu)$$ be a $$\sigma$$-finite measure space and let $$\nu$$ be a measure on $$\mathcal{B}$$ which is absolutely continuous with respect to $$\mu$$. Let $$X=\bigcup X_i$$ with $$\mu(X_i)<\infty$$. We may assume the $$X_i$$ are pairwise disjoint. For each $$i$$, let $$\mathcal{B}_i=\left\{E\in\mathcal{B}:E\subset X_i\right\}$$, $$\mu_i=\left.\mu\right|_{\mathcal{B_i}}$$ and $$\nu_i=\left.\nu\right|_{\mathcal{B_i}}$$. Then $$(X_i,\mathcal{B}_i, \mu_i)$$ is a finite measure space and $$\nu_i<<\mu_i$$. Thus for each $$i$$ there is a nonnegative $$\mu_i$$-measurable function $$f_i$$ such that $$\nu_i(E)=\int_E f_id\mu_i$$ forall $$E\in \mathcal{B}_i$$. Define $$f$$ by $$f(x)=f_i(x)$$ if $$x\in X_i$$. If $$E\subset X$$, then $$E\cap X_i\in \mathcal{B}_i$$ for each $$i$$. Thus $$\nu(E)=\sum \nu(E\cap X_i)=\sum \nu_i(E\cap X_i)=\sum \int_{E\cap X_i} f_id\mu_i=\sum \int_{E\cap X_i} fd\mu=\int_E fd\mu$$.

I have doubts:

Why $$\mathcal{B}_i$$ is a sigma-algebra? (If $$E\in \mathcal{B}_i$$, then $$E\subset X_i$$ but $${X_i}^c\subset E^c$$. i.e. $$E^c\not\in \mathcal{B}_i$$...)

• The complement of $E$ is relative to $X_i$, not to $X$. That is to say, $\cal B_i$ is a $\sigma$-algebra of subsets of $X_i$. – Umberto P. Jan 3 at 21:28
• Oh, true! Thank you – eraldcoil Jan 3 at 21:31