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. $E^c\not\in \mathcal{B}_i$...)

  • 3
    $\begingroup$ 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 $X_i$. $\endgroup$ – Umberto P. Jan 3 at 21:28
  • $\begingroup$ Oh, true! Thank you $\endgroup$ – eraldcoil Jan 3 at 21:31

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