Measure defined by integration of function w.r.t the other measure

Question

Define $\mu'(A)=\int f1_Ad\mu$ and $f$ is a nonnegative integrable function.

Prove that if $g$ is integrable w.r.t $\mu'$, then $fg$ is integrable w.r.t $\mu$ and

$$\int g\mu'=\int fg d\mu$$

Givmme some hint for it.

Answer 1

It is easy to show for characteristic functions. Then show it for simple functions. That is $\int s d \mu' = \int s f d \mu$ for simple $s$. (Integrability of $s$ follows from integrability of $f$.)

The only trick is to show that $fg$ is $\mu$ integrable.

Let $g_+(0) = \max(g(x),0)$, and similarly, $g_+(0) = \max(-g(x),0)$, Then $g_+, g_-$ are non-negative and $g = g_+-g_-$, and $|g| = g_+ +g_-$. Let $s_{n,+}, s_{n,-}$ be a sequence of simple functions $s_{n,+} \uparrow g_+$ and $s_{n,-} \uparrow g_-$.

Then the monotone convergence theorem shows that $\lim_n \int s_{n,\pm} d \mu' = \int g_{\pm} d \mu'$, and $\lim_n \int s_{n,\pm} f d \mu = \int g_{\pm} f d \mu$. Since $\int s_{n,\pm} d \mu' = \int s_{n,\pm} f d \mu $, we see that $\int g_{\pm} d \mu' = \int g_{\pm} f d \mu$. It follows from the fact that $|g|=g_++g_-$ that $|g|f$ is $\mu$ integrable, and then, since $g = g_+-g_-$, it follows that $\int g d \mu' = \int g f d \mu$.