How to show that $f$ is integrable with respect $\mu h^{-1}$ if and only if $f \circ h$ is integrable with respect to $\mu$. This is a problem of measure theory with some difficulty to me. Please any suggestion would be really appreciated. The problem is the following.
We have a measure space $(C, {\cal C}, \mu)$ and a measurable space $(X, {\cal B})$. Let $h: C \to X$ a measurable function. Define $\mu h^{-1}: {\cal B} \to \mathbb{R}$ by $\mu h^{-1}(B) = \mu (h^{-1}(B))$. We have to show the following.

*

*$\mu h^{-1}$ is a measure on $(X, {\cal B})$.

*$f$ is integrable with respect to $\mu h^{-1}$ if and only if $f \circ h$ is integrable with respect to $\mu$.

The first part is as follows: $h: C \to X$ is measurable if and only if for all $B\in {\cal B}$, $h^{-1} (B) \in {\cal C}$. Then

*

*$\mu h^{-1}(\emptyset) = \mu (h^{-1}(\emptyset)) = \mu (\emptyset) = 0.$

*$\mu h^{-1} (B) = \mu (h^{-1}(B)) \geq 0$.

*For any disjoint sequence $\{B_{n}\}$, the sequence $\{h^{-1}(B_{n})\}$ is disjoint, then
$$
\mu h^{-1} (\cup_{n} B_{n}) = \mu (h^{-1}( \cup_{n} B_{n})) = \mu (\cup_{n}h^{-1}(B_{n})) = \sum_{n} \mu (h^{-1}(B_{n})) = \sum_{n} \mu h^{-1}  (B_{n}).
$$
Thefore, $\mu h^{-1}$ is a measure on $(X, {\cal B})$. The second part is more difficult to me. I know that if $f: X \to \mathbb{R}$ is measurable, then $f \circ h: C \to \mathbb{R}$ is measurable since
$$
(f \circ h)^{-1}(-\infty, r] = h^{-1}(f^{-1} (-\infty, r]).
$$
My difficulty is to show that $f$ is integrable with respect to $\mu h^{-1}$ if and only if $f\circ h$ is integrable with respect to $\mu$.
Thank you for your help.
 A: Here's one thing that you can start off by trying.
Given any simple function $\phi = \sum_{i=1}^{n} a_i \chi_{E_i}$ from $X \to \mathbb{R}$, the integral of $\phi$ with respect to $\mu h^{-1}$ is given by
$$ \int_X \phi d(\mu h^{-1}) = \sum_{i=1}^{n} a_i (\mu h^{-1})(E_i)$$.
Now note that if we instead consider the simple function $\psi$ on $C$ given by $\psi = \sum_{i=1}^{n} a_i \chi_{h^{-1}(E_i)}$, then $\int_C \psi d\mu = \int \phi d(\mu h^{-1})$.
Using this you should be able to prove the result holds for simple functions.
After proving the result for simple functions, you can extend this by approximating
$f$ by simple functions from below.
A: Notice that for any $E \in B$,
$$\int_{X}\chi_E(x)\,d\mu_{h^{-1}}(x) = \mu_{h^{-1}}(E) = \mu(h^{-1}(E)) = \int_C \chi_{h^{-1}(E)}(c)\,d\mu(c) = \int_{C}\chi_{E}(h(c))\,d\mu(c).$$
Using approximation arguments, we obtain that for all measurable $f \colon X \to [0, \infty]$,
$$\int_X f(x) \,d\mu_{h^{-1}}(x) = \int_{C}f(h(c))\,d\mu(c).$$
Your claim follows from this.
