showing $\int_\mathbb{R}g(t) d\nu(t) = \int_\mathbb{R}g(f(s))d\mu(s)$ for $g$ a measurable simple function Given:
$f:S \to \mathbb{R}$ is a measurable function
$\nu:\mathcal{B} (\mathbb{R})\to [0,\infty] \text{ and } \nu(B) = \mu(f^{-1}(B)) $
$\nu$ is a measure.
I would like to show the following:
$$\int_\mathbb{R}g(t) d\nu(t) = \int_\mathbb{R}g(f(s))d\mu(s) $$
First I would like to that for a simple function $g: S \to [0,\infty)$. I got a hint that you should to use the linearity property for integrals of simple functions, but I don't see how. Can anyone help me showing the equation above for $g$ being a simple function?
 A: In fact $\nu$ is a measure. In order to prove the result holds when $g$ is a non-negative simple function, you need to prove this for $g=1_{A}$, where $A \in \mathcal{B(\mathbb{R})}$, because any non-negative simple function $s$ can be written of the form, $$s= \sum_{i=1}^{n}c_i1_{A_i},$$ where {$A_i$} is a measurable partition & $c_i'$s are some non-negative constants. 
Then, the result automatically follows for any simple function from the linearity.Let $g=1_{A}$. Then, $$\int_{\mathbb{R}}1_{A}d\nu = \nu(A) = \mu f^{-1} (A)=\int_{\mathbb{R}}1_{f^{-1}(A)}d\mu $$. 
By definition, we know $1_{f^{-1}(A)}(x)=1$ iff $x\in f^{-1}(A)$ iff $f(x)\in A$. 
i.e: $$1_{f^{-1}(A)}(x)= (1_{A} \circ f) (x)$$ . So, $$\int_{\mathbb{R}}1_{A}d\nu=\int_{\mathbb{R}}1_{A} \circ fd\mu.$$
The result is true when $g=1_{A}$; therefore for any non-negative simple function. 
Finally,to prove the result for any unsigned measurable function $g$, use MCT along with the approximation theorem (i.e: any unsigned measurable function $g$ can be approximated by a sequence of non-negative simple function, {$s_n$} so that $s_n$ increases to $g$).
