Let $\mu$ measure and $\mu_{f}$ the image measure of $\mu$ by $f$. Can we say that $\int (g\circ f) d\mu =\int g d\mu_{f} \: ?$

Question

Let $(X,\mathcal{M},\mu)$ be a measurable space and $f:X\rightarrow X $ be a $\mathcal{M}$-measurable function. We define the image measure of $\mu$ by $f$ as $$\mu_{f}(A):=\mu(f^{-1}(A)) \qquad A\in \mathcal{M}.$$

Question: Let $g:X\rightarrow \mathbb{R}$ be a measurable function, Can you say that
$$\int (g\circ f) d\mu =\int g d\mu_{f} \: ?$$

Answer 1

When $g$ is an indicator function $I_X$ of some measureable set $X$ you can quickly verify that

$$ \int (I_X \circ f) d\mu = \int I_{f^{-1}(X)} d\mu = \mu(f^{-1}(X)) = \mu_{f}(X) = \int I_X d\mu_f $$

Since all measurable functions can be approximated as the limit of linear combinations of indicator functions (indeed this is how Lebesgue integration is extended to arbitrary functions) the equality must also hold for all measurable functions $g$.