Let $\mu_f$ denote the probability distribution of $f$ with $f(x)=10x-1$ for $x\in(0,1/2]$ and $f(x)=1$ for $x\in[1/2,1]$. If $g:\mathbb{R}\rightarrow\mathbb{R}$ is a continuous function, what is $\int_\mathbb{R}gd\mu_f$?
For a subset $X\in\mathbb{R}$, we have $\mu_f(X)=\mu(f^{-1}(X))$. This is a sum of two values:
First value: $\mu(x\mid 0<x\leq 1/2 \text{ and } 10x-1\in X)$
Second value: $1/2$ if $1\in X$ and $0$ otherwise
Now I could think about calculating $g:\mathbb{R}\rightarrow\mathbb{R}$ using the definition of Lebesgue integral (bounding below by simple functions). But it seems complicated. How can I work it out?