calculating an integral with respect to a mesure

Question

Let $g(x) = |x|1_{[0,1]}$ let $f(x) = \frac{1}{\pi(1+x^2)}$ and $\mu$ Lebesgue mesure over $\mathbb{R}$. we define $\mu^f(A) = \mu(f^{-1}(A))$ calculate $\int g d\mu^f$ how do i know over which interval i integrate over? i assume i have to integrate over the reals but when looking at the $1_{[0,1]}$ is only non-zero if i integrate over $[0,1]$. After trying both cases i end up with a different value of the integral, a note : $\int g d\mu^f = \int g(f(x)) \mu(d(x))$

Answer 1

$\int gd\mu^{f}=\int g(f(x))dx$. $g(f(x))=|f(x)|$ if $|f(x)| \leq 1$ and $0$ otherwise. Note that $0\leq |f(x)| \leq 1$ for all $x$. Hence, $\int gd\mu^{f}=\frac 1 {\pi}\int \frac 1 {1+x^{2}}dx=1$.