Consider the measure space $(\mathbb R, \mathcal B(\mathbb R), \lambda)$ and let $\phi: \mathbb R \rightarrow \mathbb R$ be given by $\phi(x) = x^2$.
I want to show that for $f \in \mathcal M(\mathcal B(\mathbb R))$ we have $f \circ \phi \in \mathcal L^1(\lambda) \iff \int_0^{\infty} \frac {|f(x)|}{\sqrt x} \lambda (dx) < \infty$:
I've computed that the measure $\lambda \circ \phi^{-1}$ is given by $t \mapsto t^{-1/2} 1_{(0, \infty)}(t)$ (density). Should I use the transformation theorem ? I'm out of ideas how to proceed.
Also, I want to prove that $\int_{\mathbb R} f \circ \phi \ \lambda(dx) = \int_0^{\infty} \frac {f(x)} { \sqrt x} \lambda(dx) $, but this can be done by applying transformation with respect to measure ?