Let $(X,\mathcal{A})$ be a measurable space and $T :X \to X$ a measurable transformation and $\mu,\nu$ probability measures on $(X,\mathcal{A})$. Prove TFAE:
(i) $\nu(T^{-1}(E))=\mu(E)$ for all $E \in \mathcal{A}$.
(ii) $\int \phi d \mu = \int \phi \circ T d \nu$ for all $\phi \in L^1(\mu)$.
Attempt:
For the forward direction, letting $\phi \in L^1(\mu)$, I define a function
$$\psi:X \to \Bbb{R}$$
via $\psi(x)=\phi(T(x))$ for every $x \in X$. Then it suffices to prove $\int \phi d \mu = \int \psi d \nu$ and $\psi$ is clearly measurable as it is the composition of measurable functions. I need to also show $\psi \in L^1(\mu)$ right? Does it follow that
$$\int_E \vert \psi \vert d \nu = \int_{T^{-1}(E)} \vert \phi \vert d \nu = \nu(T^{-1}(E))=\mu(E).$$
Where the first equality is by change of variables and assumption of (i) and the last equality holds by just (i). Am I on the right track?
For the backwards direction should I consider as my $\phi$ just $\chi_E$ the indicator function on $E$? Then by assumption of (ii) we have
$$\int \chi_E d \mu = \int \chi_E \circ T d \nu.$$
If this is the case I see the LHS is just $\mu(E)$ and I need to play with the RHS to get $\nu(T^{-1}(E))$.