Show that $\lim_{n \to \infty} \nu(T^{-n}A) = \mu(A)$ The following is an exercise from my lecture notes on Ergodic Theory. I managed to solve part a), but I have doubts on one step in my solution of part b).
Let $(X, \mathcal{F}, \mu)$ be a probability space and $T \colon X \to X$ a transformation.
a) Let $\nu$ be a probability measure on $(X, \mathcal{F})$ that is equivalent to $\mu$ and let $f$ be the Radon-Nikodym derivative $f = \frac{d\mu}{d\nu}$. Let $P_{T, \mu}$ and $P_{T, \nu}$ denote the Perron-Frobenius operators of $T$ with respect to the measures $\mu$ and $\nu$. Profe that for any $g \in L^1(X, \mathcal{F}, \mu),
$$
p_{T, \mu}(g) = \frac{P_{T, \nu}(fg)}{f}.
$$
(I was able to show this by first considering indicator functions and then extend to simple functions by linearity and using Monotone Convergence for general functions).
b) Let $\nu$ be a probability measure on $(X, \mathcal{F})$ that is absolutely continuous with respect to $\mu$. Assume that $\mu$ is $T$-invariant and that $T$ is strongly mixing with respect to $\mu$. Prove that
$$
\lim_{n \to \infty} \nu(T^{-n}A) = \mu(A)
$$
holds for any $A \in \mathcal{F}$.
My idea for b) was to use the following fact: $T$ is strongly mixing if and only if for all $f \in L^1(X, \mathcal{F}, \mu)$ and $g \in L^\infty(X, \mathcal{F}, \mu)$,
$$
\lim_{n \to \infty} \int_X (P_T^n f) \cdot g d\mu = \int_X f d\mu \int_X g d\mu.
$$
Toward this end, i want to write
$$
\nu(T^{-n}A) = \int_{T^{-n}A} \,d\nu = \int_A P_{T, \nu}^n(1) \,d \nu = \int_A P_{T^n, \nu}(1) \,d\nu = \int_X \frac{1}{f} P_{T^n, \mu}(f) \cdot \chi_A \,d\nu \\ \stackrel{*}{=} \int_X P_{T^n, \mu}(f) \cdot \chi_A \, d\mu,
$$
where in * i would like to use the fact that $\int_X h \, d\nu = \int_X h\cdot f \, d\mu$. The problem is that I don`t know that $f$ is nonzero $\mu$-a.e. I know it is nonzero $\nu$-a.e., why writing the integral just before * is fine. If $\mu$ and $\nu$ were equivalent (as in a), then i would know that $f$ is in fact nonzero $\mu$-a.e. and it should be fine. Is there a way to fix it without that assumption?
Thanks! Let me know if I should provide additional definitions.
 A: I found a solution, so I will post it as an answer. The solution is actually pretty simple if one looks at b) as its own question (and doesn't try to use a) to solve it). For completeness sake, I will also present a proof of part a).
Proof of a). Let $B \in \mathcal{F}$ be any measurable set. Then we have
$$
\int_B P_{T, \mu}(g) \, \mathrm{d}\mu \stackrel{*}{=} 
\int_{T^{-1}B}g \, \mathrm{d}\mu \stackrel{\heartsuit}{=} 
\int_{T^{-1}B} gf\, \mathrm{d}\nu \stackrel{*}{=} 
\int_{B}P_{T, \nu}(fg) \, \mathrm{d}\nu \stackrel{\heartsuit}{=} 
\int_B \frac{P_{T,\nu}(fg)}{f} \, \mathrm{d}\mu,
$$
where in $*$ we use the definition of the Perron-Frobenius operator and in $\heartsuit$ we use the fact that $\mu = f\mathrm{d}\nu$. Note that since the two measures are equivalent, the Radon-Nikodym derivative is non-zero almost everywhere, so there is no problem dividing by $f$ in the last integral.
Since $B\in \mathcal{F}$ was arbitrary, it follows that $P_{T,\mu}(g) = \frac{P_{T, \mu}(fg)}{f}$ almost surely (with respect to either measure, since they are equivalent.
Proof of b). We will use the following characterization of strongly mixing.

T is strongly mixing if and only if for all $f \in L^1(X, \mathcal{F}, \mu)$ and $g \in L^\infty(X, \mathcal{F}, \mu)$,
$$
\lim_{n \to \infty} \int_X (P_T^nf)\cdot g \, \mathrm{d}\mu = \int_X f \, \mathrm{d}\mu \int_X g \, \mathrm{d}\mu.
$$

Then we have
$$
\lim_{n \to \infty} \nu(T^{-n}A) =
\lim_{n \to \infty} \int_{T^{-1}A} f \, \mathrm{d}\mu = 
\lim_{n \to \infty} \int_X (P_{T, \mu}^n f) \cdot \chi_A \, \mathrm{d}\mu =
\int_X f \, \mathrm{d} \mu  \int_X \chi_A \, \mathrm{d} \mu = \mu(A).
$$
