Definition of Frobenius-Perron (transfer) operator I came across two definitions for the transfer operator. The two definitions seem to me to be very similar, but I couldn't deduce one of them from the other.
The first definition is the classic one, which I have already found in several books.

Let $(X,\mu)$ be a $\sigma$-finite measure space. For $f\in L^{1}(X,\mu)$, define a signed measure $\mu_{f}$ on $X$ by
$$\mu_{f}(A)=\int_{T^{-1}(A)}f \ d\mu$$
for all measurable $A\subset X$. By the Radon-Nikodym theorem, there exists a unique element $P_{T}f\in L^{1}(X,\mu)$ such that
$$\mu_{f}(A)=\int_{A}P_{T}f \ d\mu$$
for all measurable $A\subset X$. The operator $P_{T}\colon L^{1}(X,\mu)\to L^{1}(X,\mu)$ define by
$$\int_A P_{T}f \ d\mu = \int_{T^{-1}(A)} f\ d\mu\qquad \forall f\in L^{1}(X,\mu)$$
is called the Perron-Frobenius or transfer operator.

The second definition can be found, for example, in the book An introduction to infinite ergodic theory by  J. Aaronson.

The transfer operator $\widehat{T}:L^1(X,\mu)\to L^1(X,\mu)$ is given by,
$$\int_X g (\widehat{T}f)\ d\mu = \int_X (g\circ T)f\ d\mu \qquad \forall f\in L^1(X,\mu),\ g\in L^{\infty}(X,\mu)$$
this equality is deduced from the following equality,
$$\widehat{T}f=\frac{d\nu_f \circ T^{-1}}{d\mu}\qquad \text{where } \nu_f (A)= \int_A f \ d\mu$$

So I would like to know if $P_{T}$ and $\widehat{T}$ are the same operator. It seems so, but I couldn't get to prove it.
 A: $(X,\mathscr{F},\mu)$ $\sigma$-finite, and $T:(X,\mathscr{F})\rightarrow(X,\mathscr{F})$ measurable.
Under he assumption that $\mu\circ T^{-1}\ll \mu$, suppose $P_T:L_1\rightarrow L_1$ is such that
$$\int_X(\mathbb{1}_A\circ T) f\,d\mu=\int_X\mathbb{1}_A P_T f\,d\mu$$
for anu $A\in\mathscr{F}$
To show that the $P_T$ satisfies the second definition it is enough to consider $g\in L^+_\infty$. Define $s_n=\sum^{2^n}_{k=1}2^{-n}k\mathbb{1}_{[2^{-n}k,2^{-n}(k+1))}\circ g$. It îs easy to see that $s_n\nearrow g$ as $n\rightarrow\infty$. By dominated convergence
\begin{align}
\int_X (g\circ T)\,f\,d\mu&=\lim_n \int_X\Big(\sum^{2^n}_{k=1} 2^{-n}k\mathbb{1}_{[2^{-n}k,2^{-n}(k+1))}\circ g\circ T\Big) f\,d\mu\\
&=\lim_n \sum^{2^n}_{k=1} 2^{-n}k\int_X\big(\mathbb{1}_{g^{-1}([2^{-n}k,2^{-n}(k+1)))}\circ T\big)\,f\,d\mu\\
&=\lim_n \sum^{2^n}_{k=1} 2^{-n}k\int_X \mathbb{1}_{g^{-1}([2^{-n}k,2^{-n}(k+1)))}\,P_Tf\,d\mu\\
&=\lim_n\int_X\Big(\sum^{2^n}_{k=1}2^{-n}k\mathbb{1}_{[2^{-n}k,2^{-n}(k+1))}\circ g\Big) \,P_T f\,d\mu=\int_X g\,P_T f\,d\mu
\end{align}
That the second definition implies the firs is easy: consider $g=\mathbb{1}_A$ with $A\in\mathscr{F}$.
