Show, there exists exactly one operator with $\int_A P_T(f)\, d\lambda=\int_{T^{-1}(A)}f\, d\lambda$ 

Let $T\colon\mathbb{R}\to\mathbb{R}$ be a non-singular function, i.e. a measurable function with the property that
    $$
\forall A\in\mathcal{B}: \lambda(A)=0 \implies \lambda(T^{-1}(A))=0.
$$
    Show: There exists exactly one linear operator  $P_T\colon L_{\lambda}^1\to L_{\lambda}^1$ so that for all $f\in L_{\lambda}^1$ and all $A\in \mathcal{B}$ it is
    $$
\int_A P_T(f)\, d\lambda=\int_{T^{-1}(A)}f\, d\lambda.
$$


Hello, I would really prefer to present you my own recent ideas but I do not have own ideas. To be honest, I am rather helpless. Can you pls give me help?
Greetings.
math12
New Edit:
The only thing I already know is that
$A\mapsto\int_{T^{-1}(A)}f\, d\lambda$
is a signed measure.
Now one can apply Radon-Nikodým or something like that?
 A: Consider the signed measure
$$
  \mu(A)
  =
  \int_{T^{-1}(A)}f\, d\lambda.
$$
AFTER verifying that $\mu$ satisfies the Radon-Nikodým hypothesis,
use the theorem to find a unique $g$ such that
$$
  \mu(A)
  =
  \int_A g\, d\lambda.
$$
This $g$ is exactly $P_T(f)$.
For the hypotheses, you have to show that $\mu$ is $\sigma$-finite and that $\mu \ll \lambda$. Let me know if you need help with that.

Edit:
Contrary to what I have written, you do not need to show that $\mu$ is $\sigma$-finite. This hypothesis is for $\lambda$. And by the way, $\lambda$ does not have to be Lebesgue. It just has to be $\sigma$-finite for the proof to hold.
It seems that, in order to use the theorem, both measures, $\lambda$ and $\mu$ have to be $\sigma$-finite. If you know of a proof of the Radon-Nikodým theorem that does not require $\mu$ to be $\sigma$-finite, let me know... :-)
But anyway, $\mu$ is not only $\sigma$-finite. It is finite, because $f$ has finite integral. In fact
$$
  |\mu(A)|
  =
  \left|\int_{T^{-1}(A)} f d\lambda\right|
  \leq
  \int_{\mathbb{}} |f| d\lambda
  <
  \infty
$$
Had we missed the hypothesis that $f \in L^1_{\lambda}$? ;-)
Also, $\lambda$ does not have to be Lebesgue. It just has to be $\sigma$-finite for the proof to hold.
A: It is
$$
\int_{T^{-1}(A)}f\, d\lambda=\int_{T^{-1}(A)}\, df\lambda=\int_{T^{-1}(A)}\, d\lambda_f=\int_{\mathbb{R}}\chi_{T^{-1}(A)}\, d\lambda_f=\int_{\mathbb{R}}\chi_A\circ T\, d\lambda_f \\=\int_{T^{-1}(A)}\chi_A\circ T\, d\lambda_f=\int_A\chi_A\, d\lambda_f\circ T^{-1}=\int_A\, d\lambda_f\circ T^{-1}
$$
(So the function
$$
A\longmapsto\int_{T^{-1}(A)}f\, d\lambda=\int_A \, d\lambda_f\circ T^{-1}
$$
is a measure relating to $\lambda_f\circ T^{-1}$ with density $1$.)
Let $A\in\mathcal{B}$ with $\lambda(A)=0$.
Then
$$
\lambda_f\circ T^{-1}(A)=f\lambda(T^{-1}(A))=0
$$
by assumption, i.e. $\lambda_f\circ T^{-1}\ll\lambda$.
Now use Radon-Nikodým to determine $P_T(f)$.
And do not forget to show that $P_T$ is linear (using some properties of Radon-Nikod´m-densities you probably know.)
A: $ \int_{T^{-1}(A)}fd\lambda = \int_{A}dT\#(f\lambda) $ where $ T\#\nu $ is the push-forward measure of $ \nu $ by $T$. Then by change of variable you can find that
$ T\#(f\lambda) = f\circ T^{-1}|det(DT\circ T^{-1})|^{-1}\lambda $.
Let
$P_T(f) = f(T^{-1})|det(DT\circ T^{-1})|^{-1} $, clearly linear in f. Not sure this proves uniqueness and you need some more assumptions on T, such as derivability, but perhaps this might help you.
