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?
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
$$
  |\mu(A)|
  =
  \left|\int_{T^{-1}(A)} f d\lambda\right|
  \leq
  \int_{\mathbb{}} |f| d\lambda
  &lt;
  \infty
$$
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.
$ \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.