I have a question!
The problem at hand is the following:
Let $(X, S, \mu)$ be a measure space and let $X$ be $\sigma$-finite. Let $f: X\rightarrow (0,\infty)$ be a measurable function, and define a second measure $\nu$ by:
$$\nu(A) = \int\limits_{A} f \ d\mu,$$ for $A \in S.$
Now it is to show that: $$\mu(A) = \int\limits_{A} \frac{1}{f}\ d\nu.$$
We were also given a hint by the professor's assistent to first consider the case $A \subset f^{-1}(a,\lambda * a]$, for positive $a$ and $\lambda > 1$.
Now it is very hard to show my work because I haven't made any significant progress. I first looked at the integral and tried to make some progress using the case where $f$ is a simple function, but I nowhere got a good moment to use the hint and have the feeling that I am stepping in the dark.
As this is part of an exercise, hints are also appreciated!
Thanks!
Ignore the assistant's hint. Show that $$\int g \, d\nu = \int fg \, d\mu$$ holds for all nonnegative functions $g$. This follows easily from approximations by simple functions and the monotone convergence theorem.
The original claim follows from applying this identity to $g = \frac{1}{f}$.