A question about Lebesgue measure and integration. Let $(X,A,\mu)$ be a measurable space.
$f\colon X\to [0,\infty]$ measurable.
Denote $S=\lbrace x\in X : f(x)<1\rbrace$.
Prove
a. $\mu(S)=\lim_{n\to \infty} \int_S e^{-f^n}\,d\mu$
b. Assume $X=S$ , prove that:
$$\sum_{n=1}^{\infty} \int_X f^n\, d\mu= \int_{X} \frac{f}{1-f}\,d\mu$$
I have difficulty where I'm supposed to move between integrals.
We can define the measure in terms of integral. So in a I know that it is the start but it is not clear to me how to connect between the set $S$ to the integral.
In b, this is what I did:
$\sum_{n=1}^{\infty} \int_X f^n \,d\mu$= (by a theory int of sum =sum of int for a sequence of positive measurable functions)
=$\int_X (\sum_{n=1}^{\infty}f^n)\, d\mu$=(sum of a geometry column with $f(x)<1$)= $\int_X \frac{f}{1-f}\, d\mu$.

Hint:
For (a) -
We want to write $\mu(S)$ as the limit of some integrals... So it might be useful to write $\mu(S)$ as an integral. Of course, we know how to do this: $\mu(S) = \int_S 1 \mathrm{d}\mu$.
Notice this already looks kind of like the limit we're meant to compute. If we can show
$$\lim_{n \to \infty} \int_S e^{-f^n} \mathrm{d} \mu = \int_S 1 \mathrm{d} \mu$$
then we're done!
If you play around with this, you might notice $f^n \to 0$ pointwise on $S$ (since $0 \leq f(x) < 1$ for each $x \in S$). Then $e^{-f^n} \to e^{-0} = 1$.
So the stuff we're integrating limits to $1$, which is what we need it to be!
If only we knew how to relate the limit of a sequence of integrals to the integral of the limit of the integrands...

For (b) -
You're on the exact right track! Do you know how to justify the fact that $\sum \int \text{stuff} = \int \sum \text{stuff}$? Since infinite sums are simply limits of finite sums, you'll want to use the same technique of swapping limits and integrals from part (a).

I hope this helps ^_^
