Question: For $\sigma-$finite measure space $(X,\mu)$, if $f_n \uparrow f$ almost everywhere, $\int_X f d\mu$ exists $\in [-\infty,\infty]$ and $\int_A f_n d\mu \rightarrow \int_A f d\mu$ whenever $\mu(A) < \infty$, is it true that $\int_X f_n \rightarrow \int_X f$ (where the limit may be $\pm \infty$) ?
This is non-trivial because MCT only applies for $f_n \geq 0$, whereas here the limit $f$ could be negative.
Attempt: Assuming that $f_n$ and $f$ are integrable, we have that $X = \cup_{m \geq 1} F_m$, where $\mu(F_m) < \infty$. Now, we'd be done if $\lim_n \int_{F_m} f_n d\mu = \int_{F_m} f d\mu$, so we'd be done if, observing that $\int_X f d\mu = \lim_m \int_{F_m} f d\mu$ (by integrability of $f$), we have the equality $(*)$ in:
$$ \lim_m \int_{F_m} f d\mu = \lim_m \lim_n \int_{F_m} f_n d\mu =_{(*)} \lim_n \lim_m \int_{F_m} f_n d\mu = \lim_n \int_X f_n d\mu $$
Now, if $f_n, f \geq 0$, then the above follows by this answer (the condition is that we need monotonicity in both $n$ and $m$ for the doubly-indexed sequence $\int_{F_m} f_n d\mu$.). For the general case, we can also break up $f = f^+- f^-$, and have convergence to each piece, so the result holds for $f$ integrable.
Doubts:
- Is there an easier proof of the above not needing interchanging of limits? Alternatively, can the hypothesis be relaxed in this case?
- If $f$ is not integrable, then we cannot proceed by the above argument since we no longer have $\int_X f d\mu = \lim_m \int_{F_m} f d\mu$ (which requires DCT). In that case, is the above even true?
P.S. The case for $f_n \rightarrow f$ (i.e., not increasing) was disproven here.
