$f_n \rightarrow f$ a.e. and $(\mu(A) < \infty \implies \int_A f_n d\mu \rightarrow \int_A f d\mu) \implies \int f_n \rightarrow \int f$

Question

Question: If $f_n \rightarrow f$ almost everywhere and $\int_A f_n d\mu \rightarrow \int_A f d\mu$ whenever $\mu(A) < \infty$, for $\mu$ a $\sigma-$finite measure on $X$, is it true that $\int_X f_n \rightarrow \int_X f$ (where the limit may be $\pm \infty$) ?

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. For the general case, we can also break up $f = f^+- f^-$, and have convergence to each piece.

Doubts:

1. 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?

Answer 1

This is false even if $f$ is integrable. The interchange of the two limits in your argument is not permissible.

Let $g$ be any positive bounded measurable function on $[0,\infty)$ with $\int_0^{\infty} g(t)dt=\infty$. Let $f_n=\frac g n$ and $f=0$. Then $\int_A \frac {g(t)} n dt \to 0$ if $A$ has finite Lebesgue measure but $\int_0^{\infty} f_n(t)dt=\infty$ for each $n$.

Here is an example in which each $f_n$ is integrable. Let $f_n=\frac 1 n1_{(0,n)}$ and $f=0$. In this case, $\int f_n \to 1\neq \int f$. Note that $\int_A f_n \le \frac {\mu(A)} n \to 0$ if $\mu (A) <\infty$.