Let $(S, \Sigma, \mu)$ be a measure space. Let $f_n : S \rightarrow \mathbb{R}$ be a sequence that is bounded in $\mathcal{L}^1$ meaning there exists $M > 0$ such that
$$\int_S |f_n| d\mu \leq M \quad \forall n.$$
Suppose that $f_n \rightarrow f$ a.e. where $f : S \rightarrow \overline{\mathbb{R}}$ is measurable. Prove that
$$\int_S |f| d\mu \leq M.$$
My try
- Step one (because $f_n \rightarrow f$ a.e.)
$$\int_S f d\mu = \int_S \lim_{n \rightarrow \infty} f_n d\mu$$
- Step two: (we use the Fatou's lemma to obtain):
$$ \int_S |f| d\mu = \int_S \lim_{n \rightarrow \infty} |f_n| d\mu = \int_S \liminf_{n \rightarrow \infty} |f_n| d\mu \leq \liminf_{n \rightarrow \infty} \int_S |f_n| d\mu \leq M, $$
where the first inequality follows from Fatou's lemma and the second inequality follows from the fact that $f_n$ is bounded in $\mathcal{L}^1$.
Therefore, we have shown that $\int_S |f| d\mu \leq M$, as required.
Questions
Using $\int_S f d\mu = \int_S \lim_{n \rightarrow \infty} f_n d\mu$, can I write $\int_S |f| d\mu = \int_S \lim_{n \rightarrow \infty} |f_n| d\mu$? In other words, am I using the property of a.e. correctly?
Is there any other way than Fatou's lemma to show this?
It is true that if $f_n$ converges a.e. to $f$ then $|f_n|$ converges a.e. to $|f|$.
Indeed, if $f_n(x) \rightarrow f(x)$ then $|f_n(x)| \rightarrow |f(x)|$. So we have the inclusion $$\{x \in S \mid f_n(x) \rightarrow f(x)\} \subset \{x \in S \mid |f_n(x)| \rightarrow |f(x)|\}. $$ If the first set is of complement of zero measure, so is the second one.
There may be other ways to prove your statement, but Fatou's lemma gives a quick short proof that seems hard to beat.