Question about liminf for a pointwise convergent sequence of functions. If $f_n \rightarrow f$ pointwise, then does $$\liminf \int f_n=\lim\int f_n?$$
I know that $\liminf f_n=\lim f_n$ since the sequence converges, but I'm not sure if the $(L)$ integral throws us off.
I'm trying to prove the Fatou's Reverse Lemma, and I got stuck.
EDITED: $f$ is integrable and $f_n\le f$
 A: I'm going to assume your measure space is $\mathbb{R}$ with Lebesgue measure.
Consider $f_n=n1_{[0,\frac{1}{n}]}$ where $1_A$ is the characteristic function of the set $A\subset \mathbb{R}$. Then $f_n \to f=0$ pointwise but
$$
\liminf\int_{\mathbb{R}} f_n(x)dx=1 > 0= \int_{\mathbb{R}} f(x)dx
$$
(You can even take $f_n, f\in C^{\infty}(\mathbb{R})$ so it's not a regularity issue).
In other measure spaces it might still be false: For example in $\mathbb{N}$ with counting measure take $f_n(m)= 1$ if $m=1,n$ and zero otherwise then $f_n \to f$ pointwise, where $f(1)=1$ and is zero otherwise, and
$$
\liminf \int_{\mathbb{N}} f_n(m)dm = 2 > 1 = \int_{\mathbb{N}} f(m)dm
$$
So no, in general only one inequality is true in Fatou's lemma.
With the edit it's still not true: Take $g_n=-f_n$ as above. You could put $|f_n|\leq f$ but then this is just the dominated convergence theorem.
A: If we had some guarantee that the limit on the right hand side exists, then of course this equality would be true, because for a convergent sequence $a_n = \int f_n d \mu$ we would have $\liminf a_n = \lim a_n$.
But in your case there doesn't seem to be such a guarantee. For example, consider a sequence of functions $f_n: \mathbb{R} \to \mathbb{R}$ like this: for $n$ even, $f_n=0$. And for $n$ odd, $f_n = 1_{[n, n+1]}$. Then $\int f_n d\mu$ is $0$ for n even and $1$ for n odd, so $\liminf$ on the LHS is equal to $0$, and the $\lim$ on the RHS doesn't exist.
A: A slight modification of the usual counterexample works here:
$$
f_n=\left\{\begin{array}{}n1_{(0,1/n]}&\mbox{if $n$ even}\\0&\mbox{if $n$ odd}\end{array}\right.
$$
Here, $f_n\to0$ pointwise, and
$$
\liminf_{n\to\infty}\int_{\mathbb{R}}f_n(x)\,\mathrm{d}x=0
$$
yet
$$
\limsup_{n\to\infty}\int_{\mathbb{R}}f_n(x)\,\mathrm{d}x=1
$$
so the limit does not exist.
Of course if $\lim\limits_{n\to\infty}\int_{\mathbb{R}} f_n(x)\,\mathrm{d}x$ exists, then $\liminf\limits_{n\to\infty}\int_{\mathbb{R}} f_n(x)\,\mathrm{d}x=\lim\limits_{n\to\infty}\int_{\mathbb{R}} f_n(x)\,\mathrm{d}x$ by definition.
