Weak convergence in $L^1(X,\mu)$ space I found an interesting property in some lecture notes on weak convergence: lemma 3.2 (2) on page 10: https://www.uio.no/studier/emner/matnat/math/MAT4380/v06/Weakconvergence.pdf . The idea is as follows:
Let $(X,\mu)$ be a $\sigma$ finite measure space and $(f_n)_{n\in\mathbb{N}^*}, (g_n)_{n\in\mathbb{N}^*}, f,g :X\to \overline{\mathbb{R}}$ are $\mu$ measurable functions. Prove that if :
$\bullet$ There is a function $f:X\to\overline{\mathbb{R}}$ such that $\lim\limits_{n\to\infty} f_n(x)= f(x),$ for $\mu$ - almost all $x\in X$. (pointwise convergence);
$\bullet$ There is some positive constant $M>0$ such that: $||f_n||_{\infty}:=\mathrm{ess sup}\ |f_n|<M, \forall n\in\mathbb{N}^*$; 
$\bullet$ $g_n\rightharpoonup g$ in $L^1(X,\mu)$, 
then $f_n g_n\rightharpoonup fg$ in $L^1(X,\mu)$ i.e. $\lim\limits_{n\to\infty} \displaystyle\int_{X} f_n g_n h\ d\mu=\int_{X} fgh\ d\mu, \forall h\in L^{\infty}(X,\mu)$
I tried to prove this by definition but I cannot make good estimates on $\int_{X} |g_n||f_n-f|\ d\mu$. In the lecture notes sais that this is a simple property and the proof is omitted...Any help is very welcomed. 
 A: The missing argument is a combination of Egorov's theorem and the Dunford-Pettis theorem (for the precise versions of both that we are going to use, see [Brezis, Haim, Functional analysis, Sobolev spaces and partial differential equations, Springer (2011), theorems 4.29 and 4.30 page 115 ]). Roughly speaking, the former tells us that $f_n\to f$ uniformly, up to removing arbitrarily small sets from $X$. And the latter guarantees that $|g_n|$ gives small mass to such small sets, uniformly in $n$ and only depending on the measure of the small set.

Disclaimer: as correctly pointed out by Nik Weaver in his comment, Egorov's theorem crucially requires a finite measure $\mu(X)<+\infty$. Step 1 below gives the key argument in this finite situation. Then Johannes Schürz's comment settles in step 2 the general case, based on the uniform integrability.

Step 1: assume first that $\mu(X)<+\infty$, and pick any $\epsilon>0$.
By Egorov's theorem there exists a subset $X_\epsilon\subset X$ with small complement $\mu(X\setminus X_\epsilon)\leq \epsilon$ such that $f_n\to f$ uniformly in $X_{\epsilon}$, in particular also $f_nh\to fh$. Thus by standard $L^\infty-L^1$ strong-weak convergence the term
$$
\int_{X_\epsilon} f_ng_nhd\mu\to \int _{X_\epsilon}fg hd\mu.
$$
For the remaining term (the integral on $X\setminus X_\epsilon$), the Dunford-Pettis theorem guarantees that $\{g_n\}_n$ is uniformly integrable. Given $\delta>0$, this means that $\int_{X\setminus X_\epsilon} |g_n|\leq \delta$ uniformly in $n$, as soon as $\mu(X\setminus X_\epsilon)\leq \epsilon$ is sufficiently small. Since $|f_n|_\infty\leq M$ and $h\in L^\infty$ this immediately gives
$$
\left|
\int_{X\setminus X_\epsilon} f_ng_nhd\mu
\right|\leq M\|h\|_\infty \delta
$$
Putting everything together and playing a bit with $\epsilon,\delta,n\geq n_0$, and quantifiers gives the desired result (also noticinng that $\int _{X_\epsilon}fg hd\mu\to \int _{X}fg hd\mu$ if $\epsilon\to 0$).
Step 2: assume now that $X$ has infinite measure. Since the sequence $g_n$ is $L^1$-weakly converging, it is uniformly integrable (by the Dunford-Pettis theorem) and therefore for any small $\eta>0$ there exists $X_\eta\subset X$ with $\mu(X_\eta)<+\infty$ and
$$
\int_{X\setminus X_\eta} |g_n|d\mu \leq \eta
\qquad
\forall\, n\geq 0
$$
As a consequence
$$
\left|
\int_{X\setminus X_\eta} f_n g_n hd\mu
\right|
\leq \|f_n\|_{L^\infty(X)} \|h\|_{L^\infty(X)} \|g_n\|_{L^1(X\setminus X_\eta)}\leq M\|h\|_{L^\infty(X)}\eta
$$
can be made arbitrarily small, uniformly in $n$.
The result follows next by applying step 1 on the finite measure set $X_\eta$.
PS: I was not aware of this specific statement, and it may turn out to be quite handy at some point so thank you Maxim Diana!
