If $0\leq f_{n}\leq g_{n}$, $f_{n}\longrightarrow f$, $g_{n}\longrightarrow g$, $\int_{0}^{1}g{n}\longrightarrow \int_{0}^{1}g$, then so is $f_{n}$. I am working on an exercise stating that:

If $0\leq f_{n}\leq g_{n}$ are defined on $[0,1]$, $f_{n}\longrightarrow f$, $g_{n}\longrightarrow g$ and $\int_{0}^{1}g_{n}(x)dx\longrightarrow \int_{0}^{1}g(x)dx$, then show that $\int_{0}^{1}f_{n}(x)dx\longrightarrow \int_{0}^{1}f(x)dx$ is also true.

As a first thought, I wanted to use Dominated convergence theorem, but it turned out that DCT cannot be applied here since the exercise does not really say if $\int_{0}^{1}|g(x)|dx<\infty$.
The exercise also has a hint that DCT cannot be used, so I guess it does not assume the integrability of $g$ implicitly.
I then tried to bound the integral, but what I obtained was only the following. Since $0\leq f_{n}\leq g_{n}$, we must have $\int_{0}^{1}f_{n}(x)dx\leq \int_{0}^{1}g_{n}(x)dx$, then it follows from Fatou that $$\int_{0}^{1}f(x)dx\leq \liminf_{n\rightarrow\infty}\int_{0}^{1}f_{n}(x)dx\leq \limsup_{n\rightarrow\infty}\int_{0}^{1}f_{n}(x)dx\leq \limsup_{n\rightarrow\infty}\int_{0}^{1}g_{n}(x)dx=\int_{0}^{1}g(x)dx,$$ which is not really useful..
It seems that I did not really use $f_{n}\longrightarrow f$ and $g_{n}\longrightarrow g$, but I have no idea about how to use them. Any idea?
 A: Here is a sketch for a solution which uses the assumption that $\int g <\infty$ (leave a comment if you want more details for the solution):
You have already shown $\int f \leq \liminf \int f_n$.
If we apply Fatou to
the nonnegative sequence $g_n-f_n$,
then one can obtain $\int f \geq \limsup \int f_n$
after using the other assumptions.
But what if $\int g=\infty$?
I think then the following is a counterexample.
We set $g(s):=\chi_{(0,1)} 1/s$, $g_n=\min(g,n)$,
$f_n(s)=n\chi_{[0,1/n]}$ and $f=0$.
The assumptions are satisfied,
but we have $\int f_n=1\neq 0=\int f$.
A: This is false if there are no integrability assumptions. Let $f_n=n\chi_{(0,\frac 1  n)}, g_n\equiv n, f(x)=0$ and $g(x)=\infty$ for all $x$. Then the hypothesis is satisfied. But $1=\int f_n$ for all $n$ and  $ \int f=0$.
If $g$ is integrable then an application of Fatou's Lemma to $(g_n-f_n)$ gives $\lim \sup \int f_n \leq \int f$. Another application of Fatou's Lemma gives $\int f \leq \lim \inf \int f_n$ and this completes the proof.
A: Without the hypothesis $\int g<\infty$ there are counter examples:
Let's take $g(x)=\dfrac 1x\ $ and $\ g_n(x)=g(x)\ \chi_{[\frac 1n,1]}$
Now with $f_n(x)=n(1-x)^n\ \chi_{[\frac 1n,1]}\ $ and $\ f(x)=0$
