Let $E$ be a borel-measurable set and $f_k, g_k$ two sequences of functions that are defined and measurable on $E, \forall k \in \mathbb{N}$. Also they converge pointwise $\forall x \in E: \lim \limits_{k \to \infty} f_k(x) = f(x)$ and $\lim \limits_{k \to \infty} g_k(x) = g(x)$. Also $|f_k(x)| \leq g_k(x), \forall k \in \mathbb{N}, \forall x \in E$. It is also given that $\lim \limits_{k \to \infty} \int \limits_E g_k(x)dx = \int \limits_E g(x)dx$.
I am asked to show that $\lim \limits_{k \to \infty} \int \limits_E f_k(x)dx = \int \limits_E f(x)dx$.
So far I only have been able to show:
$$\int \limits_E g(x)dx - \int \limits_E |f(x)|dx = \int \limits_E g(x)-f(x)dx = \int \limits_E
\lim \limits_{k \to \infty} (g_k(x)dx -|f_k(x)|) dx $$
$$\leq \lim \limits_{k \to \infty} \int \limits_E g_k(x) -|f_k(x)| dx = \lim \limits_{k \to \infty} \int \limits_E g_k(x) dx - \lim \limits_{k \to \infty} \int \limits_E |f_k(x)|dx = \int \limits_E g(x)dx - \lim \limits_{k \to \infty} \int \limits_E |f_k(x)|dx$$
$$ \Rightarrow \int \limits_E |f(x)|dx \geq \lim \limits_{k \to \infty} \int \limits_E |f_k(x)|dx \geq \int \limits_E \lim \limits_{k \to \infty} |f_k(x)| dx = \int \limits_E |f(x)|dx $$
So I have been using Fatou's Lemma since $|f_k|\geq 0$ but how can I show it generally, is there a trick so I am able to apply Fatou's Lemma or do I have to look for something different? I know that the first time is not a problem since $g_k(x) - f_k(x) \geq 0, \forall k \in \mathbb{N}, \forall x \in E$ but what can I do about the 2nd time I am using it?
You have to apply Fatou’s lemma differently; for this you need to find an appropriate non-negative function. First, by taking the limit $k\to\infty$, we see that $|f|\leq |g|$. Next, $|f_k-f|\leq |f_k|+|f|\leq g_k+g$, so that $\Phi_k=g_k+g-|f_k-f|$ is our desired non-negative function. By applying Fatou’s lemma to this, we get \begin{align} \int_X2g\,d\mu&=\int_X\liminf_{k\to\infty}\Phi_k\,d\mu\\ &\leq \liminf_{k\to\infty}\int_{X}\Phi_k\,d\mu\\ &=\liminf_{k\to\infty}\left(\int_Xg_k\,d\mu+\int_Xg\,d\mu-\int_X|f_k-f|\,d\mu\right)\\ &=\lim_{k\to\infty}\left(\int_Xg_k\,d\mu+\int_Xg\,d\mu\right)+\liminf_{k\to\infty}\left(-\int_X|f_k-f|\,d\mu\right)\\ &=2\int_Xg\,d\mu-\limsup_{k\to\infty}|f_k-f|\,d\mu. \end{align} If you now suppose $\int_Xg\,d\mu$ is finite, then you can cancel things and rearrange to get \begin{align} \limsup\limits_{k\to\infty}\int_X|f_k-f|\,d\mu\leq 0. \end{align} If a $\limsup$ of non-negative quantities is $\leq 0$, then it is equal to zero, and the limit exists and is zero, so $\lim\limits_{k\to\infty}\int_X|f_k-f|\,d\mu=0 $, which implies your statement by the triangle inequality $\left|\int_Xf_k\,d\mu-\int_Xf\,d\mu\right|\leq \int_X|f_k-f|\,d\mu$ (also note that $f\in L^1$ since $|f|\leq g$ and $g\in L^1$).
As you can see here, the proof goes through for an arbitrary measure space. Also, I merely used basic properties of $\liminf$ and $\limsup$ for numerical sequences. If you’re unfamiliar with them, then you should start with those simpler properties first.