Assuming that $f_n$ is a sequence of integrable function that converges pointwise to $f$ that is also integrable (so everything is defined below). Why is it the case that:
$$ \int\lim_{n\to\infty} f_n(x)\leq\lim_{n\to\infty}\int f_n(x) $$
My text gave no explanation and it looks like no one asked this question on ME. Is this suppose to be that obvious and there are no conditions on this (other than the limits being defined)? So the sequence $f_n$ can be converging from below or above and $f$ can be positive or negative and the inequality would still hold?
Your book is wrong unless you forgot an assumption. Take $f_n(x) = -(1/n) \cdot \chi_{[0, n]}(x),$ where $\chi_{[0, n]}$ is the function which is 1 on $[0, n]$ and 0 everywhere else.
Then as $n\rightarrow\infty,$ we have $f_n(x)\rightarrow 0$ for every $x.$ So the left hand side of your inequality is 0.
But the right hand side is obviously $-1,$ since the integral of $f_n$ for each $n$ is just $(-1/n) \cdot n = -1,$ as you integrate a step function which is $-1/n$ on an interval of length $n,$ and zero everywhere else.
You can also have another sort of failure, where the limit on the right hand side doesn't exist even if the $f_n$ converge pointwise to something. For this example, take $f_n(x) = (1/n) \cdot \chi_{[0, n]}(x)$ if $n$ is even and $f_n(x) = (1/n) \cdot \chi_{[0, 17n]}(x)$ if $n$ is odd. Then the integrals of $f_n$ oscillate between being 1 and 17, even if they still converge pointwise to 0.
If you know that $f_n \geq 0$, and that the limit on the right exists, then this is true and called Fatou's Lemma. This is a lemma in measure theory, and can be viewed as a consequence of the fact that if you have an increasing chain of sets $A_1 \subseteq A_2 \subseteq \cdots,$ then the length of the 'limit' of the sets (ie, the set $A = A_1 \cup A_2 \cup\cdots$) is equal to the limit of the length of $A_n.$