Monotone Convergence theorem requires non-negativiness. Just consider the functions $f_{n} = \mathbb {-1}_{[n,n+1]}$. Note that $f_{n} \uparrow 0$ , but $ \int f_n = 1$, for all $n$.
Of course, if $f_n$ are measurable, you can work with $f_n^+$ and $f_n^-$. The issue here is that, if $f_n \uparrow f$, then $f_n^- \downarrow f^-$. So you can not apply Monotone Convergence Theorem to the negative part (you would need Dominated Convergence Theorem).
In the proof of Fatou's lemma, that is exactly where the proof breaks if you don't assume that $f_{n}$ are non-negative:
Let $g_{k}:=\inf_{n \geq k} f_{n}$, then, for the negative part we have $g_{k}^-=\sup_{n \geq k} f_{n}^-$ and $\lim_n \sup f_n^- = \downarrow \lim g_k^-$. (then you would need Dominated Converge Theorem).
To correct this, you might be tempted to write:
Let $g_{k}:=\inf_{n \geq k} f_{n}^+$ and $h_{k}:=\inf_{n \geq k} f_{n}^-$. We have then $\lim_{n} \inf f_{n}^+ = \uparrow \lim g_{k}$ and $\lim_{n} \inf f_{n}^- = \uparrow \lim h_{k}$, but now the issue is that, although $f =f^+ - f^-$, we have that $\lim_{n} \inf f_{n} \neq \lim_{n} \inf f_{n}^+ - \lim_{n} \inf f_{n}^-$. So, although you can apply Fatou to the positive and negative parts separately, you can not combine the two results for $f$.
Once you have the MCT, everything else follows.
First, we can show that Fatou's lemma follows from MCT.
Proof: Suppose $f_n \geqslant 0$ and define $g_m = \inf_{k \geqslant m} f_k$. It follows that $g_m \leqslant f_n$ and $\int g_m \leqslant \int f_n$ for all $n \geqslant m$. Thus, $\int g_m \leqslant \liminf_{n \to \infty} \int f_n$. The sequence $(g_m)$ is increasing and by definition $\lim_{m \to \infty} g_m = \liminf_{n \to \infty} f_n$. By the MCT, it follows that
$$\int \liminf_{n \to \infty} f_n = \int\lim_{m \to \infty} g_m = \lim_{m \to \infty}\int g_m \leqslant \liminf_{n \to \infty} \int f_n\quad \text{(Fatou's lemma)}$$
Then we can show that DCT follows from Fatou's lemma.
Proof: We can assume WLOG that $f_n \to f$. (otherwise redefine appropriately on the measure zero set where $f_n \not\to f$). Since $|f_n| \leqslant g$, we have $g+f_n \geqslant 0$. Using Fatou's lemma, it follows that
$$\int g + \int f = \int(f+g) \leqslant \liminf_{n \to \infty}\int(g + f_n) = \int g + \liminf_{n \to \infty}\int f_n,$$
and, hence,
$$\tag{*} \int f \leqslant \liminf_{n \to \infty}\int f_n$$
Similarly, applying Fatou's lemma to $g- f_n \geqslant 0$, we get
$$\tag{**} \limsup_{n \to \infty} \int f_n \leqslant \int f$$
Together (*) and (**) imply that
$$\lim_{n \to \infty} \int f_n = \int f \quad \text{(DCT)}$$
Best Answer
Fatou's Lemma, the Monotone Convergence Theorem (MCT), and the Dominated Convergence Theorem (DCT) are three major results which answer the question "When do $\lim_{n\rightarrow\infty}$ and $\int$ commute?"
Same definition in measure theory:
Monotone Convergence Theorem: If $\{f_n:X\rightarrow[0,\infty)\}$ is a sequence of measurable functions on a measurable set $X$ such that $f_n\rightarrow f$ pointwise almost everywhere and $f_1\leq f_2\leq\cdots,$ then $$\lim_{n\rightarrow\infty}\int_{X}f_n=\int_{X}f$$
Dominated Convergence Theorem: If $\{f_n:\mathbb R\rightarrow\mathbb R\}$ is a sequence of measurable functions which converge pointwise almost everywhere to $f$, and if there exists an integrable function $g$ such that $|f_n(x)|\leq g(x)$ for all $n$ and for all $x$, then $f$ is integrable and $$\int_{\mathbb R}f=\lim_{n\rightarrow\infty}\int_{\mathbb R}f_n$$
Wait a minutes, did you say I am not using expectation, then recall the definition,
In general, if $X$ is a random variable defined on a probability space $(\Omega ,\Sigma ,P)$, then the expected value of $X$, denoted by $\mathbb{E}\:X$, is defined as the Lebesgue integral, $$\mathbb E\:X=\int_{\Omega}X(\omega)\:dP(\omega)$$
Which is nothing more than integration.