Muhammad Haris Rao
Throughout, let $(X, \mathcal{A}, \mu)$ be a measure space.
Theorem(Monotone Convergence Theorem): For $n \ge 1$, let $f_n : X \longrightarrow \mathbb{R}_{\ge 0}$ be a sequence of measurable functions with $f_n \le f_{n + 1}$ for all $n \ge 1$ and $f_n \longrightarrow f$ pointwise. Then $$ \lim_{n \to \infty} \int_X f_n \, d \mu = \int_X f \, d \mu $$
Proof. Let $\alpha \in [0, 1)$. We aim to prove that $$ \lim_{n \to \infty} \int_X f_n \, d \mu \ge \alpha \int_X f \, d \mu $$ For this, define $$ A_n = \left\{ x \in X \mid f_n(x) \ge \alpha f(x) \right\} \subseteq X $$ Since $f_{n + 1} \ge f_n$, it follows that $A_{n + 1} \supseteq A_n$ for all $n \ge 1$. Since $f_n \longrightarrow f$ pointwise, we have the limit of the sequence of sets $\{ A_n \}_{n \ge 1}$ is $$ X = \bigcup_{n \ge 1} A_n $$ Since $f_n(x) \ge \alpha f(x)$ when $x \in A_n$, we have $$ \int_{A_n} f_n \, d \mu \ge \int_{A_n} f \, d \mu $$ for all $n \ge 1$. But then because also $A_n \subseteq X$, $$ \lim_{n \to \infty} \int_X f_n \, d \mu \ge \lim_{n \to \infty} \int_{A_n} f_n \, d \mu \ge \lim_{n \to \infty} \alpha \int_{A_n} f \, d \mu $$ Let $s : X \longrightarrow \mathbb{R}_{\ge 0}$ be any simple function with $s \le \alpha f$. Write $$ s = \sum_{i = 1}^m a_i \mathbf{1}_{B_i} $$ for $\{ B_n \}_{n \ge 1} \subseteq \mathcal{A}$ disjoint, and $\{ a_n \}_{n \ge 1} \subseteq \mathbb{R}_{\ge 0}$. We have $$ \lim_{n \to \infty} \alpha \int_{A_n} f \, d \mu \ge \int_X s \, d \mu = \lim_{n \to \infty} \sum_{i = 1}^m a_i \mu \left( A_n \cap B_i \right) = \sum_{i = 1}^m a_i \mu \left( X \cap B_i \right) = \sum_{i = 1}^m a_i \mu \left( B_i \right) = \int_X s \, d \mu $$ since the sequence of sets $\{ A_n \}_{n \ge 1}$ converges to $X$. Now letting $s_m : X \longrightarrow \mathbb{R}_{\ge 0}$ be a sequence of simple functions indexed by $m \ge 1$ converging upwards to $\alpha f$, we have by applying the above inequality to each $s_m$ that $$ \alpha \int_X f \, d \mu = \lim_{m \to \infty} \int_X s_m \, d \mu \le \lim_{m \to \infty} \lim_{n \to \infty} \alpha \int_{A_n} f \, d \mu = \lim_{n \to \infty} \alpha \int_{A_n} f \, d \mu $$ So now, we have overall $$ \lim_{n \to \infty} \int_X f_n \, d \mu \ge \lim_{n \to \infty} \int_{A_n} f_n \, d \mu \ge \lim_{n \to \infty} \alpha \int_{A_n} f \, d \mu \ge \alpha \int_X f \, d \mu $$ which is the inequality we set out to show. This is true for any $\alpha \in [0, 1)$ which can only imply that $$ \lim_{n \to \infty} \int_X f_n \, d \mu \ge \int_X f \, d \mu $$ The reverse inequality is true because $f_n \le f$ for all $n \ge 1$ so we have the desired $$ \lim_{n \to \infty} \int_X f_n \, d \mu = \int_X f \, d \mu $$ $\space$$\blacksquare$
Theorem(Fatou's Lemma): If $f_n : X \longrightarrow \mathbb{R}_{\ge 0}$ is a sequence of measurable functions indexed by $n \ge 1$, then $$ \liminf_{n \to \infty} \int_X f_n \, d \mu \ge \int_X \liminf_{n \to \infty} f_n \, d \mu $$
Proof. Define $g_n : X \longrightarrow \mathbb{R}$ by $$ g_n = \inf_{k \ge n} f_k $$ Then we have $g_n \le g_{n + 1}$ for all $n \ge 1$. By monotone convergence theorem, $$ \lim_{n \to \infty} \int_X g_n \, d \mu = \int_X \lim_{n \to \infty} g_n \, d \mu = \int_X \liminf_{n \to \infty} f_n \, d \mu $$ Because $g_n \le f_n$, we have also $$ \lim_{n \to \infty} \int_X g_n \, d \mu = \liminf_{n \to \infty} \int_X g_n \, d \mu \le \liminf_{n \to \infty} \int_X f_n \, d \mu $$ Combining these two gives the desired inequality.$\blacksquare$
Theorem(Reverse Fatou's Lemma): If $f_n : X \longrightarrow \mathbb{R}_{\ge 0}$ is a sequence of measurable functions indexed by $n \ge 1$, and $f : X \longrightarrow \mathbb{R}$ is integrable and such that $f_n \le f$ for all $n \ge 1$, then $$ \limsup_{n \to \infty} \int_X f_n \, d \mu \le \int_X \limsup_{n \to \infty} f_n \, d \mu $$
Proof. Since $f_n \le f$ for all $n \ge 1$, we have a sequence of non-negative measruable functions $\{ f - f_n \}_{n \ge 1}$. For $n \ge 1$, define $g_n : X \longrightarrow \mathbb{R}_{\ge 0}$ by $$ g_n = \inf_{k \ge n} \left( f - f_n \right) = f - \sup_{k \ge n} f_n $$ By Fatou's lemma, \begin{align*} \int_X f \, d\mu - \int_X \limsup_{n \to \infty} f_n \, d \mu &= \int_X \lim_{n \to \infty} \left( f - \sup_{k \ge n} f_k \right) \, d \mu \\ &= \int_X \lim_{n \to \infty} \inf_{k \ge n} \left( f - f_k \right) \, d \mu \\ &= \int_X \liminf_{n \to \infty} \left( f - f_n \right) \, d \mu \\ &\le \liminf_{n \to \infty} \int_X \left( f - f_n \right) \, d \mu \\ &= \int_X f \, d\mu + \liminf_{n \to \infty} \left( - \int_X f_n \, d \mu \right) \\ &= \int_X f \, d\mu - \limsup_{n \to \infty} \int_X f_n \, d \mu \end{align*} Cancelling the integrals of $f$ yields $$ \limsup_{n \to \infty} \int_X f_n \, d \mu \le \int_X \limsup_{n \to \infty} f_n \, d \mu $$ which is what we wanted to show. $\blacksquare$
Theorem(Lebesgue Dominated Convergence Theorem): If $f_n : X \longrightarrow \mathbb{R}$ is a sequence of measurable functions indexed by $n \ge 1$, $g : X \longrightarrow \mathbb{R}$ is integrable and such that $f_n \le g$ for all $n \ge 1$, and $f : X \longrightarrow \mathbb{R}$ such that $f_n \longrightarrow f$ pointwise then $$ \lim_{n \to \infty} \int_X f_n \, d \mu = \int_X f \, d \mu $$
Proof. Assume for the moment that $f_n, f \ge 0$ for all $n \ge 1$. Then by Fatou's lemma, $$ \liminf_{n \to \infty} \int_X f_n \, d \mu \ge \int_X \liminf_{n \to \infty} f_n \, d \mu = \int_X \lim_{n \to \infty} f_n \, d \mu = \int_X f \, d \mu $$ By reverse Fatou's lemma, $$ \limsup_{n \to \infty} \int_X f_n \, d \mu \le \int_X \limsup_{n \to \infty} f_n \, d \mu = \int_X \lim_{n \to \infty} f_n \, d \mu = \int_X f \, d \mu $$ Combining these two, $$ \limsup_{n \to \infty} \int_X f_n \, d \mu \le \int_X f \, d \mu \le \liminf_{n \to \infty} \int_X f_n \, d \mu $$ which can only imply $$ \lim_{n \to \infty} \int_X f_n \, d \mu = \int_X f \, d \mu $$ The case where $\{ f_n \}_{n \ge 0}$ are not necessarily non-negative is proved by splitting into positive and negative parts. $\blacksquare$