Monotone Convergence Theorem (MCT)

We need to show that $\int f = \lim_{n \to \infty} \int f_n$.

We know that $\int f \ge \int f_n$ for all $n \in \mathbb{N}$, so $\int f \ge \lim_{n \to \infty} \int f_n$.

It remains to show that $\int f \le \lim_{n \to \infty} \int f_n$. Since $\int_f = \sup \{ \int \phi: \phi \textrm{non-negative simple function}, 0 \le \phi \le f \}$, it is sufficient to show that all non-negative simple functions $0 \le \phi \le f$ have $\int \phi \le \lim_{n \to \infty} \int f_n$. So let $\phi = \sum_{i=1}^k \beta_i \chi_{E_i}$ be such a function.

Let $\alpha \in (0,1)$, and for $n \in \mathbb{N}$, let $B_n = \{x \in \mathbb{R}: f_n(x) \ge \alpha \phi(X) \}$ (this is a measurable set). Therefore for all $n \in \mathbb{N}$, $\alpha \phi \chi_{B_n} \le f_n \chi_{B_n} \le f_n$, so $\alpha \int_{B_n} \phi \le \int_{B_n} f_n \le \int_\mathbb{R} f_n$ and by taking the limit $\alpha \to 1$, we get $\int_{B_n} \phi \le \int_\mathbb{R} f_n$

Since $(f_n)$ is an increasing sequence, $B_n$ are increasing.

For $x \in \mathbb{R}$, either $\phi(x)=0$ (so $0 \in B_n \forall n$) or $lim_{n \to \infty} f_n(x) = f(x) \ge \phi(x) > \alpha\phi(x)$; in either case, $x \in B_n$ for some $n \in \mathbb{N}$. Hence $\bigcup_{n=1}^{\infty} B_n = \mathbb{R}$.

Hence $\int_{B_n} \phi = \sum_{i=1}^{k} \beta_i m(E_i \cap B_n) \longrightarrow \sum_{i=1}^{k} \beta_i m(E_i) = \int_\mathbb{R} \phi$ as $n \to \infty$. Since we have $\int_{B_n} \phi \le \int_\mathbb{R} f_n$, by taking limits on both sides we get $\int_\mathbb{R} \phi \le \lim_{n \to \infty} \mathbb{R} f_n$ as required.

$\square$