The Monotone Convergence Theorem

Since $\int f_n\,d\mu$ is a nondecreasing sequence of extended reals (as $f_n$ increases), the limit $L := \lim_{n\to\infty}\int f_n\,d\mu$ exists in $[0,\infty]$. Since $f_n \le f$ for every $n$, monotonicity of the integral gives $\int f_n\,d\mu \le \int f\,d\mu$ for every $n$, hence $L \le \int f\,d\mu$.

For the reverse inequality, fix any simple measurable function $s$ with $0 \le s \le f$ and fix a constant $c \in (0,1)$. Define $E_n = \{x : f_n(x) \ge c\,s(x)\}$. Because $f_n$ increases to $f \ge s$ pointwise, the sets $E_n$ increase and their union is (almost) all of $X$: for a.e. $x$ with $s(x)=0$ this is automatic, and for a.e. $x$ with $s(x)>0$, eventually $f_n(x) > c\,s(x)$ since $f_n(x)\to f(x) \ge s(x) > c\,s(x)$.

Writing $s = \sum_k c_k \mathbf{1}_{E_k}$, we get $\int f_n\,d\mu \ge \int_{E_n} f_n\,d\mu \ge c\int_{E_n} s\,d\mu = c\sum_k c_k\,\mu(E_k \cap E_n)$. By countable additivity (continuity of measure from below, since $E_n$ increases to $X$), $\mu(E_k \cap E_n) \to \mu(E_k)$ as $n\to\infty$, so letting $n\to\infty$ gives $L \ge c\sum_k c_k\,\mu(E_k) = c\int s\,d\mu$.

Letting $c \to 1$ gives $L \ge \int s\,d\mu$ for every simple $s \le f$, and taking the supremum over all such $s$ (the very definition of $\int f\,d\mu$) gives $L \ge \int f\,d\mu$. Combined with $L \le \int f\,d\mu$ from above, we conclude $L = \int f\,d\mu$.