It is enough to prove the following

First of all by the monotonicity of the sequence we have $$ f(x)= \sup_k f_k(x) $$ hence we know that $f$ is measurable. Moreover being $f_k\le f$ for all $k$ , by the monotonicity of the integral, we immediately get $$ \sup_k \int_X f_k\, d\mu \le \int_X f(x)\, d\mu. $$

So take any simple measurable function $s$ such that $0\le s \le f$ . Given also $\alpha<1$ define $$ E_k = \{ x \in X \colon f_k(x) \ge \alpha s(x)\}. $$ The sequence $E_k$ is an increasing sequence of measurable sets. Moreover the union of all $E_k$ is the whole space $X$ since $\lim_{k\to\infty} f_k(x)=f(x) \ge s(x) > \alpha s(x)$ . Moreover it holds $$ \int_X f_k\, d\mu \ge \int_{E_k} f_k\, d\mu \ge \alpha\int_{E_k} s\, d\mu. $$ Since $s$ is a simple measurable function it is easy to check that $E\mapsto \int_E s\, d\mu$ is a measure and hence $$ \sup_k \int_X f_k\, d\mu \ge \alpha \int_X s\, d\mu. $$ But this last inequality holds for every $\alpha<1$ and for all simple measurable functions $s$ with $s\le f$ . Hence by the definition of Lebesgue integral $$ \sup_k \int_X f_k \, d\mu \ge \int_X f\, d\mu $$ which completes the proof.