Let $(\Omega, \mathcal{A}, \mu)$ be a measure space and $f_1, f_2, f_3,\dots:\Omega \to \overline{\mathbb{R}}$ be measurable functions such that $$f_n \to f\quad f_n\geq f_{n+1}\geq 0$$ Show that if $f_1$ is integrable then $$\lim_{n\to\infty}\int_\Omega f_n \;d\mu = \int_\Omega f \;d\mu$$ Show also that if $f_1$ is not integrable, this equality need not be true.
The result follows immediately from the Dominated Convergence Theorem, if we use $f_1$ as the dominating function.
The counterexample is a bit harder. Consider $\Omega = \overline{\mathbb{R}}$ such that $f:\overline{\mathbb{R}}\to\overline{\mathbb{R}}$ and we have a measure space equipped with the Borel sigma-algebra and Lebesgue measure.
Let $$f_n(x) = \frac{x}{n}$$
Then $$f_n(x)\to f(x)=0\quad as \quad n\to\infty$$
But
$$\int_{\overline{\mathbb{R}}}f_1 \;dm = \int_{\overline{\mathbb{R}}}x \;dm = \infty\neq 0 = \int_{\overline{\mathbb{R}}}f \;dm$$
Does this solution make sense to you?