Let $(X,\mathcal{F},\mu)$ be a measure space. Let $(f_{n})_{n\in\mathbb{N}}\subset\mathscr{L}^{1}(X,\mathcal{F},\mu)$ and $f_{n}\to f$ pointwise for some $f:X\to \mathbb{R}$. Assume $\exists g\in\mathscr{L}^{1}(X,\mathcal{F},\mu)$ such that $$|f_{n}|\le g, \ \forall n\in\mathbb{N}$$then we have that $f\in\mathscr{L}^{1}(X,\mathcal{F},\mu)$ and $$\lim_{ n \to \infty } \int\limits _{X}f_{n} \, d\mu =\int\limits _{X}f \, d\mu$$

Suppose $(f_{n})_{n\in\mathbb{N}}$ is a sequence of measurable functions such that $f_{n}\to f$ a.e. and let $g\in\mathscr{L}^{1}(X,\mathscr{M},\mu)$ s.t. $$|f_{n}|\le g \text{ a.e.},\forall n\in\mathbb{N}$$ then $f\in\mathscr{L}^{1}(X,\mathscr{M},\mu)$ and $$\int\limits \lim_{ n \to \infty } f_{n} \, d\mu =\lim_{ n \to \infty } \int\limits f_{n} \, d\mu\quad\text{a.e.}$$

Let $(\Omega,\mathcal{F},\mu)$ be a measure space and $(f_{n})_{n\in\mathbb{N}}\subset \mathscr{L}^{1}(\Omega,\mathcal{F},\mu)$. If $\mu(\Omega)<\infty$ and $f_{n}$ are uniformly bounded i.e. $$\exists M\in\mathbb{N}:|f_{n}|<M,\,\forall n\in\mathbb{N}$$then $f_{n}\to f$ $\mu$-a.e. implies $$\lim_{ n \to \infty } \int\limits f_{n} \, d\mu=\int\limits f \, d\mu$$