> [!theorem] > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]], and $L^1 = L^1(\mu)$ be the space of [[Integrable Function|integrable functions]]. Let $\seq{f_n} \subset L^1$ be a [[Sequence|sequence]] of integrable functions such that > - $f_n \to f$ [[Pointwise Convergence|pointwise]] [[Almost Everywhere|a.e.]] > - There exists a non-negative $g \in L^1$ such that $|f_n| \le g$ [[Almost Everywhere|a.e.]] for all $n$. > > Then $f \in L^1$ and $\int f = \limv{n}\int f_n$. > > *Proof*. First convert $f$ into a [[Measurable Function|measurable function]] by redefining it on a [[Null Set|null set]]. Since $|f| \le g$ a.e., $f \in L^1$. Suppose that $f$ is real-valued, then $g + f_n \ge 0$ a.e., and $g - f_n \ge 0$ a.e. By [[Fatou's Lemma]], > $ > \begin{align*} > \int g + f \le \liminf\int (g + f_n) &= \int g + \liminf \int f_n \\ > \int g - f \le \liminf \int (g - f_n) &= \int g - \limsup \int f_n > \end{align*} > $ > and > $ > \int f \ge \limsup \int f_n \ge \int f \ge \liminf \int f_n \ge \int f > $ > Suppose that $f$ is complex-valued, then > $ > \int\re{f} = \limv{n}\int \re{f_n} \quad > \int\im{f} = \limv{n}\int \im{f_n} > $ > and we get > $ > \begin{align*} > \int f &= \int \re{f} + i\int \im{f} \\ > &= \limv{n}\int\re{f_n} + i\limv{n}\int\im{f_n} \\ > &= \int\limv{n}f_n > \end{align*} > $