--- title: Convergence Theorems parent: Lebesgue Integral grand_parent: Measure and Integration nav_order: 2 --- # {{ page.title }} For all statements on this page, assume that $(X,\mathcal{A},\mu)$ is a measure space. {% theorem * Monotone Convergence Theorem %} For each $n \in \NN$ let $f_n : X \to \overline{\RR}$ be a measurable function. If $0 \le f_n \le f_{n+1}$ almost everywhere, then $$ \int_X \lim_{n \to \infty} f_n \, d\mu = \lim_{n \to \infty} \int_X f_n \, d\mu. $$ {% endtheorem %} Note that the pointwise limit $\lim_{n \to \infty} f_n$ always exists and is measurable by this proposition. {% lemma * Fatou’s Lemma %} For each $n \in \NN$ let $f_n : X \to \overline{\RR}$ be a nonnegative measurable function. Then $$ \int_X \liminf_{n \to \infty} f_n \, d\mu \le \liminf_{n \to \infty} \int_X f_n \, d\mu. $$ {% endlemma %} In the following proof we omit $X$ and $d\mu$ for visual clarity. {% proof %} By definition, we have $\liminf_{n \to \infty} f_n = \lim_{n \to \infty} g_n$, where $g_n = \inf_{k \ge n} f_k$. Now $(g_n)$ is a monotonic sequence of nonnegative measurable functions. By the [Monotone Convergence Theorem](#monotone-convergence-theorem) $$ \int \liminf_{n \to \infty} f_n = \lim_{n \to \infty} \int g_n. $$ For all $k \ge n$ one has $g_n \le f_k$, hence $\int g_n \le \int f_k$ by the monotonicity of the integral. This implies $$ \int g_n \le \inf_{k \ge n} \int f_k $$ for all $n \in \NN$. In the limit $n \to \infty$ we obtain $$ \lim_{n \to \infty} \int g_n \le \liminf_{n \to \infty} \int f_n $$ thereby completing the proof. {% endproof %} {% theorem * Dominated Convergence Theorem %} Let $(X,\mathcal{A},\mu)$ be a measure space. For each $n \in \NN$ let $f_n : X \to \overline{\RR}$ (or $\CC$) be a measurable function. Suppose that the pointwise limit $f = \lim_{n \to \infty} f_n$ exists almost everywhere. Suppose further that there exists an integrable function $g : X \to \overline{\RR}$ such that $\abs{f_n} \le g$ almost everywhere for all $n \in \NN$. Then the functions $f_n$ and $f$ are all integrable, and $$ \lim_{n \to \infty} \int_X f_n \, d\mu = \int_X f \, d\mu. $$ {% endtheorem %} {% proof %} TODO {% endproof %}