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---
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 %}
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