From 28407333ffceca9b99fae721c30e8ae146a863da Mon Sep 17 00:00:00 2001
From: Justin Gassner <justin.gassner@mailbox.org>
Date: Wed, 14 Feb 2024 07:24:38 +0100
Subject: Update

---
 .../lebesgue-integral/convergence-theorems.md      | 77 ++++++++++++++++++++++
 1 file changed, 77 insertions(+)
 create mode 100644 pages/measure-and-integration/lebesgue-integral/convergence-theorems.md

(limited to 'pages/measure-and-integration/lebesgue-integral/convergence-theorems.md')

diff --git a/pages/measure-and-integration/lebesgue-integral/convergence-theorems.md b/pages/measure-and-integration/lebesgue-integral/convergence-theorems.md
new file mode 100644
index 0000000..67f0996
--- /dev/null
+++ b/pages/measure-and-integration/lebesgue-integral/convergence-theorems.md
@@ -0,0 +1,77 @@
+---
+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 %}
-- 
cgit v1.2.3-54-g00ecf