Dominated Convergence Theorem: Difference between revisions

Revision as of 08:38, 18 December 2020

In measure theory, the dominated convergence theorem is a cornerstone of Lebesgue integration. It can be viewed as a culmination of all efforts, and is a general statement about the interplay between limits and integrals.

Theorem Statement

Consider the measure space $(X,{\mathcal {M}},\lambda )$ . Suppose $\{f_{n}\}$ is a sequence in $L^{1}(\lambda )$ such that

$f_{n}\to f$ a.e
there exists $g\in L^{1}(\lambda )$ such that $|f_{n}|\leq g$ a.e. for all $n\in \mathbb {N}$

Then $f\in L^{1}(\lambda )$ and $\int f=\lim _{n\to \infty }\int f_{n}$ .

Proof of Theorem

$f$ is a measurable function in the sense that it is a.e. equal to a measurable function, since it is the limit of $f_{n}$ except on a null set. Also $f\leq g$ a.e., so $f\in L^{1}(\lambda )$ .

@@ Line 1: / Line 1: @@
 In measure theory, the dominated convergence theorem is a cornerstone of Lebesgue integration. It can be viewed as a culmination of all efforts, and is a general statement about the interplay between limits and integrals.
-==Statement and proof of Theorem==
+==Theorem Statement==
-====Statement====
+Consider the measure space <math> (X,\mathcal{M},\lambda) </math>. Suppose <math>\{f_n\}</math> is a sequence in <math>L^1(\lambda)</math> such that
-Suppose <math>\{f_n\}</math> is a sequence in <math> \mathbf{L^1}</math> such that <math>f_n(x) \to f(x)</math> a.e. x, as n goes to infinity. If <math>|f_n(x)|\leq g(x) </math> and <math>g \in \mathbf{L^1}</math>, where g is integrable, then <math>f \in \mathbf{L^1}</math> and
+# <math>f_n \to f</math> a.e
-:<math>\int f_n \to \int f</math>
+# there exists <math>g \in L^1(\lambda)</math> such that <math>|f_n| \leq g</math> a.e. for all <math>n \in \mathbb{N}</math>
+Then <math>f \in L^1(\lambda)</math> and <math>\int f = \lim_{n \to \infty} \int f_n</math>.
+==Proof of Theorem==
+<math>f</math> is a measurable function in the sense that it is a.e. equal to a measurable function, since it is the limit of <math>f_n</math> except on a null set. Also <math>f \leq g</math> a.e., so <math>f \in L^1(\lambda)</math>.

Dominated Convergence Theorem: Difference between revisions

Revision as of 08:38, 18 December 2020

Theorem Statement

Proof of Theorem

Navigation menu

Search