Dominated Convergence Theorem: Difference between revisions
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Suppose <math>f</math> is integrable in <math>R</math>. Then for every <math>\epsilon>0</math>, there exist a set of finite measure B such that | Suppose <math>f</math> is integrable in <math>R</math>. Then for every <math>\epsilon>0</math>, there exist a set of finite measure B such that | ||
:<math>\int_{B^c} |f|<\epsilon</math> | :<math>\int_{B^c} |f|<\epsilon</math> | ||
*Proof of lemma: By replacing <math>f</math> with <math>|f|</math> we may assume without loss of generality that <math>f\geq 0</math> | *Proof of lemma: By replacing <math>f</math> with <math>|f|</math> we may assume without loss of generality that <math>f\geq 0</math>.Let <math>B_N</math> denote the ball of radius N centered at origin, and note that if <math>f_N(x)=f(x)\mathbf{1}_{B_N}(x)</math>, then <math>f_N\geq0 is measurable</math>, <math>f_N(x)\leq f_{N+1}(x)</math>, and <math>f_N(x)\to f(x)</math>. By the monotone convergenc theorem, we must have | ||
Revision as of 04:05, 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.
Statement and proof of Theorem
Statement
Suppose is a sequence of measurable functions such that a.e. x, as n goes to infinity. If , where g is integrable, then
and consequently
Lemma
Suppose is integrable in . Then for every , there exist a set of finite measure B such that
- Proof of lemma: By replacing with we may assume without loss of generality that .Let denote the ball of radius N centered at origin, and note that if , then , , and . By the monotone convergenc theorem, we must have