Monotone Convergence Theorem: Difference between revisions

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(Completing the proof of the Monotone Convergence theorem)
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Now, for <math> n \in \mathbb{N} </math> and some <math> t \in (0,1) </math>, define the increasing sequence of sets <math> E_n = \{x \in X \mid f_n - t\phi > 0\}. Since <math> f_n\leq \lim_{n\rightarrow +\infty} f_n </math>, we have <math> E_n \subseteq E_{n+1} </math> for <math> n \in \mathbb{N} </math>. Moreover, we have <math> X = \cup_{n=1}^{\infty} E_n </math> since <math> g = \sup_{n\in \mathbb{N}} f_n </math>.
Now, for <math> n \in \mathbb{N} </math> and some <math> t \in (0,1) </math>, define the increasing sequence of sets <math> E_n = \{x \in X \mid f_n - t\phi > 0\}. Since <math> f_n\leq \lim_{n\rightarrow +\infty} f_n </math>, we have <math> E_n \subseteq E_{n+1} </math> for <math> n \in \mathbb{N} </math>. Moreover, we have <math> X = \cup_{n=1}^{\infty} E_n </math> since <math> g = \sup_{n\in \mathbb{N}} f_n </math>.


Recall that the set function <math> \nu(A) = \int_{A} \phi \, d\mu</math> for <math> A \in \mathcal{M} </math> is a measure on <math> (X,\mathcal{M},\mu) </math>. We now see that <math> \lim_{n\rightarrow +\infty} \int f_n \geq  \lim_{n\rightarrow +\infty} \int_{E_n} f_n \geq t \cdot \lim_{n\rightarrow +\infty} \int_{E_n} \phi_n  = t \cdot \int_{X} \phi </math>, where we have used the fact that <math> \nu(X) = \nu(\cup_{n=1}^{\infty} E_n) = \lim_{n\rightarrow +\infty} \nu(E_n) </math> for the last equality. Taking the supremum over both simple functions such that <math> 0 \leq \phi \leq g </math> and <math> t \in (0,1) </math> yields the required inequality.
Recall that the set function <math> \nu(A) = \int_{A} \phi \, d\mu</math> for <math> A \in \mathcal{M} </math> is a measure on <math> (X,\mathcal{M},\mu) </math>. We now see that <math> \lim_{n\rightarrow +\infty} \int f_n \geq  \lim_{n\rightarrow +\infty} \int_{E_n} f_n \geq t \cdot \lim_{n\rightarrow +\infty} \int_{E_n} \phi_n  = t \cdot \int_{X} \phi </math>, where we have used the fact that <math> \nu(X) = \nu(\cup_{n=1}^{\infty} E_n) = \lim_{n\rightarrow +\infty} \nu(E_n) </math> for the last equality. Taking the supremum over simple functions such that <math> 0 \leq \phi \leq g </math> and over <math> t \in (0,1) </math> yields the required inequality.
 


==References==
==References==

Revision as of 20:16, 8 December 2020

Theorem

Consider the measure space and suppose is a sequence of non-negative measurable functions, such that for all . Furthermore, . Then

[1]

Proof

First we prove that .

Since for all , we have and further .

Sending on LHS gives us the result.

Then we only need to prove that . In this regard, denote (which is measurable as the limit of measurable functions) and consider a simple function so that .

Now, for and some , define the increasing sequence of sets , we have for . Moreover, we have since .

Recall that the set function for is a measure on . We now see that , where we have used the fact that for the last equality. Taking the supremum over simple functions such that and over yields the required inequality.

References

  1. Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, second edition, §2.2