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