(New page: MA_598R_pweigel_Summer_2009_Lecture_4 4.14) <math>\left(X,\mathcal{F},\mu\right)</math> be a finite measure space, and <math>f</math> a measurable extended real-valued function define...) |
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Hence <math> \sum_{k=1}^{\infty}\mu\{|f|\ge k\} \le \int{|f|d\mu} < \infty</math> | Hence <math> \sum_{k=1}^{\infty}\mu\{|f|\ge k\} \le \int{|f|d\mu} < \infty</math> | ||
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Latest revision as of 17:11, 5 July 2009
MA_598R_pweigel_Summer_2009_Lecture_4
4.14) $ \left(X,\mathcal{F},\mu\right) $ be a finite measure space, and $ f $ a measurable extended real-valued function defined on $ X $. Show that $ f\in L(\mu) $ if and only if
$ \sum_{k=1}^{\infty}{\mu\{\left|f\right|\ge k\}} < \infty $
Solution:
"$ \Leftarrow $"
$ \int{|f|d\mu} = $
$ = \int{\lim_{n\rightarrow\infty}\sum_{k=0}^{n}|f|\mathcal{X}_{\{k\le |f| \le k+1\}}d\mu } $
$ = \lim_{n\rightarrow\infty}\int{\sum_{k=0}^{n}|f|\mathcal{X}_{\{k\le |f| \le k+1\}}d\mu } $ by the Monotone Convergence Theorem
$ = \lim_{n\rightarrow\infty}\sum_{k=0}^{n}\int{|f|\mathcal{X}_{\{k\le |f| \le k+1\}}d\mu } $ because the sum is finite
$ = \sum_{k=0}^{\infty}\int{|f|\mathcal{X}_{\{k\le |f| \le k+1\}}d\mu } $
$ \le \sum_{k=0}^{\infty}\int{(k+1)\mathcal{X}_{\{k\le |f| \le k+1\}}d\mu } $
$ = \sum_{k=0}^{\infty}\left( \int{(k+1)\mathcal{X}_{\{k\le |f| \le k+1\}}d\mu } + \int{(k+1)\mathcal{X}_{\{k+1\le |f|\}}d\mu} - \int{(k+1)\mathcal{X}_{\{k+1\le |f|\}}d\mu} \right) $
$ = \sum_{k=0}^{\infty}\left( \int{(k+1)\mathcal{X}_{\{k\le |f|\}}d\mu } - \int{(k+1)\mathcal{X}_{\{k+1\le |f|\}}d\mu} \right) $
$ = \sum_{k=0}^{\infty}\int{(k+1)\mathcal{X}_{\{k\le |f|\}}d\mu } - \sum_{k=0}^{\infty}\int{(k+1)\mathcal{X}_{\{k+1\le |f|\}}d\mu} $
$ = \sum_{k=0}^{\infty}\int{(k+1)\mathcal{X}_{\{k\le |f|\}}d\mu } - \sum_{k=1}^{\infty}\int{(k)\mathcal{X}_{\{k\le |f|\}}d\mu} $
$ = \int{\mathcal{X}_{\{0\le |f|\}}d\mu } + \sum_{k=1}^{\infty}\int{\mathcal{X}_{\{k\le |f|\}}d\mu} $
$ = \mu(X) + \sum_{k=1}^{\infty}\mu\{|f|\ge k\} $
$ < \infty $
"$ \Rightarrow $"
$ \int{|f|d\mu} = $
$ = \int{\lim_{n\rightarrow\infty}\sum_{k=0}^{n}|f|\mathcal{X}_{\{k\le |f| \le k+1\}}d\mu } $
$ = \lim_{n\rightarrow\infty}\int{\sum_{k=0}^{n}|f|\mathcal{X}_{\{k\le |f| \le k+1\}}d\mu } $ by the Monotone Convergence Theorem
$ = \lim_{n\rightarrow\infty}\sum_{k=0}^{n}\int{|f|\mathcal{X}_{\{k\le |f| \le k+1\}}d\mu } $ because the sum is finite
$ = \sum_{k=0}^{\infty}\int{|f|\mathcal{X}_{\{k\le |f| \le k+1\}}d\mu } $
$ \ge \sum_{k=0}^{\infty}\int{k\mathcal{X}_{\{k\le |f| \le k+1\}}d\mu } $
$ = \sum_{k=0}^{\infty}\left( \int{k\mathcal{X}_{\{k\le |f| \le k+1\}}d\mu } + \int{k\mathcal{X}_{\{k+1\le |f|\}}d\mu} - \int{k\mathcal{X}_{\{k+1\le |f|\}}d\mu} \right) $
$ = \sum_{k=0}^{\infty}\left( \int{k\mathcal{X}_{\{k\le |f|\}}d\mu } - \int{k\mathcal{X}_{\{k+1\le |f|\}}d\mu} \right) $
$ = \sum_{k=0}^{\infty}\int{k\mathcal{X}_{\{k\le |f|\}}d\mu } - \sum_{k=0}^{\infty}\int{k\mathcal{X}_{\{k+1\le |f|\}}d\mu} $
$ = \sum_{k=0}^{\infty}\int{k\mathcal{X}_{\{k\le |f|\}}d\mu } - \sum_{k=1}^{\infty}\int{(k-1)\mathcal{X}_{\{k\le |f|\}}d\mu} $
$ = \int{0\mathcal{X}_{\{0\le |f|\}}d\mu } + \sum_{k=1}^{\infty}\int{\mathcal{X}_{\{k\le |f|\}}d\mu} $
$ = \sum_{k=1}^{\infty}\mu\{|f|\ge k\} $
Hence $ \sum_{k=1}^{\infty}\mu\{|f|\ge k\} \le \int{|f|d\mu} < \infty $
-Ben Bartle
