(New page: Suppose we know the conclusion of problem 8, Problem 8 Let <math>(X,A,mu)</math> be a finite measure space. If f is <math> A </math>measurable, let En = fx 2 X : n 1 � jf(x)j < ng:...) |
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Problem 8 | Problem 8 | ||
| − | Let <math>(X,A, | + | Let <math>(X,A,m)</math> be a finite measure space. If f is <math> A </math>measurable, let |
| − | + | <math>E_n = {x \in X : n-1 \leq |f(x)| < n}</math>. Then | |
| − | + | <math>f \in L^1(X) if and only if sum_{n=1}^{\infty}nm(X) < \inf.</math> | |
| + | |||
| + | |||
| + | .<math>(\Rightarrow)</math> First we apply Tchebyshev to <math>E_n</math> and find that | ||
| + | |||
| + | <math> (n-1)\left|\{x \in X \mid |f(x)| \geq n-1 \}\right| \leq \int_{E_n}|f|</math> | ||
| + | |||
| + | or rather | ||
| + | |||
| + | <math>(n-1) m(E_n) \leq \int_{E_n}|f|</math> | ||
| + | |||
| + | Since we have that <math>m(E_n)</math> is finite we can move it to the other side of the inequality. | ||
| + | |||
| + | <math>nm(E_n) \leq \int_{E_n}|f| + m(E_n)</math> | ||
| + | |||
| + | Since this is true for all <math>n</math> we take sums on both sides and note that the <math>E_n</math> are disjoint. | ||
| + | |||
| + | <math>\sum_{n=1}^{\infty}nm(E_n) \leq \sum_{n=1}^{\infty}\int_{E_n}|f| + \sum_{n=1}^{\infty}m(E_n)</math> | ||
| + | |||
| + | or | ||
| + | |||
| + | <math>\sum_{n=1}^{\infty}nm(E_n)\leq \int_{X}|f| + m(X)</math> | ||
| + | |||
| + | And we are in a finite measure space so <math>m(X) < \infty</math> and since <math>f \in L^1</math> we have <math>\int_{X}|f| < \infty</math>. | ||
| + | |||
| + | Thus we have that <math>\sum_{n=1}^{\infty}nm(E_n) < \infty</math>. | ||
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| + | <math>(\Leftarrow)</math> Since <math>|f|< n</math> in each <math>E_n</math> we have that | ||
| + | |||
| + | <math>\int_X|f| = \sum_{n=1}^{\infty}\left(\int_{E_n}|f|\right) \leq \sum_{n=1}^{\infty}\left( n m(E_n)\right)< \infty</math> | ||
| + | |||
| + | In other words, <math> f \in L^1</math>. | ||
Revision as of 22:44, 10 July 2008
Suppose we know the conclusion of problem 8,
Problem 8 Let $ (X,A,m) $ be a finite measure space. If f is $ A $measurable, let $ E_n = {x \in X : n-1 \leq |f(x)| < n} $. Then $ f \in L^1(X) if and only if sum_{n=1}^{\infty}nm(X) < \inf. $
.$ (\Rightarrow) $ First we apply Tchebyshev to $ E_n $ and find that
$ (n-1)\left|\{x \in X \mid |f(x)| \geq n-1 \}\right| \leq \int_{E_n}|f| $
or rather
$ (n-1) m(E_n) \leq \int_{E_n}|f| $
Since we have that $ m(E_n) $ is finite we can move it to the other side of the inequality.
$ nm(E_n) \leq \int_{E_n}|f| + m(E_n) $
Since this is true for all $ n $ we take sums on both sides and note that the $ E_n $ are disjoint.
$ \sum_{n=1}^{\infty}nm(E_n) \leq \sum_{n=1}^{\infty}\int_{E_n}|f| + \sum_{n=1}^{\infty}m(E_n) $
or
$ \sum_{n=1}^{\infty}nm(E_n)\leq \int_{X}|f| + m(X) $
And we are in a finite measure space so $ m(X) < \infty $ and since $ f \in L^1 $ we have $ \int_{X}|f| < \infty $.
Thus we have that $ \sum_{n=1}^{\infty}nm(E_n) < \infty $.
$ (\Leftarrow) $ Since $ |f|< n $ in each $ E_n $ we have that
$ \int_X|f| = \sum_{n=1}^{\infty}\left(\int_{E_n}|f|\right) \leq \sum_{n=1}^{\infty}\left( n m(E_n)\right)< \infty $
In other words, $ f \in L^1 $.
