| Line 4: | Line 4: | ||
Let <math>(X,A,m)</math> be a finite measure space. If f is <math> A </math> measurable, let | 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>E_n = \{x \in X : n-1 \leq |f(x)| < n \}</math>. Then |
<math>f \in L^1(X)</math> if and only if <math>\sum_{n=1}^{\infty}nm(X) < \infty.</math> | <math>f \in L^1(X)</math> if and only if <math>\sum_{n=1}^{\infty}nm(X) < \infty.</math> | ||
Revision as of 22:47, 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) < \infty. $
.$ (\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 $.
