Difference between revisions of "Jets7.3 Old Kiwi" - Rhea

Revision as of 22:55, 10 July 2008

Suppose we know the conclusion of problem 8,

Problem 8 Let $$ X $$ be a finite measure space. If $$ f $$ is measurable, let

$E_n = \{x \in X : n-1 \leq |f(x)| < n \}$ . Then

$f \in L^1$ if and only if $\sum_{n=1}^{\infty}nm(E_n) < \infty.$

First, if $$ m(X)=/infty $$ , it's done. Hence let's suppose that $m(X)<\infty$

@@ Line 8: / Line 8: @@
 <math>f \in L^1</math> if and only if <math>\sum_{n=1}^{\infty}nm(E_n) < \infty.</math>
+First, if <math> m(X)=/infty </math>, it's done. Hence let's suppose that <math> m(X)<\infty </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>.
-<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>.