Difference between revisions of "Team 1: The Jets most magnificent solutions Old Kiwi" - Rhea

Latest revision as of 17:17, 11 July 2008

@@ Line 1: / Line 1: @@
-.<math>(\Rightarrow)</math> First we apply Tchebyshev to <math>E_n</math> and find that
+[[Jets7.1 _Old Kiwi| Solution to 7.1]]
-<math> (n-1)\left|\{x \in X \mid |f(x)| \geq n-1 \}\right| \leq \int_{E_n}|f|</math>
+[[Jets7.2 _Old Kiwi| Solution to 7.2]]
-or rather
+[[Jets7.3 _Old Kiwi| Solution to 7.3]]
-<math>(n-1) m(E_n) \leq \int_{E_n}|f|</math>
+[[Jets7.4 _Old Kiwi| Solution to 7.4]]
-Since we have that <math>m(E_n)</math> is finite we can move it to the other side of the inequality.
+[[Jets7.5 _Old Kiwi| Solution to 7.5]]
-<math>nm(E_n) \leq \int_{E_n}|f| + m(E_n)</math>
+[[Jets7.6 _Old Kiwi| Solution to 7.6]]
-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.
+[[Jets7.7 _Old Kiwi| Solution to 7.7]]
-<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>
+[[Jets7.8 _Old Kiwi| Solution to 7.8]]
-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>.