Difference between revisions of "4.14 MA598R in progress" - Rhea

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-Suppose <math>f\in L^{1}(X,d\mu)</math>. Prove that for each <math>\varepsilon</math> there exists a <math>\delta >0</math> such that <math>\int_{E}|f|d\mu\leq\varepsilon</math> whenever <math>\mu (E) <\delta</math>
-Proof:
-Case 1: <math>0\leq f \leq M</math> (<math>f</math> is bounded)
-Given <math>\varepsilon >0</math>, choose <math>\delta < \frac{\varepsilon}{M}</math>
-<math>\therefore \int_{E}fd\mu \leq M\int_{E}d\mu = M\mu(E) < M\delta < \varepsilon</math>
-Case 2: <math>f\geq 0</math>
-Set <math>f(x)\left\{\begin{array}{lr} f(x) & \text{if } f(x)\leq n\\ n & \text{otherwise}\end{array}\right.</math>, <math>n\in \mathbb{N}</math>
-Then each <math>f_{n}</math> is bounded and converges to <math>f</math> pointwise.
-By the Monotone Convergence Theorem, there exists an <math>N\in \mathbb{N}</math> such that <math>\int_{X}f_{N} > \int_{X}f - \varepsilon/2</math>
-<math>\Rightarrow \int_{X}f-f_{N} < \varepsilon/2</math>
-Choose <math>\delta < \frac{\varepsilon}{2N}</math>. Now, if <math>\mu(E)<\delta</math>, then
-<math>\int_{E}fd\mu = \int_{E}(f-f_{N})d\mu + \int_{E}f_{N}d\mu</math>
-<math>< \int_{X}(f-f_{N})d\mu + N\mu(E) < \frac{\varepsilon}{2} + \frac{N\varepsilon}{2N} = \varepsilon</math>

Latest revision as of 12:03, 6 July 2009