(New page: 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 <...) |
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Case 2: <math>f\geq 0</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. | + | Set <math>f(x)\left\{\begin{array}{lr} f(x) & \text{if } f(x)\leq n\\ n & \text{otherwise}\end{array}\right.</math> |
Then each <math>f_{n}</math> is bounded and converges to <math>f</math> pointwise. | Then each <math>f_{n}</math> is bounded and converges to <math>f</math> pointwise. | ||
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Choose <math>\delta < \frac{\varepsilon}{2N}</math>. Now, if <math>\mu(E)<\delta</math>, then | 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_{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> | <math>< \int_{X}(f-f_{N})d\mu + N\mu(E) < \frac{\varepsilon}{2} + \frac{N\varepsilon}{2N} = \varepsilon</math> | ||
Revision as of 16:58, 5 July 2009
Suppose $ f\in L^{1}(X,d\mu) $. Prove that for each $ \varepsilon $ there exists a $ \delta >0 $ such that $ \int_{E}|f|d\mu\leq\varepsilon $ whenever $ \mu (E) <\delta $
Proof: Case 1: $ 0\leq f \leq M $ ($ f $ is bounded)
Given $ \varepsilon >0 $, choose $ \delta < \frac{\varepsilon}{M} $
$ \therefore \int_{E}fd\mu \leq M\int_{E}d\mu = M\mu(E) < M\delta < \varepsilon $
Case 2: $ f\geq 0 $
Set $ f(x)\left\{\begin{array}{lr} f(x) & \text{if } f(x)\leq n\\ n & \text{otherwise}\end{array}\right. $
Then each $ f_{n} $ is bounded and converges to $ f $ pointwise.
By the Monotone Convergence Theorem, there exists an $ N\in \mathbb{N} $ such that $ \int_{X}f_{N} > \int_{X}f - \varepsilon/2 $
$ \Rightarrow \int_{X}f-f_{N} < \varepsilon/2 $
Choose $ \delta < \frac{\varepsilon}{2N} $. Now, if $ \mu(E)<\delta $, then
$ \int_{E}fd\mu = \int_{E}(f-f_{N})d\mu + \int_{E}f_{N}d\mu $
$ < \int_{X}(f-f_{N})d\mu + N\mu(E) < \frac{\varepsilon}{2} + \frac{N\varepsilon}{2N} = \varepsilon $
