| Line 12: | Line 12: | ||
Case 2: <math>f\geq 0</math> | Case 2: <math>f\geq 0</math> | ||
| − | Set <math> | + | Set <math>f_n(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. | ||
Latest revision as of 09:39, 24 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_n(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 $
$ \Box $
(Proof by Robert)
