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 February 24th, 2014, 04:39 AM #1 Member   Joined: Aug 2012 Posts: 32 Thanks: 0 Lebesgue integration question Lebesgue integral question: Consider measure space $(X, \epsilon, \lambda)$ with $f$ a summable function. I want to show that if $\int_{X}|f|d\lambda= 0$ then $|f|= 0$ a.e. and therefore $f= 0$ a.e. My proposed proof is as follows: Assume that $|f| > 0$ for some $x \in X$. Since $|f|$ is positive measurable function it follows that it is the limit of an increasing sequence of simple functions which I will show as $|f|= \sum_{i=1}^{k}C_{i,n}\chi_{E_{i},n}$: Using Beppoi Levi I know that the limit of the integral is the integral of the limit which gives the following: $\int_{X}|f|d\lambda= \int_{X}\lim\limits_{n \rightarrow \infty}\sum_{i=1}^{k}C_{i,n}\chi_{E_{i},n} = \lim\limits_{n \rightarrow \infty}\int_{X}\sum_{i=1}^{k}C_{i,n}\chi_{E_{i},n} = \lim\limits_{n \rightarrow \infty}\sum_{i=0}^{k}c_{i,n}\lambda(E_{i,n}) = 0$ This is only possible if for any $n \in \mathbb{N}$ we have $\sum_{i=1}^{k}c_{i,n}\lambda(E_{i,n}) = 0$. $\therefore$ for all $n$ and $i$ it follows that either $c_{i,n}= 0$ or $\lambda(E_{i,n})= 0$. Since we assumed that $|f|$ is nonzero for some $x \in X$ it follows that there is some $c_{i,n}\chi_{E_{i,n}} > 0$ where $c_{i,n} > 0$ therefore there must be some $E_{i,n}$ such that $\lambda(E_{i,n})= 0$ and which gives $f(x) > 0$ for $x \in E_{i,n}$. It follows that $|f|= 0$ a.e. and therefore $f= 0$ a.e. Is this proof fine? Thanks for assistance.

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