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Math Help - Lebesgue integration question

  1. #1
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    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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  2. #2
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    Re: Lebesgue integration question

    Quote Originally Posted by Johnyboy View Post
    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}:
    The number of summands for each simple function may be different. So the $k$ may depend on your $n$.

    I think it is better to try doing something like this. First, it is sufficient to just assume that f\geq 0 and \smallint f ~ d\lambda = 0. Let E be the set were f is positive, we want to show that E is a set of measure zero. One way of doing this is to use one of the limit theorems on measures. Define E_n to be the set were f\geq 1/n and argue that each E_n has measure zero, from here it will follow that E has measure zero.
    Thanks from mash
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    Re: Lebesgue integration question

    Thanks for the response. Is the following fine?

    \lambda (E_{n}) = \int_{X}\chi_{E_{n}}d\lambda \leq \int_{X}\chi_{X} d\lambda \leq \int_{X}nf\chi_{X}d\lambda = n\int_{X}fd\lambda = 0

    It follows that that since E_{n} is an increasing sequence that \lambda(\bigcup_{n=0}^{\infty}E_{n}) = \lim\limits_{n \rightarrow \infty}\lambda(E_{n}) and therefore \lambda(\bigcup_{n=0}^{\infty}E_{n}) = 0 so \lambda(E) \leq \lambda(\bigcup_{n=0}^{\infty}E_{n}) = 0 therefore \lambda(E) = 0. So f = 0 a.e.
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  4. #4
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    Re: Lebesgue integration question

    Quote Originally Posted by Johnyboy View Post
    Is the following fine?

    \lambda (E_{n}) = \int_{X}\chi_{E_{n}}d\lambda \leq \int_{X}\chi_{X} d\lambda \leq \int_{X}nf\chi_{X}d\lambda = n\int_{X}fd\lambda = 0
    I think you rather mean,
    $$f\geq f\cdot \chi_{E_n} \geq \tfrac{1}{n} \chi_{E_n} \implies \smallint f \geq \smallint \tfrac{1}{n} \chi_{E_n} = \tfrac{1}{n}\lambda(E_n) $$
    Now since $\smallint f = 0$ it follows that $\lambda(E_n) = 0$ for every n.

    It follows that that since E_{n} is an increasing sequence that \lambda(\bigcup_{n=0}^{\infty}E_{n}) = \lim\limits_{n \rightarrow \infty}\lambda(E_{n}) and therefore \lambda(\bigcup_{n=0}^{\infty}E_{n}) = 0 so \lambda(E) \leq \lambda(\bigcup_{n=0}^{\infty}E_{n}) = 0 therefore \lambda(E) = 0. So f = 0 a.e.
    As you said $E_n$ is increasing and $E = \bigcup_{n\geq 1} E_n$ is the set where $f>0$, futhermore since $E_n$ has finite measure (this is important) we have that,
    $$ \lambda(E) = \lim ~ \lambda(E_n) = 0 $$
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