1. ## Lebesgue integration question

Lebesgue integral question:

Consider measure space $\displaystyle (X, \epsilon, \lambda)$ with $\displaystyle f$ a summable function.
I want to show that if $\displaystyle \int_{X}|f|d\lambda = 0$ then $\displaystyle |f| = 0$ a.e. and therefore $\displaystyle f = 0$ a.e.
My proposed proof is as follows:

Assume that $\displaystyle |f| > 0$ for some $\displaystyle x \in X$.
Since $\displaystyle |f|$ is positive measurable function it follows that it is the limit of an increasing sequence of simple functions which I will show as $\displaystyle |f| = \sum_{i=1}^{k}C_{i,n}\chi_{E_{i},n}$:

Using Beppoi Levi I know that the limit of the integral is the integral of the limit which gives the following:
$\displaystyle \int_{X}|f|d\lambda = \int_{X}\lim\limits_{n \rightarrow \infty}\sum_{i=1}^{k}C_{i,n}\chi_{E_{i},n} = \lim\limits_{n \rightarrow \infty}\int_{X}\sum_{i=1}^{k}C_{i,n}\chi_{E_{i},n} \\ = \lim\limits_{n \rightarrow \infty}\sum_{i=0}^{k}c_{i,n}\lambda(E_{i,n}) = 0$

This is only possible if for any $\displaystyle n \in \mathbb{N}$ we have $\displaystyle \sum_{i=1}^{k}c_{i,n}\lambda(E_{i,n}) = 0$.
$\displaystyle \therefore$ for all $\displaystyle n$ and $\displaystyle i$ it follows that either $\displaystyle c_{i,n} = 0$ or $\displaystyle \lambda(E_{i,n}) = 0$.

Since we assumed that $\displaystyle |f|$ is nonzero for some $\displaystyle x \in X$ it follows that there is some $\displaystyle c_{i,n}\chi_{E_{i,n}} > 0$ where $\displaystyle c_{i,n} > 0$ therefore there must be some $\displaystyle E_{i,n}$ such that $\displaystyle \lambda(E_{i,n}) = 0$ and which gives $\displaystyle f(x) > 0$ for $\displaystyle x \in E_{i,n}$.

It follows that $\displaystyle |f| = 0$ a.e. and therefore $\displaystyle f = 0$ a.e.

Is this proof fine? Thanks for assistance.

2. ## Re: Lebesgue integration question

Originally Posted by Johnyboy
Since $\displaystyle |f|$ is positive measurable function it follows that it is the limit of an increasing sequence of simple functions which I will show as $\displaystyle |f| = \sum_{i=1}^{k}C_{i,n}\chi_{E_{i},n}$:
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 $\displaystyle f\geq 0$ and $\displaystyle \smallint f ~ d\lambda = 0$. 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 $\displaystyle E_n$ to be the set were $\displaystyle f\geq 1/n$ and argue that each $\displaystyle E_n$ has measure zero, from here it will follow that E has measure zero.

3. ## Re: Lebesgue integration question

Thanks for the response. Is the following fine?

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

It follows that that since $\displaystyle E_{n}$ is an increasing sequence that $\displaystyle \lambda(\bigcup_{n=0}^{\infty}E_{n}) = \lim\limits_{n \rightarrow \infty}\lambda(E_{n})$ and therefore $\displaystyle \lambda(\bigcup_{n=0}^{\infty}E_{n}) = 0$ so $\displaystyle \lambda(E) \leq \lambda(\bigcup_{n=0}^{\infty}E_{n}) = 0$ therefore $\displaystyle \lambda(E) = 0$. So $\displaystyle f = 0$ a.e.

4. ## Re: Lebesgue integration question

Originally Posted by Johnyboy
Is the following fine?

$\displaystyle \lambda (E_{n}) = \int_{X}\chi_{E_{n}}d\lambda \leq \int_{X}\chi_{X} d\lambda \leq \int_{X}nf\chi_{X}d\lambda = n\int_{X}fd\lambda = 0$
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 $\displaystyle E_{n}$ is an increasing sequence that $\displaystyle \lambda(\bigcup_{n=0}^{\infty}E_{n}) = \lim\limits_{n \rightarrow \infty}\lambda(E_{n})$ and therefore $\displaystyle \lambda(\bigcup_{n=0}^{\infty}E_{n}) = 0$ so $\displaystyle \lambda(E) \leq \lambda(\bigcup_{n=0}^{\infty}E_{n}) = 0$ therefore $\displaystyle \lambda(E) = 0$. So $\displaystyle f = 0$ 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,
$$\lambda(E) = \lim ~ \lambda(E_n) = 0$$