# Lebesgue integration question

• February 24th 2014, 04:48 AM
Johnyboy
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.
• February 24th 2014, 01:01 PM
ThePerfectHacker
Re: Lebesgue integration question
Quote:

Originally Posted by Johnyboy
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.
• February 25th 2014, 06:54 AM
Johnyboy
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.
• February 25th 2014, 10:26 PM
ThePerfectHacker
Re: Lebesgue integration question
Quote:

Originally Posted by Johnyboy
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.

Quote:

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,
$$\lambda(E) = \lim ~ \lambda(E_n) = 0$$