Lebesgue integration question

Lebesgue integral question:

Consider measure space with a summable function.

I want to show that if then a.e. and therefore a.e.

My proposed proof is as follows:

Assume that for some .

Since is positive measurable function it follows that it is the limit of an increasing sequence of simple functions which I will show as :

Using Beppoi Levi I know that the limit of the integral is the integral of the limit which gives the following:

This is only possible if for any we have .

for all and it follows that either or .

Since we assumed that is nonzero for some it follows that there is some where therefore there must be some such that and which gives for .

It follows that a.e. and therefore a.e.

Is this proof fine? Thanks for assistance.

Re: Lebesgue integration question

Re: Lebesgue integration question

Thanks for the response. Is the following fine?

It follows that that since is an increasing sequence that and therefore so therefore . So a.e.

Re: Lebesgue integration question

Quote:

Originally Posted by

**Johnyboy** Is the following fine?

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

is an increasing sequence that

and therefore

so

therefore

. So

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,

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