Integrability of g, Nonnegativity & Infinite Points: Is it True?

[kala]

I can't make up my mind if this statement is true or not: is it true that if we assume g is integrable and g\geq0 on [a,b], then if g(x)\geq0 for an infinite number of points x is in [a,b] then \int g >0.

I can't figure out if its true or false, i thought that i had a counter example:
if g>0 at a single point i.e. g(a)=1 and g=0 otherwise, then g is integrable and non negative, and the set of discontinuities must be finite.
I don't know if this is exactly right. Any help please?

 

[sutupidmath]

I can't make up my mind if this statement is true or not: is it true that if we assume g is integrable and g\geq0 on [a,b], then if g(x)\geq0 for an infinite number of points x is in [a,b] then \int g >0.

I can't figure out if its true or false, i thought that i had a counter example:
if g>0 at a single point i.e. g(a)=1 and g=0 otherwise, then g is integrable and non negative, and the set of discontinuities must be finite.
I don't know if this is exactly right. Any help please?

What do you mean with g(x)>0 for an infinite number of points x in [a,b]? Are u saying here, that we are assuming that for some points in [a,b] g(x) might not be greater than 0 or?

 

[adriank]

I'm assuming you mean if on [a, b], g is integrable and nonnegative, and g(x) > 0 for an infinite number of points x in [a, b], then \int_{[a, b]} g > 0.

Your example only has g(x) > 0 for a finite number of points, so it is not a counterexample.

However, here is a counterexample: On [0, 1], let g(x) = 1 if x = 1/n for some positive integer n, and 0 otherwise. Then g(x) = 1 at infinitely many points, but it is integrable and the integral is zero. Reason: each point in S = {1/n | n is a positive integer} is an isolated point of S.

 

[HallsofIvy]

No, it is not true. For example, if f(x)= 1 for x rational and f(x)= -1 for x irrational, then f(x) is "positive for an infinite number of points" but its (Lebesque) integral, from x=0 to x=1, is -1 since f(x)= -1 except on a set of measure 0.

 

[adriank]

Two issues I have with that argument: First, it doesn't directly address the problem, since there is the assumption that g(x) ≥ 0 everywhere. Second, he didn't specify which kind of integrability we're talking about, so I just assumed we're talking about Riemann-integrable functions, and yours is not one of them.

 

[JG89]

The statement is true if you make it \int^b_ag(x) \ge 0

To see this evaluate the integral with Riemann sums and we make the distance between each point of subdivision equal. So let h = \frac{b-a}{n}. Then we have \int^b_ag(x) = \lim_{n \rightarrow \infty} \sum\limits_{i=1}^n h(g(x_i))

The sum is greater than or equal to zero because each term in the sum is nonnegative.

 

[adriank]

Yes, but that's trivial.

 

[JG89]

It was similar to what the OP was asking. There is no harm done in adding what I added.

 

[rochfor1]

You can easily make HallsOfIvy's function nonnegative, just make it 0 at every irrational number. Come on, you could make that small modification.

 

[adriank]

You could, and the (Lebesgue) integral would be zero. However, that is not Riemann integrable (which is what I was assuming).

 

[rochfor1]

Well, working in this situation with the Riemann integral is rather delicate, because even the integrability of such a function is difficult to establish. Sorry if you meant Riemann, I just always think Lebesgue.

 

[adriank]

I think what I was really trying to say was that the statement is false even for Riemann-integrable functions, nothing more.