# Is this statmenet true

1. Dec 4, 2008

### 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$$\geq$$0 on [a,b], then if g(x)$$\geq$$0 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?

2. Dec 4, 2008

### sutupidmath

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?

3. Dec 4, 2008

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.

Last edited: Dec 4, 2008
4. Dec 5, 2008

### HallsofIvy

Staff Emeritus
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.

5. Dec 5, 2008

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.

6. Dec 6, 2008

### 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.

7. Dec 7, 2008

Yes, but that's trivial.

8. Dec 7, 2008

### JG89

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

9. Dec 7, 2008

### 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.

10. Dec 7, 2008

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

11. Dec 7, 2008

### 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.

12. Dec 7, 2008