Integrability Proof

1. Oct 27, 2013

Yagoda

1. The problem statement, all variables and given/known data Let h(x) = 0 for all x in [a,b] except for on a set of measure zero. Show that if $\int_a^b h(x) \, dx$ exists it equals 0.

We are given the hint that a subset of a set of measure zero also has measure 0.

2. Relevant equations
We've discussed the Lebesgue integrability criterion: A bounded function f is Riemann integrable if and only if f the points of discontinuity on [a,b] are a null set.

3. The attempt at a solution
First, is there a case where this integral would not exist? It seems like if h is not 0 only on a null set then it would be bounded and thus integrable by the criterion.
Second, I understand intuitively that the integral is 0, but I am having trouble formalizing it. If h is equal to a non-zero $k$ only at one point then we could set the norm of the partition, $||P|| < \epsilon/k$ and we would have $|\sigma - 0| < \epsilon$. This could be modified for a finite number of non-zero points, by letting $k$ be the max of the absolute values of the non-zero points. But what if there are countably infinite non-zero points?
I also haven't used the hint. Is this to be applied over partitions of [a,b]?

2. Oct 27, 2013

brmath

First, are you trying to show this is Riemann integrable or Lebesgue integrable? We know it is Lebesgue integrable.

Second, if f is non-zero on a set of measure 0 that does not mean it is bounded. Let [a,b] = [0,1] and f(x) = 1/x where x = 1/n and f(x) = 0 otherwise. Quite unbounded.

However, your subset hint may be useful. What do you know about a set of measure 0 on [a,b]? Can it be uncountable? If countable could it be the limit of a sequence of finite sets?

3. Oct 28, 2013

Yagoda

I am trying to show Riemann integrability. We haven't covered the Lebesgue integral, just the criterion.

Yes, I do see now that being non-zero only on a null set, does not imply boundedness. If the Riemann integral does exist, the function must be bounded though, correct?

Well, the null set must be countable or finite. I'm not sure what you mean by "the limit of a sequence of finite sets," though.

4. Oct 28, 2013

brmath

First, there are some unbounded functions which are Riemann integrable. For example on [0,1] $x^{-1/2}$ is Riemann integrable. So unboundedness alone does not disqualify a function.

The second thing that is useful is that if $\int_a^bg(x)dx$ exists, and h(x) = g(x) except at one point where h(x) = k (k is any constant), then $\int_a^bh(x)dx$ exists and = $\int_a^bg(x)dx$. The same is true if h(x) differs from g(x) just at a finite number of points. If this wasn't proved in your class, you probably have to prove it for this assignment (or at least point to a reference for it).

So if you pick a set of measure 0 which consists of a finite number of points, and h is bounded on that set, you should be able to easily prove your result.

Which brings us to the set which is not finite. Let's work on the interval [0,1], although what I say translates to any interval. Let S$_n$ be the set {1,1/2,1/3,...1/n}. Now consider h(x) = 0 outside S$_n$, and has finite values on S$_n$, what can you conclude about $\int_a^bh(x)dx$ ?

Let S =$\lim_{n \rightarrow \infty}S_n$ . That is what I meant by "the limit of a sequence of finite sets". Suppose h(1/n) = 1 for all 1/n in S, and 0 otherwise. This will not be Riemann integrable. The proof would arise out of the need for the sup of every sequence Riemann sums to converge to the same number as the inf if h is to be integrable.(1)

However, if h(1/n) = a$_n$ where lim$_{n \rightarrow \infty}a_n = 0$ then h is integrable, and your result will hold. You would need to prove both that h is integrable and that its integral is 0 (2).

All of this is doable, but loading (1) and (2) onto a student who is just starting the study integrals is a lot to ask. If you tackle it, the approach is in the part I put in bold.

This business of infinite sets of measure 0 gums up the use of the Riemann integral for functions which are discontinuous on such sets. The inadequacies of the Riemann integral set off a little competition in defining better integrals. The Lebesgue integral nicely avoided this kind of problem and has some other overwhelming advantages, so it was eventually adopted as a standard.

5. Oct 28, 2013

Yagoda

Thanks for the explanation. I feel like in the text and in class we've barely scratched the surface of what is needed for a rigorous proof, but this helps. I'll see where I get.

The following question after this one is to show that if $f(x) = g(x)$ almost everywhere then their Riemann integrals are equal. I was hoping to use the result of this question to show that.

6. Oct 28, 2013

brmath

You are doing quite well at your proofs.

For the next problem, you should assume the result of this problem even if you haven't proved it. You can be explicit that you yourself haven't proved it, if you haven't. But someone has, so you are okay.

Actually, I'm a little suspicious of that problem. It is possible for f = g almost everywhere, and for f to be Riemann integrable and g not. So you should specifically add the hypothesis that they are both integrable.

7. Oct 28, 2013

Dick

It is certainly not true that an unbounded function can be Riemann integrable. If it's unbounded above then over any finite partition at least one of the intervals the function has no maximum. So all of the upper sums are undefined. Which kind of negates the definition of Riemann integrable. Functions like x^(-1/2) only have an IMPROPER integral, defined as a limit of normal Riemann integrals. To Yagoda, I would say look over the ingredients of your discussion of the Lebesgue integrability criterion and see if there are parts you can use to attack this.

Last edited: Oct 28, 2013
8. Oct 29, 2013

brmath

Dick,

Why does one make a distinction between a proper Riemann integral and one that results from the convergence of a sequence of Riemann integrals and which is then called improper? I understand you can't get an upper sum, but it seems reasonable to me to call the function integrable if its integral can be defined as the limit of ordinary Riemann integrals.

9. Oct 29, 2013

Dick

Because if h(x) is integrable in the proper Riemann sense then you can assume it's bounded. Actually, I'm not sure this is any use, but that's what I was thinking.

10. Oct 29, 2013

brmath

I imagine a certain amount of this is probably just vocabulary -- i.e. Riemann integrable was defined as the sup and inf of the Riemann sums converging to the same number, which implies boundedness. And I suspect that the improper integral terminology is traditional or historic rather than essential. One can after all just say that a function is bounded.

I am sorry for not being aware of this distinction, and possibly misleading Yagoda, who no doubt has to please a professor, who may in turn be a stickler for terminology.