## Jordan Content Integrability Condition

We wanted a better necessary and sufficient condition for integrability than Riemann’s condition, and now we can give a result halfway to our goal. We let be a bounded function defined on the -dimensional interval .

The condition hinges on defining a certain collection of sets. For every we define the set

of points where the oscillation exceeds the threshold value . The first thing to note about is that it’s a closed set. That is, it should contain all its accumulation points. So let be such an accumulation point and assume that is *isn’t* in , so . So there must exist a neighborhood of so that and this means that for *any* point , and so none of these points can be in . But if this is the case, then can’t be an accumulation point of , and so we must have .

And now for our condition. The function is integrable if and only if the Jordan content for every .

We start by assuming that for some , and we’ll show that Riemann’s condition can’t hold. Given a partition of we calculate the difference between the upper and lower sums

where is the part of the sum involving subintervals which contain points of and is the rest. The intervals in have total length , and in these intervals we must have

because if the difference were less then the subinterval would be a neighborhood with oscillation less than and thus couldn’t contain any points in . Thus we conclude that , and the difference between the upper and lower sums is at least as big that. This happens no matter what partition we pick, and so the upper and lower integrals must also differ by at least this much, violating Riemann’s condition. Thus if the function is integrable, we must have .

Conversely, take an arbitrary and assume that . For this to hold, there must exist a partition so that . In each of the subintervals not containing points of , we have for all in the subinterval. Then we know there exists a so that we can subdivide the subinterval into smaller subintervals each with a diameter less than , and the oscillation on each of these subintervals will be less than . We will call this refined partition .

Now, if is finer than , we can again write

where again contains the terms from subintervals containing points in and is the remainder. In all of these latter subintervals we know the difference between the maximum and minimum values of is less than , and so

For , on the other hand, we let and be the supremum and infimum of on all of , and we find

Thus we conclude that

Since this inequality is valid for any , we see that Riemann’s condition must hold.

[…] Lebesgue’s Condition At last we come to Lebesgue’s condition for Riemann-integrability in terms of Lebesgue measure. It asserts, simply enough, that a bounded function defined on an -dimensional interval is Riemann integrable on that interval if and only if the set of discontinuities of has measure zero. Our proof will go proceed by way of our condition in terms of Jordan content. […]

Pingback by Lebesgue’s Condition « The Unapologetic Mathematician | December 15, 2009 |

Thanks for your blog.

It seems that you assume J_\epsilon is always Lebesgue measurable. What guarantees that?

Comment by Dan Gluck | October 23, 2015 |

Sorry, Dan, it’s been a while since I wrote this, and I was never a pro analyst. Can you point out where you think that assumption is being used?

Comment by John Armstrong | October 25, 2015 |

How to prove that “if the difference were less then the subinterval would be a neighborhood with oscillation less than ε and thus couldn’t contain any points in J_ε”? If x belonged to the interior of the subinterval, the subinterval would be a neighborhood of x with oscillation less than ε, but do we see that if x belongs to the frontier of the subinterval, then x cannot belong to J_ε?

Comment by Mayan Rainbow | November 20, 2015 |