# Under what circumstances is a characteristic function Riemann integrable?

Let $$E$$ be a subset of the interval [a,b]. My question is, under what circumstances is the characteristic function $$1_E$$ Riemannn integrable on $$[a,b]$$?

Now a function is Riemann integrable if and only if its set of discontinuities is of Lebesgue measure zero. And the set of discontinuities of $$1_E$$ is equal to the boundary of $$E$$. So this is equivalent to asking, under what circumstances does the boundary of a set $$E$$ have measure zero? $$E$$ having measure zero isn't a strong enough condition, because a set of measure zero could have a boundary of positive measure. So what condition does $$E$$ need to satisfy?

And what is the Sigma algebra generated by sets with Riemann integrable characteristic functions?

• don't you mean the boundary, not the closure? For $E=[0,1/2]$ the characteristic function should be Riemann integrable, although its closure (which is $E$ itself) has most definetely positive lebesque measure – Enkidu Oct 31 '18 at 13:23
• Don't have the answer, but yes the rationals in [a, b] are a set of Lebesgue measure zero whose closure is [a, b]. – Tom Collinge Oct 31 '18 at 13:25
• well, the obvious answer you already gave yourself: if and only if the boundary of your set contains an open set. adn that should be actually equivalent to the boundary of your set being descrete. however, since that seems a little bit fiddely with definitions i am not too keen on posting that answer, but I will do so as soon as i think of a nice argument. – Enkidu Oct 31 '18 at 13:35
• @Enkidu You're saying that the characteristic function of a set is Riemann integrable if and only if its boundary has a non-empty open subset? But the boundary of $[0,1]$ is just the set containing $0$ and $1$, which has no non-empty open subset, yet it's characteristic function is Riemann integrable. – Keshav Srinivasan Oct 31 '18 at 13:42
• sorry, meant NOT Riemann integrable – Enkidu Oct 31 '18 at 13:48

This Wikipedia article answers my question:

An indicator function of a bounded set is Riemann-integrable if and only if the set is Jordan measurable.

I don't know what the proof of that is though.

Well lets just consider $$X=\delta E$$. If this contains an open set, it is obviously not integrable, since your incontinuities contain an open set, which contains an interval, and hence has positive lebesque measure.

In the other direction, assume the characteristic function is not Riemann integrable, i.e. your boundary does not have measure zero, and since it is closed, it is actually measurable! Hence a generating measurable set with positive measure is contained in $$X$$. But those generating sets are precisely the open sets!

Also, pls forgive me if I forgot some intricacies with the definition of the lebesque sigma algebra. been some time since I did measure theory.

• "assume the characteristic function is not Riemann integrable, i.e. your boundary has measure zero," Don't you have things backwards? If a characteristic function of a set is not Riemann integrable then its boundary will have positive measure. – Keshav Srinivasan Oct 31 '18 at 13:58
• no, I want to show, that if it is NOT integrable, then the boundary contains an open set. (converse of first line) also, should most definitely drink less tea (makes me make typos) – Enkidu Oct 31 '18 at 14:00