Riemann Integrable <-> Continuous almost everywhere?

[holomorphic]

Riemann Integrable <--> Continuous almost everywhere?

I ran across a statement somewhere in the forums saying that a function is Riemann-integrable iff it is continuous almost everywhere, i.e. if its set of discontinuities has measure 0.

Is that right?

What about the case of a function f:R->R which is, say, f(x) = c everywhere except on an interval [a,b], where f(x) = 1 if x is rational and f(x) = 0 otherwise... Doesn't the set of discontinuities have measure 0? But my intuition says this function is not integrable.

Is there an argument that explains either why the statement is false, or why the function I defined above is actually Riemann-integrable? (OR, why the set of discontinuities doesn't have measure 0... which would be a surprise).

 

[mathman]

The function you described (1 at rationals, 0 otherwise) is discontinuous everywhere, not just at the rationals. Every irrational number can be expressed as a sequence of rationals, so the function is discontinuous at the irrationals as well.

 

[holomorphic]

The function you described (1 at rationals, 0 otherwise) is discontinuous everywhere, not just at the rationals. Every irrational number can be expressed as a sequence of rationals, so the function is discontinuous at the irrationals as well.

No, no, that's not the function I described. The function I described is
f(x) = c for x not in [a,b]
f(x) = 1 if x is in Q between a and b
f(x) = 0 if x is not in Q between a and b

 

[holomorphic]

The function you described (1 at rationals, 0 otherwise) is discontinuous everywhere, not just at the rationals. Every irrational number can be expressed as a sequence of rationals, so the function is discontinuous at the irrationals as well.

Though I see how this argument applies to the function I described anyway--so the set of discontinuities has a positive measure.

 

[Office_Shredder]

This sounds pretty plausible. You do Riemann integration by first picking your sub-intervals on the real line, and then making Riemann sums. If the set of discontinuities is of measure zero, then that says that when you pick your subdivisions, the set of discontinuities will be inside of subdivisions of arbitrarily small measure. That part is going to go to zero anyway, so you get convergence.

On the other hand, if the set of discontinuities is of positive measure, then no matter how much you try to pick your subdivisions, subdivisions of measure at least e for some small value of e will have discontinuities. You don't really have a chance of working around it

This is basically just extending why you can integrate a function with one discontinuity: when you take various Riemann sums, the discontinuity can screw up the value that you add near the point, but because the discontinuity is contained inside of a box of smaller and smaller size, eventually it goes away in the limit anyway

Obviously you need some boundedness condition to avoid things like 1/x

 

[dimitri151]

What about the case of a function f:R->R which is, say, f(x) = c everywhere except on an interval [a,b], where f(x) = 1 if x is rational and f(x) = 0 otherwise... Doesn't the set of discontinuities have measure 0? But my intuition says this function is not integrable.

Is there an argument that explains either why the statement is false, or why the function I defined above is actually Riemann-integrable? (OR, why the set of discontinuities doesn't have measure 0... which would be a surprise).

The function you describe is not Reimann integrable but it is integrable by the generalized Reimann integral. This is the Reimann integral with a gauge. Read "A Modern Theory of Integration" by Bartle, pg12,29. The integral is zero on the Dirichlet rational characteristic function part (on [a,b]).

 

[disregardthat]

It's a theorem in my book, I could write down the proof if you want.