First I do not consider this homework and as a mentor I am discussing it now with the other mentors, but for now it will remain here. As is often the case MIT comes to the rescue:a riemann integrable function is continuous almost everywhere. if g is continuous at c, the value of g at c is determined by its values at all rational points. so....????
The problem with Riemann integrability of Dirichlet's function is in the necessary discretion in the evaluation of the function within the subintervals of the Riemann sums. Since both the rationals and irrationals are dense in the interval one can choose to only evaluate [itex]f(x^*_k)[/itex] at rational values or only evaluate at irrational values and you'd get non-convergence of the limits. By making measure the issue instead of density, Lebesgue integration circumvents this problem....Now the question is why is the Dirichlet Function not Riemann Integrable - but is Lebesgue Integrable? What condition of the theorem is broken?
Ah - I see your issue now - you did not actually say g(x) was continuous - I assumed it due to its the usual case in Riemann integration. I will need to think a bit.The original problem was not a homework problem! I deliberately made a point that it is Riemann integrable. I am well aware that if it was only Lebesgue integrable, then it would be obviously undecided to say g=f anywhere except at the rationals. My own thought about proving it true is to use the equivalence of Darboux and Riemann integrability and use upper Darboux and lower Darboux sums to squeeze g to the point where it equals f almost everywhere. I just am not sure how to flesh it out.
Yes that is what I showed but forgot that the OP did not say g was continuous - just integrable. This one, as the above shows, is proving harder than I thought.can you show g=f wherever g is continuous? (assuming that for every point c of the domain, there is a sequence of rationals converging to c.)
No, the Dirichlet function is discontinuous everywhere, since the rationals are dense.But wait - the Dirichlet function has that property - it is continuous at irrational points but discontinuous at the rationals - it would seem a counter example disproving the theorem. It has defeated me. Someone else with more knowledge needs to look at it I think.
I think taking @mathwonk's hint would make the argument crystal clear. That is: at points where g is continuous it must equal f.It appears that the theorem about Lebesgue and Riemann integral should apply here. g is Riemann integrable therefore continuous almost everywhere. Since f=g on a dense set, while f is continuous, this should imply f=g almost everywhere. Is that a complete argument?
I am somewhat lost here. The Riemann integral was defined before measure theory was invented, so talking about "measure zero" and "almost everywhere", while perfectly sensible when talking about Lebesgue integrals, makes no sense when talking about Riemann integrals. Just check out https://en.wikipedia.org/wiki/Riemann_integral.A cool example: Let f be zero on the unit interval Let g be the function which is one on the Cantor set and zero on the rest of the unit interval. Since the Cantor set has measure zero, g is Riemann integrable and equals f almost everywhere,
Now let g be one on a Cantor set of positive measure - e.g by removing middle 1/4's the rather than middle thirds. This is a again a closed set with empty interior but it has positive measure. g does not equal f a.e. and it is not Riemann integrable.
Yes but its interpretation using measure theory had to await until it was mathematicaly defined and investigated.The Riemann integral was defined before measure theory was invented,
If by the Dirichlet function you mean the function on the unit interval which is zero on the rationals and 1 on the irrationals then it is discontinuous at every point. So it is discontinuous on a set of measure 1 not zero.Yes but its interpretation using measure theory had to await until it was mathematicaly defined and investigated.
Why I am sitting back and not commenting is I am hoping someone can explain how a function can be continuous at all on dense set of measure zero. There is obviously something in the wording of the theorem I quoted. The following theorem suggests the answer is no to the original question because of it only being defined on a set of measure zero - and the set is dense in the reals - A function f : [a, b] → R is Riemann integrate iff it is bounded and the set S(f) = {x ∈ [a, b] | f is not continuous at x} has measure zero. Yet the Dirichlet function is a counter example to that statement - as it must be if you think about it. Consider any partition - if the set is dense its maximum and minimum can always be different hence can not be Riemann Integrable. There is obviously something I am missing, something in the detail of the theorem. I am waiting for someone to spot it. If you studied the theorems proof and thought hard you could probably figure it out - but I do not feel like spending the time doing that. My feeling for what it is worth is its really saying one can ignore those values and make them anything you like to avoid the partition issue. But I do not know - its just a guess.
Thanks
Bill
Can you expand on this please? I have proven myself in the deep past the rationals have measure 0 - and there are many proofs of that around eg (see example 2.3):So it is discontinuous on a set of measure 1 not zero
Can you expand on this please? I have proven myself in the deep past the rationals have measure 0 - and there are many proofs of that around but as mentioned in the link below: The measure of a point is zero and the rational number set is a countable union of point sets of measure zero so the whole thing has measure zero.So it is discontinuous on a set of measure 1 not zero
It is discontinuous on both the rationals and the irrationals. Every irrational is the limit of a Cauchy sequence of rationals. Every rational is the limit of a Cauchy sequence of irrationals. You do not need to know the measure of the rationals.Can you expand on this please? I have proven myself in the deep past the rationals have measure 0 - and there are many proofs of that around but as mentioned in the link below: The measure of a point is zero and the rational number set is a countable union of point sets of measure zero so the whole thing has measure zero.
Others have wondered about this strange property as well - dense and of measure zero eg:
https://www.quora.com/Why-are-the-rational-numbers-dense-and-of-measure-zero-at-the-same-time
It seems counter intuitive.
Thanks
Bill