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Let f(x) be a bounded continuous function on [0,1]. Let g(x)=f(x) on all rational points in [0,1]. Let g(x) be Riemann integrable on [0,1]. Does g(x)=f(x) almost everywhere in the interval? If so - proof? If not -counterexample.
mathwonk said: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...?
bhobba said:...Now the question is why is the Dirichlet Function not Riemann Integrable - but is Lebesgue Integrable? What condition of the theorem is broken?
mathman said: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.
mathwonk said: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.)
bhobba said: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.
mathman said: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.lavinia said: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.
Svein said:The Riemann integral was defined before measure theory was invented,
bhobba said: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
lavinia said:So it is discontinuous on a set of measure 1 not zero
lavinia said:So it is discontinuous on a set of measure 1 not zero
bhobba said: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
lavinia said: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.
bhobba said:Are you getting at the measure of the irrationals is 1 over the interval [0,1]. Sure. Are you saying if we do not define, or define it, at the rationals, it doesn't matter which, the measure of the real line between 0 and 1 is the measure of the irrationals ie one. Again sure - just different language to what I said before. But how does that resolve the original query. Under the usual definition of the Riemann Integral it means if it is discontinuous at the rationals it is not Riemann integrable. Thus one could say it can't be discontinuous at the rationals because its Riemann Integrable. - so my argument since the rationals are dense it must be continuous at those points - hence f=g? Have I got your argument correct?
My suspicion, and I have never seen a development along those lines, Riemann Integration can be defined in a more general way as detailed previously.
Thanks
Bill
lavinia said:So the measure of the rationals doesn't come into the argument since the Dirichlet function is discontinuous everywhere.The original post is answered with Mathwonk's hint which is that at points where g is continuous it must equal the function f. If g is Riemann integrable it is continuous except possibly on a set of measure zero. So it equals f almost everywhere. However g does not have to equal f exactly since for instance one can take g to be f everywhere except at a single irrational number.
lavinia said:Not sure if this is what you were asking.
The Riemann Integral is a mathematical concept used to calculate the area under a curve represented by a function. It is a way to find the total amount of a quantity over a given interval by dividing it into smaller and smaller parts and then adding them together.
The Riemann Integral is a specific type of integral that is defined by a limit of sums. Other types of integrals, such as the Lebesgue Integral, are defined using different methods and may give different results for certain functions.
When two functions are equal Almost Everywhere, it means that they are equal at every point except for a set of points with measure zero. In other words, the two functions may have slightly different values at a few points, but they are essentially the same function.
In the context of the Riemann Integral, Almost Everywhere equality means that if two functions are equal Almost Everywhere, then they will have the same Riemann Integral. This is because the Riemann Integral only considers the values of the functions at a finite number of points, so any differences at points with measure zero will not affect the overall result.
Some common examples of functions that are equal Almost Everywhere include the function f(x) = x, which is equal to the function g(x) = x^2 at all points except x = 0, and the function h(x) = sin(x), which is equal to the function k(x) = cos(x) at all points except x = π/2 + nπ for any integer n. These functions are essentially the same, but they may have different values at a few isolated points.