| Author | Message | ||
| David Franklin Veteran poster Post Number: 790 |
Am I right in thinking no such function f: | ||
| Yatir Halevi Veteran poster Post Number: 888 |
I believe that you can construct a proof that if f is riemann integrable then f has at least one continuous point... But it will take me some time to find my proof from last year...unless someone remembers how to do it...? Yatir | ||
| Yatir Halevi Veteran poster Post Number: 891 |
Until I find my proof... hint: it has to do something with: https://nrich.maths.org/discus/messages/114352/116089.html Yatir | ||
| Yatir Halevi Veteran poster Post Number: 893 |
Ok! Found it... These are the steps of the proof: (Using the definitions in the link I gave  1) Prove that f is continuous in xÎ[a,b] iff wf(x)=0 2) Lets define V={1£i£n: Mi-mi³e} Show that if 3) Prove what I wrote in the link 4) Now prove the if f is integrable then it has a continuous point in [a,b] 5) And as a bonus prove that if f is integrable then the set of it's continuous points is dense in [a,b] Yatir | ||
| David Franklin Veteran poster Post Number: 794 |
Thanks Yatir for the reply - I think this is a bit above what I'm used to so far! (1) If wf(x)=0, then we have |x-y| < e Þ |f(x)-f(y)| £ sup{|f(a)-f(b)| : a,bÎ(x+e,x-e)}, and as e®0 we can make this arbitrarily small since its infimum is zero. I'm not too sure how to continue though since I want something along the lines of ""a>0 $e: |x-y| < e Þ |f(x)-f(y)| < a". Although I can make |f(x)-f(y)| arbitrarily small as e®0, I can't see how to guarantee that it's also less than a (since we can make a arbitrarily small too). The other way round, we have f cts Þ "a>0 $e: |x-y| < e Þ |f(x)-f(y)| < a, so |f(y)-f(z)| £ |f(y)-f(x)|+|f(x)-f(z)| < 2a and I have the same problem: just because I can make |f(y)-f(z)| arbitrarily small by varying a, I don't see why this should imply that the infimum over all e is 0. (2) If x isn't in that union of intervals, it's in some open interval (xj-1, xj). Taking e small enough such that x-e > xj-1 and x+e < xj, we have sup{|f(a)-f(b)| : a,bÎ(x+e,x-e)} £ Mj - mj < e, so wf(x)<e... but this is confusing me - now I can take the infimum over all e to get wf(x)=0? (3) Can't we just use the lemma that states "if f is integrable, then "e>0, $ a dissection I think I'll stop there for now since I'm a bit confused and not sure I'm approaching the proof in the right way. Could you explain what's going on? | ||
| Bhishan Jacelon Prolific poster Post Number: 227 |
(To add some context to this thread, Lebesgue proved in his PhD thesis (1902) that a bounded function on [a,b] is Riemann integrable if and only if the set of points where it is discontinuous is of measure zero.) | ||
| Yatir Halevi Veteran poster Post Number: 898 |
Yeah...it was a bit above what I was used to when I first encountered it last year as well Lets start it step at a time... 1) To see whats going on, try defining (in your notation... I am used to epsilon-delta) (If you want to prove that f is continuous at x) Wf(x,e)=sup{|f(z)-f(y)|:z,yÎ(x-e,x+e)} Now... Let a>0 what can you conclude from the definition of the infimum? That there exists a e such that.... And for the other way around... |f(y)-f(z)| is always positive, so if you can make it arbitrarily small what can you say about its value? Could you reach a contradiction if its value was some strictly positive number? Yatir | ||
| David Franklin Veteran poster Post Number: 796 |
Thanks a lot Yatir - just so you know I'm not ignoring this thread and will come back to it when I've got some more time on my hands. Incidentally, I'm used to epsilon-delta notation too but I think I must've got in a bit of a muddle and convinced myself that my epsilons from one thing were going to become my deltas from another, hence the alpha I put in as well. | ||
| Yatir Halevi Veteran poster Post Number: 907 |
Sure. Take your time. Yatir |