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This is a well known result in mathematics, but it's my first time attempting to prove it. I'm following the second book of Analysis from Folland. Below are the notations used and the theorem, from the text.

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Now we have the following exercise, which is supposed to led us to prove the part b of this theorem.

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I managed to prove the part a of this exercise, but part b is hard to do. The best idea I have so far is to write for each $x\in[a,b]$, $$\lim_{\delta\to 0}\sup_{|x-y|\leq\delta}f(y) = \lim_{n\to\infty}\sup_{|x-y|\leq 1/n}f(y)$$

and note that $\sup_{|x-y|\leq 1/n}f(y) = M_i$ for some convenient partition chosen. In fact, for each $n\in\mathbb{N}$ we may choose the partition $P_n$ so it there is always a $M_i$ equal to $\sup_{|x-y|\leq 1/n}f(y)$. The problem is that there is uncountable sup's for $H$ while there is countable for $G_{P_n}$. So I don't know how this approach will (if it will) work. I need some help here, thanks.

Integral
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  • $$\overline{I}-\underline{I}=\int H-\int h=\int (H-h)$$ but the function $H-h$ is zero except for a set of measure zero (from part (a)). Then $\int(H-h)=0$. – Pp.. Jan 20 '15 at 16:24
  • You can't integrate $H$ if is not measurable. That's the point showing $H=G$ ae, for Lebesgue measure is complete and $G$ is measurable, therefore $H$ is measurable and then we can think about integrating $H$. But you can't integrate before showing $H=G$ ae. – Integral Jan 20 '15 at 16:27
  • Ah, your problem is showing $H$ measurable? Notice that $ G_{p_k}$ decreases to $G=H$ a.e. And $G_{p_k}$ are measurable. Notice that $\liminf$ of measurables is measurable. – Pp.. Jan 20 '15 at 16:34
  • Noting $G_{P_k}$ decreases to $G$ is ok, but is not clear that $G=H$ ae, that is the problem. – Integral Jan 20 '15 at 16:36
  • Man $H=f=G$ a.e. Look at all continuous points of $f$. – Pp.. Jan 20 '15 at 16:37
  • Could you prove that as an answer? It's not clear what $H = f$ ae. Note that we could have $f$ discontinuous everywhere. – Integral Jan 20 '15 at 16:41
  • No, you can't have $f$ dicontinuous everywhere. You are assuming that its set of discontinuities has measure zero. – Pp.. Jan 20 '15 at 16:44
  • No, I'm not. We are trying to show that this set has measure zero if an only if $f$ is Riemann integrable. You can assume this in one way but not in the other. And in this problem in particular, we just assumed $f$ is bounded, nothing else. – Integral Jan 20 '15 at 16:46
  • It looks clear that $H = G$ ae, despite the measure of the set of discontinuities of $f$. You can see this problem independently of the theorem. In fact, I just showed the theorem to show the notations used, you can take this notations and just forget the theorem to make this exercise. – Integral Jan 20 '15 at 16:56
  • The proof of $f$ Riemman integrable implies Lebesgue integrable already contains the proof that the set of discontinuities has measure zero. Notice that the set of discontinuities is contained in the set where $G\neq g$. You only need to use the exercise to prove the other direction, in which you assume the set of discontinuities is small. – Pp.. Jan 20 '15 at 16:59
  • The exercise is not assuming this, you are. – Integral Jan 20 '15 at 17:02
  • The proof that $H=G$ ae does not depend on the measure of the discontinuity points of $f$. – Integral Jan 20 '15 at 17:03
  • Then you have everything you need to prove solve the exercise. Do you see it? What is your question then? – Pp.. Jan 20 '15 at 17:04
  • I don't how to prove that $H=G$ ae. I need a proof not relying on the measure of the set of discontinuity points. This measure is not mentioned on the exercise therefore we can't assume it is zero, $\varepsilon$ or anything. – Integral Jan 20 '15 at 17:07
  • $H(x)$ and $G(x)$ are both equal to $\limsup_{y\to x}f(y)$, by definition. Just write down the definition of $G(x)$, for a fixed $x$. You can forget about the whole partition. All it matters is the interval that contains $x$. As soon as you write it down you see they are the same thing. An interval $|x-y|<\delta$ can be put inside an interval $(t_{j-1},t_j]$ and the other way around too. Doing these you show $H(x)\leq G(x)$ and $G(x)\leq H(x)$ respectively. – Pp.. Jan 20 '15 at 17:25

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