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Here is a problem I have been working on recently:

Let $f \colon[a,b] \to \mathbb{R}$ be continuous, differentiable on $[a,b]$ except at most for a countable number of points, and $f^{\prime}$ is Lebesgue integrable, then the fundamental theorem of calculus holds, i.e. $\forall x,y \in [a,b]$ we have $$f(y) = f(x) + \int_x^yf'(t)\,dt.$$

The proof I have at the moment is somewhat indirect: I can show that the FTC holds by proving that $f$ is $AC([a,b])$ (http://en.wikipedia.org/wiki/Absolute_continuity). The way I can prove this is showing that the Luzin's N-property (http://en.wikipedia.org/wiki/Luzin_N_property) is satisfied. I have spent quite a lot of time looking for a direct proof, but nothing seems to work! Can anyone help me?

Here is a summary of useful things (which I'll update if I figure out something else interesting!):

  1. Thm: If $u \colon [a,b] \to \mathbb{R}$, is continuous and differentiable everywhere on [a,b], with $u' \in L^1$, then the FTC holds. (This is a well know result for the Riemann integral, maybe a little less know in the context of the Lebesgue integral.. At least I've never heard of this result before looking for it! A proof can be found in Rudin's Real & Complex Analysis)
  2. Applying 1. we can prove the result in the case the set of non differentiability is finite.
  3. (not really useful!) I can also prove it if the set is countable but with only a finite number of accumulation points. This can be done using 2. and adding and subtracting terms to work with telescopic series.

Thank you in advance for your help!

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  • $\begingroup$ According to Wikipedia, to apply Luzin's N-property, we need to know that $f$ is of bounded variation. Do we know that? $\endgroup$ – PhoemueX Oct 10 '14 at 5:27
  • $\begingroup$ We can prove that, or we can use a different theorem that relates the N property with AC, namely: u is AC iff u is continuous, diff a.e. with derivative in L^1 + the N-property. (note: this use that the interval is closed!) :) Anyway, you proof looks interesting, I upvoted to thank your efforts but I need time to read it carefully before accepting it eheh, :D $\endgroup$ – user67133 Oct 10 '14 at 17:51
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    $\begingroup$ You can be interested to a recent work by J.Koliha A Fundamental Theorem of Calculus for Lebesgue Integration, The American Mathematical Monthly, Vol. 113, No. 6 (Jun. - Jul., 2006), pp. 551-555. First he proves a preparatory result $$|f(b)-f(a)| \le \int_a^b |f'(t)| \,dt$$ Then $\,|f(v)-f(u)| \le \int_u^v |f'(t)| \,dt \,$ for any subinterval $\,[u,v]\,$ of $\,[a,b]\,$. Since the Lebesgue integral is absolutely continuous, so is $f\,$, and $\,\int_a^b f'(t) \, dt = f(b)- f(a)\,$ by a well known result. $\endgroup$ – Tony Piccolo Oct 11 '14 at 10:38
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    $\begingroup$ It can be useful to remember that, for the gauge integral, the assumption on the integrability of $f'$ is not required. $\endgroup$ – Tony Piccolo Oct 11 '14 at 10:42
  • $\begingroup$ I'll give it a look, thanks! :) $\endgroup$ – user67133 Oct 11 '14 at 16:33
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The following is a combination of a proof in the book "Principles of mathematical analysis" by Dieudonne of a version of a mean value theorem and of the proof of the Theorem (Theorem 8.21) in Rudin's book "Real and Functional Analysis" that you also cite.

The proof actually yields the stronger statement that it suffices that $f$ is differentiable from the right on $\left[a,b\right]$ except for an (at most) countable set $\left\{ x_{n}\mid n\in\mathbb{N}\right\} \subset\left[a,b\right]$.

Let $\varepsilon>0$ be arbitrary. As in Rudin's proof, there is a lower semicontinuous function $g:\left[a,b\right]\to\left(-\infty,\infty\right]$ such that $g>f'$ and $$ \int_{a}^{b}g\left(t\right)\, dt<\int_{a}^{b}f'\left(t\right)\, dt+\varepsilon. $$ Let $\eta>0$ be arbitrary. Define \begin{eqnarray*} F_{\eta}\left(x\right) & := & \int_{a}^{x}g\left(t\right)\, dt-f\left(x\right)+f\left(a\right)+\eta\left(x-a\right),\\ G_{\eta}\left(x\right) & := & F_{\eta}\left(x\right)+\varepsilon\cdot\sum_{\substack{n\in\mathbb{N}\\ x_{n}<x } }2^{-n}. \end{eqnarray*} With these definitions, $F_{\eta}$ is continuous with $F_{\eta}\left(a\right)=0=G_{\eta}\left(a\right)$.

Furthermore, if $z_{n}\uparrow z$, then $F_{\eta}\left(z_{n}\right)\to F_{\eta}\left(z\right)$ and $$ \varepsilon\cdot\sum_{\substack{m\in\mathbb{N}\\ x_{m}<z_{n} } }2^{-m}\leq\varepsilon\cdot\sum_{\substack{m\in\mathbb{N}\\ x_{m}<z } }2^{-m}, $$ which yields $$ \limsup_{n\to\infty}G_{\eta}\left(z_{n}\right)\leq G_{\eta}\left(z\right).\qquad\left(\dagger\right) $$

For $x\in\left[a,b\right)$ there are two cases:

  1. $x=x_{n}$ for some $n\in\mathbb{N}$. By continuity of $F_{\eta}$, there is some $\delta_{x}>0$ such that $F_{\eta}\left(t\right)>F_{\eta}\left(x\right)-\varepsilon\cdot2^{-n}$ holds for all $t\in\left(x,x+\delta_{x}\right)$. For those $t$, we derive \begin{eqnarray*} G_{\eta}\left(t\right)-G_{\eta}\left(x\right) & = & F_{\eta}\left(t\right)-F_{\eta}\left(x\right)+\varepsilon\cdot\sum_{\substack{m\in\mathbb{N}\\ x\leq x_{m}<t } }2^{-m}\\ & > & -\varepsilon\cdot2^{-n}+\varepsilon\cdot\sum_{\substack{m\in\mathbb{N}\\ x\leq x_{m}<t } }2^{-m}\geq0. \end{eqnarray*}

  2. $x\notin\left\{ x_{n}\mid n\in\mathbb{N}\right\} $. By assumption, this implies that $f$ is differentiable from the right at $x$, which means $$ \frac{f\left(t\right)-f\left(x\right)}{t-x}\xrightarrow[t\downarrow x]{}f'\left(x\right)<f'\left(x\right)+\eta. $$ Together with $g\left(x\right)>f'\left(x\right)$ and with the lower semicontinuity of $g$, we see that there is some $\delta_{x}>0$ such that $$ f\left(t\right)-f\left(x\right)<\left(f'\left(x\right)+\eta\right)\cdot\left(t-x\right)\text{ and }g\left(t\right)>f'\left(x\right)\qquad\forall t\in\left(x,x+\delta_{x}\right). $$ Hence, for each $t\in\left(x,x+\delta_{x}\right)$, we get \begin{eqnarray*} G_{\eta}\left(t\right)-G_{\eta}\left(x\right) & = & \varepsilon\cdot\sum_{\substack{m\in\mathbb{N}\\ x\leq x_{m}<t } }2^{-m}+F_{\eta}\left(t\right)-F_{\eta}\left(x\right)\\ & \geq & F_{\eta}\left(t\right)-F_{\eta}\left(x\right)\\ & = & \int_{x}^{t}\underbrace{g\left(s\right)}_{>f'\left(x\right)}\, ds-\left[f\left(t\right)-f\left(x\right)\right]+\eta\left(t-x\right)\\ & > & f'\left(x\right)\cdot\left(t-x\right)-\left[f'\left(x\right)+\eta\right]\left(t-x\right)+\eta\left(t-x\right)=0. \end{eqnarray*}

In summary, for each $x\in\left[a,b\right)$ there is some $\delta_{x}>0$ such that $G_{\eta}\left(t\right)>G_{\eta}\left(x\right)$ for all $x\in\left(x,x+\delta_{x}\right)$. Using $G_{\eta}\left(a\right)=0$, we see $G_{\eta}\left(t\right)\geq0$ for $t\in\left[a,a+\delta_{a}\right)$. Define $$ \varrho:=\sup\left\{ t\in\left(a,b\right)\mid G_{\eta}|_{\left[a,t\right)}\geq0\right\} . $$ It is easy to see that the supremum is actually attained. Using $\left(\dagger\right)$, we also see $G_{\eta}|_{\left[a,\varrho\right]}\geq0$.

If we had $\varrho<b$, the above would yield $G_{\eta}\geq0$ on $\left[a,\varrho\right]\cup\left(\varrho,\varrho+\delta_{\varrho}\right)=\left[a,\varrho+\delta_{\varrho}\right)$, in contradiction to maximality of $\varrho$. Hence, $\varrho=b$ which yields \begin{eqnarray*} 0 & \leq & G_{\eta}\left(b\right)\\ & = & \varepsilon\cdot\sum_{\substack{m\in\mathbb{N}\\ x_{m}<b } }2^{-m}+\int_{a}^{b}g\left(t\right)\, dt-f\left(b\right)+f\left(a\right)+\eta\left(b-a\right)\\ & \leq & \varepsilon+\int_{a}^{b}f'\left(t\right)\, dt+\varepsilon-f\left(b\right)+f\left(a\right)+\eta\left(b-a\right). \end{eqnarray*} Letting $\varepsilon\to0$ and then $\eta\to0$, we conclude $$ f\left(b\right)-f\left(a\right)\leq\int_{a}^{b}f'\left(t\right)\, dt. $$ Now apply the above argument to $-f$ instead of $f$ (note that $-f$ fulfills all assumptions). This yields $$ f\left(b\right)-f\left(a\right)\geq\int_{a}^{b}f'\left(t\right)\, dt $$ and hence $$ \int_{a}^{b}f'\left(t\right)\, dt=f\left(b\right)-f\left(a\right). $$ It is clear that the same argument yields $$ f\left(y\right)-f\left(x\right)=\int_{x}^{y}f'\left(t\right)\, dt $$ for all $x,y\in\left[a,b\right]$.

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  • $\begingroup$ Alright, awesome proof, thanks a lot! $\endgroup$ – user67133 Oct 11 '14 at 16:32
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although not stated this way in books ,I believe it is true after reading page 98 of 'The Integrals of Lebesgue, Denjoy , Perron , and Henstock (Graduate Studies in Mathematics Volume 4 ) by Russell A. Gordon '

LEMMA 6.14. Let F : [a, b] --> R be measurable and let E subset of [a, b] be measurable. If F is differentiable at each point of E, then µ* (F (E)) <=∫E |F'|dx

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  • $\begingroup$ use tex to write math functions $\endgroup$ – Nebo Alex Apr 29 '16 at 21:23

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