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I am studying Measure Theory using the book by Elstrodt and I have been stuck on an exercise for quite a while now.

Let $f: \mathbb{R} \to \mathbb{R}$ be monotonically increasing and $F: \mathbb{R} \to \mathbb{R}, F(x) := \int_0^x f(t) dt$ ($x \in \mathbb{R}$) (Riemann integral). Show that

$$\lim \limits_{h \downarrow 0} \frac{1}{h} (F(x+h) - F(x)) = \lim \limits_{y \downarrow x} f(y), \quad \lim \limits_{h \downarrow 0} \frac{1}{h} (F(x) - F(x-h)) = \lim \limits_{y \uparrow x} f(y).$$

Also show that this implies the following: For every countable subset $A \subset\mathbb{R}$ there is a continuous function $g: \mathbb{R} \to \mathbb{R}$, which is non-differentiable exactly at $x \in A$.

Proof attempt:

Since $f$ is monotone, $f(t) \leq f(x+h)$ and by the properties of the Riemann integral this implies $\int_x^{x+h} f(t) dt \leq f(x+h) \cdot h$. Hence,

$$\lim \limits_{h \downarrow 0} \frac{1}{h} (F(x+h) - F(x)) = \lim \limits_{h \downarrow 0} \frac{1}{h} \int_x^{x+h} f(t) dt \leq \lim \limits_{h \downarrow 0} f(x+h) = \lim \limits_{y \downarrow x} f(y).$$

  1. So in order to prove the first part, I have to show the reverse inequality. How can I do that?
  2. Could anyone give me a hint how to prove the second part about the function $g$?

Thanks a lot!

DerivativesGuy
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1 Answers1

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Let $R = \lim_{y \downarrow x}f(y) = \lim_{h \downarrow 0}f(x+h)$. By definition, for an arbitrary $\varepsilon > 0$, there exists a corresponding $\delta > 0$ such that for $x < y < x+\delta$, we have $|f(y) - R| < \varepsilon$. In particular, for $x < y < x +\delta$, we have $f(y) > R - \varepsilon.$

Now, for $0 < h < \delta$ (so that $x < x+h < x+\delta$), we have $$\frac{1}{h}(F(x+h)-F(x)) = \frac{1}{h}\int_{x}^{x+h}f(t) \geq \frac{1}{h}\int_{x}^{x+h}(R-\varepsilon) = R - \varepsilon.$$ So for any $\varepsilon > 0$ there exists a corresponding $\delta > 0$ so that $0 < h < \delta$ implies $$R - \varepsilon <\frac{1}{h}(F(x+h)-F(x)) < R + \varepsilon,$$ the second inequality following from an analogous argument. It is then clear that $$\lim_{h \downarrow 0} \frac{1}{h}(F(x+h)-F(x)) = R = \lim_{y \downarrow x}f(y)$$ as desired.

Here's a hint for the second part: if the left and right limits $\lim_{y \downarrow x} f(y)$ and $\lim_{y \uparrow x}f(y)$ are equal, then the function $F$ is differentiable at $x$. So consider a monotone function $f$ which has jumps at each of the points $x \in A$, and is continuous outside of $A$. Taking the corresponding $F$ should give you the desired $g$; I leave the details to you.

DerivativesGuy
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ckefa
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  • Thanks for your answer, but by reverse inequality I meant $\lim \limits_{h \downarrow 0} \frac{1}{h} (F(x+h) - F(x)) = \lim \limits_{h \downarrow 0} \frac{1}{h} \int_x^{x+h} f(t) dt \geq \lim \limits_{h \downarrow 0} f(x+h) = \lim \limits_{y \downarrow x} f(y) $ – DerivativesGuy Jan 07 '25 at 21:34
  • @DerivativesGuy Sorry for the misunderstanding. I have changed the answer to address this instead. – ckefa Jan 07 '25 at 21:40
  • No problem and thanks again. It makes a lot of sense... I don't know why I didn't think of that. Still need to think about the second part a bit more and might come back on that later. – DerivativesGuy Jan 07 '25 at 21:50
  • For the second part we can just use an enumeration of the points $x_n \in A$ and define $p(x_n) := 2^{-n}$ and $f(x) := \sum \limits_{y \in A \cap ]0,x]} p(y)$ for $x \geq 0$ and $f(x) := -\sum \limits_{y \in A \cap ]x,0]} p(y)$ for $x<0$. Then $f$ should be monotonically increasing and discontinuous exactly at $x \in A$ (see here https://math.stackexchange.com/questions/4983499/why-is-this-function-discontinuous-exactly-at-x-in-a). Does that make sense? – DerivativesGuy Jan 09 '25 at 09:04
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    @DerivativesGuy That looks good! – ckefa Jan 09 '25 at 12:33