# Define $g$ by $g(q(n))=2^{-n}$ and $f(x)=\sum_{r\in \mathbb{Q}:r<x}g(r)$. Prove $F(x):=\int_0^x f(y)\operatorname{dy}$ is not differentiable at $q$

Let $$g:\mathbb{Q}\to \mathbb{R}$$ by $$g(q(n))=2^{-n}$$ for some bijection $$q:\mathbb{N}\to\mathbb{Q}$$ and $$f:[0,1]\to \mathbb{R}$$ by $$f(x)=\sum_{r\in \mathbb{Q}:r.

Prove that for every rational number $$q\in \mathbb{Q}\cap [0,1]$$ the function $$F:[0,1]\to \mathbb{R}$$ defined by $$F(x):=\int_0^x f(y)\operatorname{dy}$$ is not differentiable at $$q$$

So I assume for contradiction that $$F$$ is differentiable at some $$q\in \mathbb{Q}$$ I've already proven previously that $$f$$ is discontinuous at rationals and is strictly monotone increasing. Since $$f$$ is monotone increasing it is riemann integrable.

Thus $$F'(x)=(\int_0^x f)'(x)=f(x)$$ by fundamental theorem of calculus.

Then $$\lim_{x\to q}\frac{F(x)-F(q)}{x-q}=F'(q)$$ is finite by assumption.

But $$\lim_{x\to q}\frac{F(x)-F(q)}{x-q}=F'(q)=\lim_{x\to q}\frac{\int_0^x f-\int_0^q f}{x-q}=\lim_{x\to q}\frac{\int_x^qf}{x-q}$$ which is by L'hopitals rule, $$\lim_{x\to q} f(x)-f(q)=f(q)$$

From here I'm not really sure what to do. I'm not sure if what I've done is correct, but I am trying to show that $$f$$ is continuous at a rational number.

• There is a strange sum over rationals in your question. Jul 12, 2020 at 17:42
• @JCAA It doesn't look strange to me. $f$ is a form of the cumulative distribution function of the measure that puts mass $2^{-n}$ at the rational number $q(n)$. Jul 12, 2020 at 17:50

It seems to me that what's going on here is already present in the seemingly simpler case of the function $$H(x)=\int_0^x h(t)dt$$, where $$h(t)=0$$ if $$t<0$$ and $$h(t)=1$$ for $$t\ge 0$$. That is, $$H(x)=\begin{cases}0&x<0\\x&x\ge0.\end{cases}$$ The claim is, that $$H$$ is not differentiable at $$0$$. The finite difference quotients $$(H(h)-0)/(h-0)$$ are equal to $$0$$ or $$1$$ according to whether $$h<0$$ or $$h>0$$, and so on. Note that $$h$$ is discontinuous at $$x=0$$, so a naive application of the fundamental theorem of calculus at $$x=0$$ is not justified.
The $$F$$ in the problem at hand is a superposition of this $$H$$: $$F(x)=\sum_n 2^{-n}H(x-q(n))$$, and for each rational $$q\in[0,1]$$ there is a unique $$n$$ such that $$q(n)=q$$, and the term $$2^{-n}H(x-q(n))$$ is not differentiable at $$x=q$$. The difference quotient $$(F(q+h)-F(q))/h$$ evaluates to $$0$$ or $$2^{-n}$$, depending on whether $$h<0$$ or $$h>0$$, and so on.
• Is what I have done completely wrong? I dont think I used L'hopital rule correctly. I was hoping I could show that $\lim_{x\to q} f(x)=f(q)$ which would contradict discontinuity at $q$. Jul 12, 2020 at 19:30