# Increasing function on R that is discontinuous on the rationals

The question (Folland's Real Analysis, 3.5.30) asks to produce an increasing function on $$\mathbb{R}$$ whose set of discontinuities is the rationals. My train of thought is as follows:

Let $$f_0$$ be the identity. We want to create a jump at every rational number, while keeping the function increasing. Create $$f_1$$ by cutting $$f_0$$ at each integer, and halving the slope of each resulting line segment, fixing the left endpoints. This results in a sort of slanted stair that is still increasing and discontinuous exactly at the integers. Given $$f_{n-1}$$, we create $$f_n$$ by cutting each segment of $$f_{n-1}$$ into $$n$$ parts and halving the slope of each resulting segment, again fixing the left endpoints. Note: this is equivalent to making cuts at all rational points $$\frac{m}{n!}$$ at the $$n^\text{th}$$ step, with $$m$$ and $$n$$ not necessarily coprime.

The limit of such sequence of functions exists. It's easy to prove that any rational point is eventually a left endpoint, and is thus kept fixed by all successive functions in the sequence. For an irrational point $$x$$ we can make two sequences $$a_n$$ and $$b_n$$ of rationals converging from the left and right respectively, such that $$a_n=\frac{m}{n!} and $$b_n=\frac{m+1}{n!}>x$$. Given $$\epsilon$$, we can then choose $$N$$ large enough with $$a_N$$ and $$b_N$$ on the same segment of $$f_{N-1}$$ (so the jump from $$f_{N-1}(a_N)$$ to $$f_{N-1}(b_N)$$ is small), with this jump being less than $$\epsilon$$. Since we can pinpoint the value of $$f_n(x)$$ between arbitrarily close values, the limit $$f(x)=\lim_n f_n(x)$$ exists.

It's clear that the function is discontinuous at all rationals since it is constructed to be, and it is increasing since each $$f_n$$ is increasing.

Can I say that $$f$$ is continuous at each irrational? My gut feeling is that, since any $$\delta$$-ball around an irrational contains a rational to the left and one to the right, we cannot necessarily say $$f$$ is continuous there. However we know by theorem 3.23 in the book that the set of discontinuities of an increasing function is countable. Is there a contradiction somewhere? If so, can the construction be tweaked or is a different construction necessary?

Thank you.

• Your construction looks promising. A function can be continuous at $x$ even if every neighbourhood of $x$ contains a point of discontinuity (e.g., let $f(x)$ be $0$ if $x$ is rational and be $x$ if $x$ is irrational, then $f$ is continuous at $0$). Try constructing the $\epsilon$-$\delta$ proof that your function is continuous at irrational numbers. Nov 21, 2018 at 23:55
• What I'm thinking is this: let $\epsilon>0$, assume there exists $\delta$ such that in a $\delta$-ball around $x$, $|f(x)-f(y)|<\epsilon$. Take two such rationals $p$ and $q$; then we have $|f(p)-f(q)|<2\epsilon$ by triangle inequality. Is this a problem? Nov 21, 2018 at 23:56
• I don't think so. Write the $\epsilon$-$\delta$ argument down and check it for yourself. Nov 22, 2018 at 0:01
• I upvoted the question since I think your approach will work. Would you be interested in another function having the required properties? [I have one for which the properties are somewhat simple to show.] Nov 22, 2018 at 4:35
• Sure, go ahead! Nov 22, 2018 at 7:17

There is a simpler construction:

If we take a numeration $$(q_n)_{n \in \mathbb{N}}$$ of $$\mathbb{Q}$$, we can define $$f(x) := \sum_{k=1}^\infty \frac{1}{2^k} 1_{[q_k, \infty)}(x).$$ The convergence is uniform and this function is monotone increasing by construction. Since all $$1_{[q_n,\infty)}$$ are continuous in all points of $$\mathbb{R} \setminus \mathbb{Q}$$, the limes is this too: If $$x$$ is irrational, then we can take $$\varepsilon >0$$ so small that $$(x-\varepsilon,x+\varepsilon)$$ doesn't contain the points $$q_1,\ldots,q_n$$. So $$|f(x)-f(y)| \le \sum_{k=n}^\infty 2^{-k} = 2^{1-n}.$$

Fix $$n$$. For any $$\max_{q_i < q_n,i=1,\ldots n} q_i < x < q_n < y < \max_{q_i > q_n,i =1 \ldots, n} q_i$$we have $$f(x) + 2^{-n} < f(y)$$. Thus $$f$$ is discontinuous in any point of $$\mathbb{Q}$$.

• I'm having such a hard time visualizing a function constructed like this. The proof is sound though.
– Rchn
Nov 22, 2018 at 10:38
• In fact, you cannot really visualize this function. The key point is that we only add a jump of height $2^{-n}$ in every step. Define $g(x) = \sum_{k=1}^\infty \frac{1}{2^k} 1_{[1/k,\infty)}$. This function has in $x=0$ the same behaviour (and can be visualized): In every neighboorhoud $(-\varepsilon,\varepsilon)$ we have infinite many jump-discontinuities. However, $g$ is continuous in $x=0$! Nov 22, 2018 at 20:49
• @p4sch How did you get that $f(x)+2^{-n}<f(y)$? Nov 19, 2020 at 13:11
• Since $x < q_n < y$, the sum defining $f(y)$ counts the jump occuring from $q_n$ with height $2^{-n}$, while $f(x)$ does not count this jump. Of course, there are other rational numbers between $q_n < y$ and thus it is a strict inequality. Nov 20, 2020 at 10:45
• @p4sch Great example! Just out of curiosity: Define the inverse of $f$ by $f^{-1}(y)=\inf\{x:f(x)>y\}$. Then $f^{-1}$ should be continuous on $(0,1)$ with flat regions. But the length of each flat region seems arbitrarily small. Can we say $f^{-1}$ is strictly increasing everywhere? Apr 21 at 2:49