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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 $a_n=\frac{m}{n!}<x$ 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.

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    $\begingroup$ 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. $\endgroup$
    – Rob Arthan
    Nov 21, 2018 at 23:55
  • $\begingroup$ 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? $\endgroup$
    – Luca S.
    Nov 21, 2018 at 23:56
  • $\begingroup$ I don't think so. Write the $\epsilon$-$\delta$ argument down and check it for yourself. $\endgroup$
    – Rob Arthan
    Nov 22, 2018 at 0:01
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    $\begingroup$ 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.] $\endgroup$
    – coffeemath
    Nov 22, 2018 at 4:35
  • $\begingroup$ Sure, go ahead! $\endgroup$
    – Luca S.
    Nov 22, 2018 at 7:17

1 Answer 1

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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 $\mathbb{Q}$.

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  • $\begingroup$ I'm having such a hard time visualizing a function constructed like this. The proof is sound though. $\endgroup$
    – Rchn
    Nov 22, 2018 at 10:38
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    $\begingroup$ 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 $x=0$! $\endgroup$
    – p4sch
    Nov 22, 2018 at 20:49
  • $\begingroup$ @p4sch How did you get that $f(x)+2^{-n}<f(y)$? $\endgroup$
    – user330477
    Nov 19, 2020 at 13:11
  • $\begingroup$ 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. $\endgroup$
    – p4sch
    Nov 20, 2020 at 10:45
  • $\begingroup$ @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? $\endgroup$
    – Ecthelion
    Apr 21 at 2:49

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