# Construct a monotone function which has countably many discontinuities

I read in a textbook, which had seemed to have other dubious errors, that one may construct a monotone function with discontinuities at every point in a countable set $C \subset [a,b]$ by enumerating the points as $c_1, c_2, \dots$ and defining $f(x) = \sum_{c_n < x}2^{-n}$. However, if seems that if we let $[a,b] = [0,1], C = \mathbb{Q} \cap [0,1]$, then $f(x)$ is constant $1$ everywhere except 0, an apparent counterexample.

So my question is: how does one construct a monotonic function which has discontinuities precisely on a countable set $C$? Further, are there any relatively easy-to-visualize constructions?

• "then f(x) is constant 1 everywhere except 0" - I don't think this claim is correct. In fact, $f$ will be strictly smaller than 1 everywhere. Oct 2, 2011 at 19:14
• What are the summation limits and index on the example? Oct 2, 2011 at 19:15
• I see. This depends on the enumeration. Great! Question answered. Oct 2, 2011 at 19:26
• Since $\sum_{n=10}^\infty 2^{-n}$ is so small, you can see almost exactly what your function will look like if you plot it (with Matlab, say) using only the ten first rationals in your enumeration. Oct 2, 2011 at 19:38

The construction is correct. I’ll use your example as an illustration. Let $\{q_n:n\in\omega\}$ be an enumeration of $\mathbb{Q}\cap [0,1]$, and let $f(x)=\sum\limits_{q_n<x}2^{-n}$ for $x\in [0,1]$.
First consider what happens at some $q_m$: $$\lim\limits_{x\to {q_m}^-}f(x) = \sum_{q_n<q_m}2^{-n}=f(q_m),$$ because as $x$ moves up towards $q_m$, $\{q_n:q_n<x\}$ includes more and more of the rationals less than $q_m$. Thus, $f$ is continuous from the left at $q_m$, but for every $x>q_m$ we have $$f(x)=\sum_{q_n<x}2^{-n} \ge \sum_{q_n\le q_m}2^{-n} = f(q_m)+2^{-m},$$ so $f$ jumps by at least $2^{-m}$ at $q_m$. In fact $$\lim_{x\to {q_m}^+}f(x) = f(q_m)+2^{-m},$$ and the jump is exactly $2^{-m}$.
At each irrational $a \in [0,1]$, however, $f$ is easily seen to be continuous: $$\lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) = \sum_{q_n<x}2^{-n}=f(x).$$
• @Bean: That it does, very definitely. So you get a bonus: since there are $2^\omega$ enumerations of a countably infinite set, you get $2^\omega$ examples for a given set of points of discontinuity for the price of one. :-) Oct 2, 2011 at 19:47