# Making pathological functions using measure theory

18 Dec 2016Over the past year I’ve been studying measure theory and functional analysis at UC Berkeley. One of the best parts of this subject is getting a glimmer of *just how weird infinity is*. **Assuming some measure theory knowledge**, we’re going to do the following today.

**Construct a nondecreasing function $f:\mathbb{R}\to\mathbb{R}$
with the property that $f$ is discontinuous at $x\in\mathbb{R}$
if and only if $x\in\mathbb{Q}$.**

*Pretty pathological right?* We just need to count infinities.

We will define a function $H$ which is continuous from the left at ever $x \in [0,1]$ but is only continuous from the right at the irrationals. To build such a function we need the folliowing scaffolding.

First let $\psi: \mathbb{N} \to \mathbb{Q} \cap [0,1]$ be a bijection enumerating the rationals in the interval $[0,1]$. Then define $B(x) = {n : \psi(n)< x}$ or equivalently $B(x) = \psi^{-1}([0,x))$. Finally let $\mu: P(\mathbb{N}) \to \overline{\mathbb{R}}$ be the counting measure. We additionally define a measure $\nu: P(\mathbb{N}) \to \mathbb{R}$ such that

The measure $\nu$ has the additional property that $\nu \ll \mu$ and

We claim that the function $H(x) = \nu(B(x))$ has the properties of $f$ in the statement of the problem. We will first show that for every $x \in [0,1]$ the function $H(x)$ is left continuous. Take a sequence of $x_k \to x$ from the left, we can then rearrange the sequence to be strict monotonic. It follows that if $k > m$ then $B(x_m) = {n: \psi(n) < x_m < x_k < x} \subset {n: \psi(n) < x_k < x} = B(x_k)$. By the finiteness of $\nu$ we have that by upward measure continuity

Note that if $ n \in \bigcup B(x_k)$ there is an $K$ so that $\psi(n) < x_k < x $ so any $n$ with $\psi(n) < x$ is in $\bigcup B(x_k).$

Next we claim that $H$ is only right continuous only when $x$ is irrational. Take a sequence of $x_k \to x$ from the right and rearrange the sequence to be strict montonic. It follows that if $k > m$ then $B(x_m) = {n: \psi(n) < x_m} \supset {n: \psi(n) < x_k < x_m} = B(x_k)$. By finiteness of $\nu$ and downard measure continuity

If $x$ is irrational then $m \in {n: \psi(n) < x_k\ \forall k}$ implies that $\psi(m) < x$ and if $\psi(m) < x$ then $\psi(m) < x_k$ for all $k$ so ${n: \psi(n) < x_k\ \forall k} = B(x)$ and $H(x_k) \to H(x)$ from the right. If $x$ is rational then $x= \psi(q)$ for some $q \in \mathbb{N}.$ Thus $x < x_k \forall k$ implies that ${n: \psi(n) < x_k\ \forall k} = B(x) + {q} = D$. It follows that $\nu(D) = \nu(B(x)) + 2^{-q} > H(x)$. So $H(x_k) \to H(x) + 2^{-q} \neq H(x)$ from the right, and so $H$ is not right continuous at the rationals.

We have thus shown that for any $x \in [0,1] \setminus \mathbb{Q}$, any sequence $x_k \to x$ has the property $\lim H(x_k) = x$ from the left and the right, and if $x \in [0,1] \cap \mathbb{Q}$ then if $x_k \to x$, $\lim H(x_k)$ does not exist. Therefore $H$ is continuous at every irrational and discontinuous at every rational.