# If $A\subseteq\mathbb R$ is closed and $f:A\to\mathbb R$ is right-continuous, is there a right-continuous extension of $f$ to $\mathbb R$?

Let $(\Omega,\tau)$ be a normal topological space, $A\subseteq\Omega$ be $\tau$-closed and $f:A\to\mathbb R$ be $\tau$-continuous. By Tietze's extension theorem, $f$ can be extended to a $\tau$-continuous function on $\Omega$.

Is there an analogue result for the case where $\Omega=\mathbb R$ and $f$ is only right-continuous?

I couldn't find any reference for that.

Consider $\Bbb R_\ell$, the lower-limit topology on $\Bbb R$. (This has as basis elements $[a,b)$ for all $a<b$.) You can check that $f\colon \Bbb R\to\Bbb R$ is right-continuous if and only if $f\colon\Bbb R_\ell\to\Bbb R$ is a continuous map of topological spaces. It's a standard exercise that $\Bbb R_\ell$ is normal. Thus, your desired result follows from the usual formulation of the Tietze extension theorem. (And similarly for left-continuous functions.)

By the way, since the lower-limit topology is finer than the usual, if $A$ is closed in $\Bbb R$, then $A$ is closed in $\Bbb R_\ell$.

• Will the extension preserve monotonicity? – 0xbadf00d Jul 8 '17 at 12:28
• What do you mean, precisely? – Ted Shifrin Jul 8 '17 at 14:20
• If $A\subseteq\mathbb R$ is closed and $f:A\to\mathbb R$ is nondecreasing and right-continuous, is there a nondecreasing right-continuous extension of $f$ to $\mathbb R$? – 0xbadf00d Jul 8 '17 at 14:38
• Not sure. You'll have to do a careful proof of Urysohn and Tietze to build that in. – Ted Shifrin Jul 8 '17 at 16:06
• If $A=[a,b]$, we can simply put $$\tilde f(x):=\begin{cases}f(a)&\text{, if }x\le a\\f(x)&\text{, if }x\in(a,b)\\f(b)&\text{, if }x\ge b\end{cases}\;.$$ The same is possible for $A=[a,\infty)$ and $A=(-\infty,b]$. And it's also easy to construct such an extension, if $A$ is dense. I don't have the time to study Tietze's proof in detail, but thank you anyway for your response. – 0xbadf00d Jul 8 '17 at 20:20

$\mathbb R$ \ $A=\cup G$ where $G$ is a family of pairwise-disjoint non-empty open intervals. (I include open half-lines and $\mathbb R$ itself among the open intervals.)

For $g=(a,b)\in G$ with $-\infty < a<b \leq \infty$ we have $a\in A.$ For $x\in g$ let $f(x)=f(a).$

For $g=(-\infty,b)$ with $b\leq \infty ,$ for $x\in g$ let $f(x)=0.$

Not as sophisticated as Ted Shifrin's elegant A but it works.