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.
 A: 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 $\Bbb R_\ell$.
A: $\mathbb R$ \ $A=\cup G$ 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.
