Definition: Let $[a,b]\subset \mathbb{R}$ and let $B$ be a Banach space. Function $f:[a,b]\to B$ is regulated if there exist $f^+(t_0)=\lim_{t \to t_0^+}{f(t)}$ and $f^-(t_0)=\lim_{t \to t_0^-}{f(t)}$, for every $t_0\in [a,b]$.
Characterization: Function $f:[a,b]\to B$ is regulated if and only if there exist a sequence of step functions (piecewise constant functions) $s_n:[a,b]\to B$ that converges to $f$ uniformly.
I tried proving that if $f$ is regulated, then sequence $(s_n)_{n\in \mathbb{N}}$ exists by defining $(s_n)_{n\in \mathbb{N}}$ in the following way:
$s_1$ is constant on $[a,b]$ and $s_1(t)=\frac{f^+(a)+f^-(b)}{2}$ and similarly:
Let $n\in \mathbb{N}$ and $t_0,\dots ,t_n \in [a,b]$, $t_k=a+k\frac{b-a}{n}, k\in \{1, \dots ,n\}$. For all $k\in \{1, \dots ,n-1\}$ $s_n$ is constant on $[t_k,t_{k+1}]$ and $s_n(t)=\frac{f^+(t_{k-1})+f^-(t_k)}{2}, t\in [t_k,t_{k+1}]$.
And I don't know what to do now. I'm not sure this idea is good and I don't know how to prove that this converges uniformly to $f$.
As for the proof of the other implication, I'm not sure I know how to do that either. I guess we could say that if there exists $(s_n)_{n\in \mathbb{N}}$ that converges uniformly to $f$, then $(\forall t\in [a,b]) \lim_{n\to\infty} {s_n(t)}=f(t)$. So we have: $$f^+(t_0)=\lim_{t\to t_0^+}{\lim_{n\to\infty}{s_n(t)}}=\lim_{n\to\infty}{\lim_{t\to t_0^+}{s_n(t)}}$$ (limits can switch places because $(s_n)$ converges uniformly?)
$\lim_{t\to t_0^+}{s_n(t)}$ should exist because $s_n$ is a step function, but how to prove that the double limit exists?