I have seen in this question howto prove whether sin$(1/x)$ is not regulated. But i'm not quite sure why it's correct. Since the definition of a regulated function is as follows:
This means that the negation of this definition is:
$f$ is not regulated if $\forall \phi \in S[a,b] \text{ there exists } \epsilon : ||f-\phi||_{\infty} \gt \epsilon$
But the answer in the question above gives that $\epsilon=1$. How are we sure that for any sequence of step functions, the condition $||f-\phi||_{\infty} \gt \epsilon$ is met? I mean isn't the condition met only when sin$(1/x)$ is equal to $1$ (or $-1$) and $\phi(x)=0$? What if I choose a step function that does not equal $0$ ever? Then how is the absolute difference going to be bigger or equal to $1$?
Well the idea is that you have sequences of points $\{x_n \}$, $\{ y_n\}$ in $[0,1]$ such that $x_n,y_n\rightarrow 0$, $f(x_n)\equiv 1$ and $f(y_n)\equiv-1$. In particular for any step function $\phi$, has to be equal to some constant $c$ on some interval $(0,\delta)$. In particular:
$$ \Vert f-\phi \Vert_\infty\geq \begin{cases} \vert c-f(x_n) \vert & c\leq0 \\ \vert c-f(y_n) \vert & c>0 \end{cases}\geq 1 $$
I think this proof shows in general that you can't approximate a function with an essential discontinuity point by steps functions uniformly.