I am needing such a "result", which I do not know if it is true, for another question, Convolution with Gaussian question.
Let $f\in L^\infty(\mathbb R)$ be of bounded variation on any interval $[a,b]$. Let $g$ be integrable such that $\int g=1$. Then $$\lim_{\epsilon\to 0^+}\int f(\epsilon t) g(t)\,dt=\frac{f(0^+)+f(0^-)}{2}.$$
Is this theorem true? If so how do we prove it?
No, this is not so. For example, let $$g(x)=\begin{cases}1,\quad & x\in [0,1] \\ 0,\quad & \text{otherwise}\end{cases}$$ Then $$ \int_{\mathbb{R}} f(\epsilon t)g(t)\,dt = \int_{0}^1 f(\epsilon t)\,dt \to f(0^+) $$ The correct statement is: let $A= \int_{-\infty}^0 g$, $ B=\int_0^\infty g$; then
$$ \lim_{\epsilon\to0^+}\int_{\mathbb{R}} f(\epsilon t)g(t)\,dt = Af(0^-)+Bf(0^+) $$ To prove this, deal with the positive and negative half-lines separately. Further, for each of them split integration according to the size of $t$.
Given $\delta>0$, for sufficiently large $M$, $$ \left|\int_0^M (f(\epsilon t)-f(0^+)) g(t)\,dt\right| < \delta $$ because the first factor is bounded and the second has small integral.
Also, $$ \int_0^M (f(\epsilon t)-f(0^+)) g(t)\,dt \to 0\quad \text{as } \ \epsilon\to 0^+ $$ because the first factor tends to zero uniformly. Thus, $$ \int_0^\infty f(\epsilon t) g(t)\,dt \to Af(0^+)$$