Let the indicator function be defined as $$I(x) \triangleq \begin{cases} 1, & \quad x \geq 0 \\ 0, & \quad x < 0 \end{cases}$$ and $I_{\nu}(x) \in [0,1]$ be a continuous approximation of $I(x)$ with the following properties: $$\lim_{\nu \rightarrow \infty} I_{\nu}(x) = I(x), \quad \forall x$$ $$\lim_{x \rightarrow - \infty} I_{\nu}(x) = 0, \quad \forall \nu$$ $$\lim_{x \rightarrow \infty} I_{\nu}(x) = 1, \quad \forall \nu.$$
1) How do I show that $${\textrm{e-lim}}_{\nu \to \infty} I_{\nu}(x) = I(x)$$ i.e., $I_{\nu}(x)$ epi-converges to $I(x)$?
2) Is it necessary for I(x) to be lower-semicontinuous? If yes, can I just redefine the indicator as $$I(x) \triangleq \begin{cases} 1, & \quad x > -\epsilon \\ 0, & \quad x \leq -\epsilon \end{cases}$$ for some sufficiently small $\epsilon>0$?
