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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 $\epsilon>0$?

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    $\begingroup$ Since you did not get answers, it might be a good idea to add the definition of epi-convergence. $\endgroup$ Mar 10, 2014 at 12:34

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