Show that $\Phi_\epsilon(x-a)\cdot\Phi_\epsilon(b-x)$ converges to an index function $I\{a\le x\le b\}$

Question

Let $$f(x)=I\{a\le x\le b\}$$ and $$f_\epsilon(x)=\Phi_\epsilon(x-a)\cdot\Phi_\epsilon(b-x),$$ where $\Phi_\epsilon(z)$ is the normal cumulative distribution with mean zero and variance $\epsilon^2$. Then as $\epsilon\to0$, $$f_\epsilon(x)\to f(x).$$

Is the above stamtement correct and how to prove it?

Answer 1

When mean=variance=$0$, normal distribution degenerates into single point distribtution, that is, $\Phi_\epsilon(x-a)\to I\{x\ge a\}$ and $\Phi_\epsilon(b-x)\to I\{b\ge x\}$, as $\epsilon\to0$. So $$\Phi_\epsilon(x-a)\cdot\Phi_\epsilon(b-x)\to I\{x\ge a\}\cdot I\{b\ge x\}=I\{a\le x\le b\}.$$