We recall that $\mu ( x ) = \frac { 1 } { \sqrt { \pi } } e ^ { - x ^ { 2 } }$ is a measure, and that for $A \in \operatorname{Int}$, $\mu(A) = \int \mathbb{1}_A(x) \mu(x) dx$. We denote $d(x, A) = \inf\{|x - y| : y \in A\}$. Let $A \subset \mathbb { R }$. 15a. Show that for all $x , y \in \mathbb { R }$, we have $$\exp \left( \frac { 1 } { 2 } d ( x , A ) ^ { 2 } - x ^ { 2 } \right) \mathbb { 1 } _ { A } ( y ) \exp \left( - y ^ { 2 } \right) \leqslant \exp \left( - \frac { ( x + y ) ^ { 2 } } { 2 } \right)$$ 15b. We assume that $A \in \operatorname { Int }$ and that $\mu ( A ) > 0$. Deduce that $$\int \exp \left( \frac { 1 } { 2 } d ( x , A ) ^ { 2 } \right) \mu ( x ) d x \leqslant \frac { 1 } { \mu ( A ) }$$
We recall that $\mu ( x ) = \frac { 1 } { \sqrt { \pi } } e ^ { - x ^ { 2 } }$ is a measure, and that for $A \in \operatorname{Int}$, $\mu(A) = \int \mathbb{1}_A(x) \mu(x) dx$. We denote $d(x, A) = \inf\{|x - y| : y \in A\}$. Let $A \subset \mathbb { R }$.
15a. Show that for all $x , y \in \mathbb { R }$, we have
$$\exp \left( \frac { 1 } { 2 } d ( x , A ) ^ { 2 } - x ^ { 2 } \right) \mathbb { 1 } _ { A } ( y ) \exp \left( - y ^ { 2 } \right) \leqslant \exp \left( - \frac { ( x + y ) ^ { 2 } } { 2 } \right)$$
15b. We assume that $A \in \operatorname { Int }$ and that $\mu ( A ) > 0$. Deduce that
$$\int \exp \left( \frac { 1 } { 2 } d ( x , A ) ^ { 2 } \right) \mu ( x ) d x \leqslant \frac { 1 } { \mu ( A ) }$$