Find a measurable function $g:\mathbb{R}\to\mathbb{R}$ s.t. $\mathbb{E}(g(\mathcal{N}(0,1)))=2$

Question

Let $X\sim\mathcal{N}(0,1)$ a random variable. Find a measurable function $g:\mathbb{R}\to\mathbb{R}$ such that $\mathbb{E}(g(X))=2$.

Answer 1

We know that the density function of $X$ is

$f_X(s)={1\over\sqrt{2\pi}}e^{-s^2\over 2}$. Let $g:\mathbb{R}\to\mathbb{R}$ a function defined by $$ \\ g(s)=\sqrt{2\pi}e^{s^2\over 2}\cdot\textbf{1}_{\{0\leq s\leq 2\}}\ \ $$ $s\mapsto \textbf{1}_{\{0\leq s \leq 2\} }$ is a measurable function because it's a linear combination of indicators of borel sets, $s\mapsto\sqrt{2\pi}e^{s^2\over 2}$ is a measurable function because it's continuous, and $g$ is a measurable function because it's a composition of measurable functions. Thus, $$ \\ \mathbb{E}(g(X))=\int_{-\infty}^\infty g(s)f_X(s)ds=\int_0^2ds=2 \ $$