Let $X\in L_2$ be a random variable and $g$ a positive real function. Let $I$ be an interval and $b>0$, and suppose that $\forall x\in I\ g(x)>b$. I have to show that:
$$\operatorname E(g(X))\geq \operatorname P(X\in I) b$$
Well,
$$\operatorname E(g(X))=\int_\Omega g(X)\ d\operatorname P\geq \int_\Omega g(X)\chi_{X^{-1}(I)}d\operatorname P$$
Where $\chi_{X^{-1}(I)}$ is the characteristic function of $X^{-1}(I)$.
$$\geq b\int_\Omega\chi_{X^{-1}(I)}d\operatorname P=b\operatorname P(X^{-1}(I))=b\operatorname P(X\in I)$$
If this is correct, then where did I use $X\in L_2$?