In this particular case, you should only apply that $EX=\int_\Omega\,w^2\,P(dw)=\int_{\Omega}\,w^2\,w\,dw$ by applying the next theorem:
Theorem. Let $(X,\mathcal{A})$ be a measurable space, and $\mu$ a measure on it and $f:X\rightarrow [0,\infty]$ a measurable function. Then \begin{align} \varphi:\mathcal{A}&\rightarrow [0,\infty]\newline A&\mapsto \int_{A}\,f\,d\mu \end{align} is a measure over $(X,\mathcal{A})$, and for every measurable function $g:X\rightarrow [0,\infty]$ we have $\int_{\Omega}\,g\,d\varphi=\int_{\Omega}\,g\cdot f\,d\mu$
Proof: For showing that it is a measure, we only have to show $\sigma$-additivity. So, let $\{E_n\}$ be a denumerable family of disjoint sets of $\mathcal{A}$, then $\varphi\left(\bigcup E_n\right)=\int_{\bigcup E_n}\,f\,d\mu=\int_{\Omega}\,\chi_{\bigcup E_n}\cdot f\,d\mu$ $=\int_{\Omega}\,\sum\chi_{E_n}\cdot f\,d\mu=\sum\int_{\Omega}\chi_{E_n}\cdot f\,d\mu=\sum\int_{E_n}f\,d\mu=\sum\varphi(E_n)$ thanks to Lebesgue's Monotone Convergence Theorem. Finally, the second part is true by the previous part for characteristic functions, so it will be also true for simple functions. And, so by Lebesgue's Monotone Convergence Theorem for arbitrary measurable functions.