Let $(\Omega, \mathcal F, \mathbb P)$ be a probability measure space. Let $X$ and $Y$ be two independent random variables on it. Then $\mathrm E [XY] = \mathrm E [X] \mathrm E [Y].$
I tried to do it in a measure theoretical way. By definition, we have $$\mathrm E [XY] = \int_{\Omega} X(\omega) Y(\omega)\ \mathrm d \mathbb P(\omega)$$
Consider the random vector $(X,Y) : \Omega \longrightarrow \mathbb R^2$ defined by $\omega \mapsto (X(\omega), Y(\omega)).$ Then for any Borel measurable function $f : \mathbb R^2 \longrightarrow \mathbb R$ we have $$\tag{1}\int\int_{\mathbb R^2} f(x,y)\ \mathrm d \mathbb P_{(X,Y)} (x,y) = \int_{\Omega} f(X(\omega), Y(\omega))\ \mathrm d \mathbb P(\omega)$$ where $\mathbb P_{(X,Y)}$ is a probability measure on $\mathbb R^2$ defined by $\mathbb P_{(X,Y)} (E) = \mathbb P ((X,Y)^{-1} (E)),$ for any $E \in \mathcal B_{\mathbb R^2}.$ Now define $f : \mathbb R^2 \longrightarrow \mathbb R$ by $f(x,y) = xy,\ x, y \in \mathbb R.$ Then $(1)$ becomes $$\int_{\Omega} X(\omega) Y(\omega)\ \mathrm d \mathbb P (\omega) = \int\int_{\mathbb R^2} xy\ \mathrm d \mathbb P_{(X,Y)} (x,y).$$ Now $\mathbb P_{(X,Y)} (A \times B) = \mathbb P (X \in A, Y \in B) = \mathbb P (X \in A) \mathbb P (Y \in B) = \mathbb P_X(A) \mathbb P_Y(B),$ for any $A,B \in \mathcal B_{\mathbb R}.$ From here can we conclude that $\mathbb P_{(X,Y)} = \mathbb P_X \times \mathbb P_Y\ $? For that we need to show that $$\tag{*} \mathbb P_{(X,Y)} (E) = \int\int_{E} \mathrm d \left (\mathbb P_X \times \mathbb P_Y \right ) (x,y)$$ for any $E \in \mathcal B_{\mathbb R^2}.$ I am trying to apply Dynkin $\pi$-$\lambda$ theorem here. We consider the collections $\mathcal C : = \left \{A \times B\ |\ A, B \in \mathcal B_{\mathbb R^2} \right \}$ and $\mathcal D : = \left \{E \subseteq \mathbb R^2\ |\ (\ast)\ \text{holds} \right \}.$ Then clearly $\mathcal C$ is a $\pi$-system and $\mathcal D$ is a $\lambda$-system and hence $\mathcal B_{\mathbb R^2} = \sigma (\mathcal C) \subseteq \mathcal D,$ which is what we wanted to show. So we have $$\begin{align*} \mathrm E[XY] & = \left (\int_{\mathbb R} x\ \mathrm d \mathbb P_{X} (x) \right ) \left (\int_{\mathbb R} y\ \mathrm d \mathbb P_{Y} (y) \right ) \\ & = \left (\int_{\Omega} X(\omega)\ \mathrm d \mathbb P (\omega) \right ) \left (\int_{\Omega} Y(\omega)\ \mathrm d \mathbb P (\omega) \right ) \\ & = \mathrm E[X] \mathrm E[Y]. \end{align*}$$
Is my argument fine? Any suggestion for correction or improvement would be greatly appreciated.
Thanks for your time.