The volume as a probability measure: Stuck in the proof I have troubles proving the fact that if we define $Vol(A)=\idotsint 1_{A}(x_1,\dots,x_n) \,dx_1 \dots dx_n$ and  $\mathbb{P}(B)=\frac{Vol(A\bigcap B)}{Vol(A)}$ then $\mathbb{P}(\bigcup\limits_{n=1}^{\infty} B_{n})= \sum_{n=1}^{\infty} P(B_{k})$ if $B_{1}, B_{2}\dots$ are disjoint. I'm starting with multivariable calculus, so I don't know how to formalize this. I know that this happens because the function $1_{A}=1$ iff what we want to integrate is part of the set, and it makes sense that is the sum because we are not "counting twice" the same points, because the fact that our $B_{k}$ sets are disjoints, but as I already said, I don't know how to work with integrals in $\mathbb{R}^{n}$ so it is hard to me to write this proof in a formal way. I will appreciate any help
 A: \begin{align*}
{\rm Vol}(A){\mathbb P}(\cup_{n=1}^{\infty}B_{n})
&={\rm Vol}(A \cap (\cup_{n=1}^{\infty}B_{n})) \\
&={\rm Vol}(\cup_{n=1}^{\infty}(A \cap B_{n})) \\
&=\int_{{\mathbb R}}\int_{{\mathbb R}} \ldots \int_{{\mathbb R}}{\bf 1}_{\cup_{n=1}^{\infty}(A \cap B_{n})}(x_{1},x_{2},\ldots,x_{n}){\rm d}x_{1}{\rm d}x_{2} \ldots {\rm d}x_{n} \\
&=\int_{{\mathbb R}}\int_{{\mathbb R}} \ldots \int_{{\mathbb R}}\sum_{n=1}^{\infty}{\bf 1}_{A \cap B_{n}}(x_{1},x_{2},\ldots,x_{n}){\rm d}x_{1}{\rm d}x_{2} \ldots {\rm d}x_{n} \\
&=\int_{{\mathbb R}}\int_{{\mathbb R}} \ldots \int_{{\mathbb R}}\uparrow \lim_{N \uparrow \infty}\sum_{n=1}^{N}{\bf 1}_{A \cap B_{n}}(x_{1},x_{2},\ldots,x_{n}){\rm d}x_{1}{\rm d}x_{2} \ldots {\rm d}x_{n} \\
&=\uparrow \lim_{N \uparrow \infty}\int_{{\mathbb R}}\int_{{\mathbb R}} \ldots \int_{{\mathbb R}}\sum_{n=1}^{N}{\bf 1}_{A \cap B_{n}}(x_{1},x_{2},\ldots,x_{n}){\rm d}x_{1}{\rm d}x_{2} \ldots {\rm d}x_{n} \\
&=\uparrow \lim_{N \uparrow \infty}\sum_{n=1}^{N}\int_{{\mathbb R}}\int_{{\mathbb R}} \ldots \int_{{\mathbb R}}{\bf 1}_{A \cap B_{n}}(x_{1},x_{2},\ldots,x_{n}){\rm d}x_{1}{\rm d}x_{2} \ldots {\rm d}x_{n} \\
&=\sum_{n=1}^{\infty}{\rm Vol}(A \cap B_{n}).
\end{align*}
In the 6th, we used Monotone convergence theorem.
