# 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

• Just try to make use of the definition first. In the middle of the proof, you probably need to make use of that fact that $1_{E \cup F} = 1_E + 1_F$ if $E, F$ are disjoint. – BGM Feb 18 at 17:23
• @BGM Thanks! But, I think that my biggest problem is, How to "open the sum" with this kind of integrals, as I already Said, I'm starting with multivariable calculus so I Don't know very well this integrals – BlueRedem1 Feb 18 at 17:26
• I think you really need some measure theory for this, e.g. the monotone convergence theorem. You also have to state precisely what properties you are assuming the sets $B_n$ to have, e.g. Borel sets. – Nate Eldredge Feb 18 at 19:50

\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.