Iverson bracket - infinite additivity for pairwise disjoint sets Suppose we have a sequence of mutually (pairwise) disjoint sets/events $B_1, B_2, B_3, ... $
EDIT: The sets $B_i$ ($i=1,2,3,\dots$) are subsets of some set $\Omega$.
For the Iverson bracket, is the following property true for every $w \in Omega$
$\sum_{n=1}^\infty  [w \in B_n] = [w \in \bigcup\limits_{n=1}^{\infty} B_{n}] \tag{*}$
I think it is true, and I think I know how to prove it.
Seems simple. The sum on the left is a series which contains
a) either only zeroes OR
b) all zeroes except for a single one
And then it becomes obvious that the LHS series is convergent and its sum is either zero or one.
OK, then the bracket on the RHS, it's either zero or one, and it's equal to one exactly when the sum in the LHS is equal one (that follows from a set theoretic argument).
Is the property $(*)$ valid? And is this proof OK?
 A: The property (*) is valid and the reasoning is comprehensible. What is missing is a suitable setting of the situation to make the proof valid. Here is a proposal:

Let $X$ be a set with $B_j\subseteq X, j\geq 1$, so that the sets $B_j$ are pairwise disjoint. If $w\in X$ the following is valid
\begin{align*}
\color{blue}{\sum_{n=1}^{\infty}\left[w\in B_n\right]=[w\in\bigcup_{n=1}^{\infty} B_n]}\tag{1}
\end{align*}

Proof: Two cases are to consider.

*

*Case 1: $w\in \bigcup_{n=1}^{\infty}B_n$. We obtain
\begin{align*}
  &w\in \bigcup_{n= 1}^{\infty}B_n\quad\Rightarrow\quad [w\in \bigcup_{n=1}^{\infty}B_n]=1\\
\end{align*}
We also have
\begin{align*}
w\in \bigcup_{n= 1}^{\infty}B_n\quad&\Rightarrow\quad \exists n_0\in\mathbb{N}: w\in B_{n_{0}}\\
&\Rightarrow\quad \sum_{n=1}^{\infty}[w\in B_n]=[w\in B_{n_{0}}]=1
\end{align*}


*Case 2: $w\in X\setminus\bigcup_{n=1}^{\infty}B_n$. We obtain
\begin{align*}
&w\in X\setminus\bigcup_{n=1}^{\infty}B_n\quad\Rightarrow\quad[w\in\bigcup_{n=1}^{\infty}B_n]=0\\
\end{align*}
We also have
\begin{align*}
  w\in X\setminus\bigcup_{n= 1}^{\infty}B_n\quad&\Rightarrow\quad\sum_{n=1}^{\infty}[w\in B_n]=\sum_{n=1}^{\infty}0=0
\end{align*}
and the claim (1) follows.
