Properties of the Characteristic/Indicator Function

Question

Let $B_1,B_2,...$ be a countable family of disjoint subsets of $\Bbb R^d$.

For any set $E \in \Bbb R^d$, let $\chi_E (x)=1$ if $x \in E$ and $\chi_E (x)=0$ otherwise.

Is it true that $\chi_{\bigcup B_n}(x) = \sum\chi_{B_n}(x)$ for all $x \in \Bbb R^d$?

If not, is this statement true almost everywhere?

Thanks very much for your help.

Answer 1

Hint: Since the sets $B_i$ are disjoint by assumption, we have for each point $x_0 \in \mathbb{R}^d$ that $$x_0 \in \cup_{n} B_n \Leftrightarrow \exists\, n_0: x_0 \in B_{n_0}$$ where the RHS implies (due to disjointness) that $x_0 \notin B_n$ for all $n\neq n_0$. Substituting into the characteristic function yields the required equality.

Answer 2

The statement is not true in general, since $\chi$ takes the values $0,1$ but $\sum\chi$ takes values on (potentially) all of $\mathbb N$. The equality only holds if the $B_n$ are disjoint. Otherwise, suppose $x\in B_1\cap B_2$. Then $\chi_{\bigcup B_n}(x)=1$, but $\sum \chi_{B_n}(x)\geq \chi_{B_1}(x)+\chi_{B_2}(x)=2$. We have a non-$0,1$ value for the sum if and only if we have a nontrivial element of $B_m\cap B_n$ for some $m,n$.

Answer 3

You have $$ \chi_{\bigcup B_n}(x) = \bigvee \chi_{B_n}(x) $$ where $\vee$ is the operation which gives the maximum of its arguments.