$1_{\cup_n A_n} \le \sum_n 1_{A_n}$

Let $\{A_n\}$ a sequence of subsets of $\Omega$. Show:

$1_{\cup_n A_n} \le \sum_n 1_{A_n}$

If the sequence of sets are mutually disjoint, the inequality holds with equality. Now suppose not mutually disjoint. There exists $i, j \in \mathbb{N}$ such that $A_i \cap A_j \ne \emptyset$. Hence, the inequality would instead hold with strict equality.

How am I supposed to show that this inequality holds in general? Induction?

You may simply show that it holds for every $x\in\Omega$. If $x\notin A_n$ for all $n\in\mathbb N$ then $x\notin\bigcup_{n=1}^\infty A_n$. Therefore, in this case $$1_{\bigcup_{n=1}^\infty A_n}(x)=0\leq\sum_{n=1}^\infty 1_{A_n}(x),$$ since every function $1_{A_n}$ is non-negative and therefore, so is their sum. If, on the other hand, $x\in A_{n_0}$ for some $n_0\in\mathbb N$, then $x\in\bigcup_{n=1}^\infty A_n$. Therefore $$1_{\bigcup_{n=1}^\infty A_n}(x)=1=1_{A_{n_0}}(x)\leq\sum_{n=1}^{\infty}1_{A_n}(x),$$ where the last inequality again holds because every summand is non-negative.
• Thank you, that "show that it holds for every $x \in \Omega$ is just right on the bulleye. – Daniel Nov 1 '12 at 17:09