Upper limit of an indicator and indicator of an upper limit of sets

Question

I am to prove the following equality $$\limsup \mathbb 1_{A_n} = \mathbb1_{\limsup A_n}$$ I don't know how to connect two different limits. I wanted to use "standard" methods which are used in set theory: $x \in ...$ and then from right side to the left or vice versa.

Answer 1

First, prove that $$\large1_{\sup\limits_{k\ge n} A_k} = \sup\limits_{k\ge n} 1_{A_k}.$$ Once this is proved, using a similar logic, we have $$\large1_{\inf\limits_{n \in \Bbb{N}} B_n} = \inf\limits_{n \in \Bbb{N}} 1_{B_n}.$$

• $\large1_{\sup\limits_{k\ge n} A_k}(x) = 1 \iff x \in \sup\limits_{k\ge n} A_k \iff \exists k \ge n: x \in A_k \iff \exists k \ge n :1_{A_k} (x) = 1 \iff \sup\limits_{k\ge n} 1_{A_k}(x) = 1.$
• $\large1_{\sup\limits_{k\ge n} A_k}(x) = 0 \iff x \notin \sup\limits_{k\ge n} A_k \iff \forall k \ge n: x \notin A_k \iff \forall k \ge n :1_{A_k} (x) = 0 \iff \sup\limits_{k\ge n} 1_{A_k}(x) = 0.$

We set $B_n = \sup\limits_{k\ge n} A_k$, and

• replace $\sup\limits_{k\ge n}$ with $\inf\limits_{n \in \Bbb{N}}$
• change $\exists k\ge n$ to $\forall n\in \Bbb{N}$ and;
• change $\forall k\ge n$ to $\exists n\in \Bbb{N}$ in the above paragraph.

Then you'll be above to decude the second equality. Combine the first two equalities, and you'll recover the equality in the question.