Indicator function for limsup, liminf

Question

If $A_i$ is a sequence of sets, define$$\liminf_i A_i = \bigcup_{j = 1}^\infty \bigcap_{i = j}^\infty A_i, \quad \limsup_i A_i = \bigcap_{j = 1}^\infty \bigcup_{i = j} A_i.$$Given a set $D$ define the function $\chi_D$ by $\chi_D(x) = 1$ if $x \in D$ and $\chi_D(x) = 0$ if $x \notin D$. Does it follow that$$\chi_{(\liminf A_i)}(x) = \liminf_i \chi_{A_i}(x), \quad \chi_{(\limsup A_i)}(x) = \limsup_i \chi_{A_i}(x)$$for each $x$?

Answer 1

Yes, it does follow. Notice that $x \in \lim \inf_{i \to \infty} A_i$ iff there is some $j$ such that $x \in A_{i}$ for all $j \le i$. In this case we clearly have $\lim \inf_{i \to \infty} \chi_{A_i} (x) = 1$. Conversely, if $\lim \inf_{i \to \infty} \chi_{A_i}(x) = 1$, then there is no infinite subsequence $(A_{i_k})_{k}$ such that $x \not \in A_{i_k}$ for all $k$. Hence there is some maximal $j-1$ such that $x \not \in A_{j-1}$. In other words: For all $i \ge j$ we have $x \in A_i$ and hence $x \in \lim \inf_{i \to \infty} A_i$.

The result for $\lim \sup$ can be proved in much the same way (or by observing the relation between $\lim \inf$ and $\lim \sup$ and applying the result for $\lim \inf$).