# Thread: Characteristic Function

1. ## Characteristic Function

Let $(A_n)$ be a sequence of measurable sets in $X$.Define $\liminf_j A_j = \bigcup^\infty _{n=1} \bigcap^\infty_{j=n} A_j$ and $\limsup_j A_j = \bigcap^\infty _{n=1} \bigcup^\infty_{j=n} A_j$.

Prove or disprove
$\chi_{\liminf_j A_j} = \liminf_j \chi_{A_j}$ and $\chi_{\limsup_j A_j} = \limsup_j \chi_{A_j}$ .

Can anyone help me to initialize the proof?

2. Originally Posted by problem
Let $(A_n)$ be a sequence of measurable sets in $X$.Define $\liminf_j A_j = \bigcup^\infty _{n=1} \bigcap^\infty_{j=n} A_j$ and $\limsup_j A_j = \bigcap^\infty _{n=1} \bigcup^\infty_{j=n} A_j$.

Prove or disprove
$\chi_{\liminf_j A_j} = \liminf_j \chi_{A_j}$ and $\chi_{\limsup_j A_j} = \limsup_j \chi_{A_j}$ .

Can anyone help me to initialize the proof?
A characteristic function can only take the values 0 or 1. Therefore for each x in X, $\textstyle\limsup_j \chi_{A_j}(x)$ must be 0 or 1. Then it's just a matter of definition-chasing:

\begin{aligned}x\in \limsup_j A_j &\Longrightarrow\ \forall n\ \exists j\geqslant n\text{ such that }x\in A_j\\ &\Longrightarrow\ \forall n\ \exists j\geqslant n\text{ such that }\chi_{A_j}(x)=1\\ &\Longrightarrow\ \textstyle\forall n\ \sup_{j\geqslant n}\chi_{A_j}(x)=1\\ &\Longrightarrow\ \textstyle\limsup_j \chi_{A_j}(x)=1.\end{aligned}

\begin{aligned}x\notin \limsup_j A_j &\Longrightarrow\ \exists n\ \forall j\geqslant n\ x\notin {\textstyle\bigcup_{j\geqslant n}}A_j\\ &\Longrightarrow\ \exists n\ \forall j\geqslant n\ \chi_{A_j}(x)=0\\ &\Longrightarrow\ \ldots\ \Longrightarrow\ \textstyle\limsup_j \chi_{A_j}(x)=0.\end{aligned}

Therefore $\chi_{\limsup_j A_j} = \textstyle\limsup_j \chi_{A_j}$

Similar argument for the liminf.