# Convergence of sets is same as pointwise convergence of their indicator functions

Let $A_1,A_2,\ldots$ be subsets of $\Omega$. Prove that $A_n\to A$ if and only if $I_{A_n}(\omega)\to I_A(\omega)$ for every $\omega\in\Omega$ (so that convergence of sets is the same as pointwise convergence of their indicator functions).

Note: $I_A(\omega)=1$ if $\omega\in A$, and $0$ if $\omega\notin A$. Use in the proof that $$\operatorname{lim\;inf}\limits_n\; x_n=\bigvee_{k=1}^\infty\bigwedge_{n=k}^\infty x_n\quad\text{ and }\quad\operatorname{lim\;sup}\limits_n\; x_n=\bigwedge_{k=1}^\infty\bigvee_{n=k}^\infty x_n.$$

Thank you very much!

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What definition of convergence of sets are you using? –  Robert Israel Jun 25 '12 at 5:02
For Robert Israel, The sequence (An) converges to A if lim inf An = lim sup An = A. Thank you –  ConfusedFan Jun 25 '12 at 6:11
Thanks, Zev Chonoles. We've only had two class meetings so far so I only know a few concepts about probability. I'm really having a hard time thinking where to start since I'm not even familiar with how a sequence of real numbers converge (which is required to be used in the proof). –  ConfusedFan Jun 25 '12 at 6:27
It helps to think of the problem in the following way. The liminf of a sequence of sets is the set of points that occur in every single $A_n$ for large enough $n>N$, where $N$ depends on the point in question. Alternately the limsup is the set of points that occur infinitely often, in that they belong to some infinite sequence of $A_{n_k}$. Now rewrite these definitions in terms of indicator function convergence. –  Alex R. Jun 25 '12 at 7:52

Let $\sigma=\langle A_n:n\in\Bbb N\rangle$ be a sequence of subsets of some set $\Omega$. A point $\omega\in\Omega$ is eventually in $\sigma$ if there is an $n_0\in\Bbb N$ such that $\omega\in A_n$ for all $n\ge n_0$, i.e., if $\omega$ is in each member of a ‘tail’ of the sequence. The point $\omega$ is frequently in $\sigma$ if for each $m\in\Bbb N$ there is an $n\ge m$ such that $\omega\in A_n$, i.e., if $\omega$ is in infinitely many members of the sequence. These terms provide an easy way to think and talk about the liminf and limsup of a sequence of sets: $\liminf_nA_n$ is the set of points of $\Omega$ that are eventually in $\sigma$, and $\limsup_nA_n$ is the set of points of $\Omega$ that are frequently in $\sigma$. This is quite easy to verify from the definitions. For example, $$\liminf_{n\in\Bbb N}A_n=\bigcup_{n\in\Bbb N}\bigcap_{k\ge n}A_k\;,\tag{1}$$ so $\omega\in\liminf_nA_n$ iff there is an $n\in\Bbb N$ such that $\omega\in\bigcap_{k\ge n}A_k$, which is the case iff $\omega\in A_k$ for each $k\ge n$: in short, $\omega\in\liminf_nA_n$ iff $\omega$ is eventually in $\sigma$. Similarly, $$\limsup_{n\in\Bbb n}A_n=\bigcap_{n\in\Bbb N}\bigcup_{k\ge n}A_k\;,\tag{2}$$ so $\omega\in\limsup_n A_n$ iff for each $n\in\Bbb N$ $\omega\in\bigcup_{k\ge n}A_k$, which is the case iff $\omega\in A_k$ for some $k\ge n$: $\omega\in\limsup_nA_n$ iff $\omega$ is frequently in $\sigma$.

It’s easy to check that $$\liminf_{n\in\Bbb N}I_{A_n}(\omega)=\bigvee_{n\in\Bbb N}\bigwedge_{k\ge n}I_{A_k}(\omega)\text{ for all }\omega\in\Omega$$ and $$\limsup_{n\in\Bbb N}I_{A_n}(\omega)=\bigwedge_{n\in\Bbb N}\bigvee_{k\ge n}I_{A_k}(\omega)\text{ for all }\omega\in\Omega$$ are simply restatements of $(1)$ and $(2)$ in terms of indicator functions. (E.g., $\omega$ is eventually in $\sigma$ iff $I_{A_n}(\omega)$ is eventually $1$.) Thus, the following statements are equivalent:

\begin{align*}&\lim_{n\in\Bbb N}A_n\text{ exists}\tag{3}\\&\liminf_{n\in\Bbb N}A_n=\limsup_{n\in\Bbb N}A_n\tag{4}\\&\liminf_{n\in\Bbb N}I_{A_n}(\omega)=\limsup_{n\in\Bbb N}I_{A_n}(\omega)\text{ for all }\omega\in\Omega\tag{5}\end{align*}

To finish the proof, you need only show that $(5)$ is equivalent to

$$\lim_{n\in\Bbb N}I_{A_n}(\omega)\text{ exists for each }\omega\in\Omega\tag{6}$$

and then show that the limit in $(6)$ is the indicator function of the limit in $(3)$.

It’s all just a matter of translating between two ways of saying the same thing: $\omega\in A$ iff $I_A(\omega)=1$, and $\omega\notin A$ iff $I_A(\omega)=0$.

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