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Please help me prove this:

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
    
Since you are new, I want to let you know some things about the site. In order to get the best possible answers, it is helpful if you say what your thoughts on the problem are so far; this will prevent people from telling you things you already know, and help them write answers at an appropriate level. Also, people are much happier to help those who show they've tried the problem themselves first. If this is homework, please add the [homework] tag; people will still help, so don't worry. –  Zev Chonoles Jun 25 '12 at 5:06
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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

1 Answer 1

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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