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I'm having some trouble with the formatting here, so here's a screenshot of the problem we were given:

Let $\{ A_n \}_{n \geq 1}$ be a sequence of sets. Show that $$ \lim\sup\limits_{n\to \infty} \mathbf{1}_{A_n} - \lim\inf\limits_{n\to \infty} \mathbf{1}_{A_n} = \mathbf{1}_{\{\lim\sup_{n \to \infty} A_n \backslash \lim\inf_{n \to \infty} A_n\}} $$ and conclude that $A_n \to A$ if and only if $\mathbf {1}_{A_n}(\omega) \to \mathbf {1}_A(\omega)$ for all $\omega \in \Omega$. [$A \backslash B = A \cap B^c$.]

The problem is, that in the book from which the problem is taken from (Probability Essentials by Jean Jacod), we aren't given much help with understanding the notation. Earlier there is a proof where he defines $\lim \sup_{n \to \infty} A_n$ but I have no idea what $\lim \sup_{n\to \infty} \mathbf{1}_{\ldots}$ could mean.

Is this some common notation?

Thanks.

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3 Answers 3

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${\mathbf 1}_{A_n}$ is a function from $\Omega$ to $\{0,1\}$, defined by ${\mathbf 1}_{A_n} (\omega) = 1$ if $\omega \in A_n$, ${\mathbf 1}_{A_n} (\omega) = 0$ if $\omega \notin A_n$ (indicator function). $\lim \sup _{n \to \infty } {\mathbf 1}_{A_n }$ is defined pointwise: for each $\omega \in \Omega$, it is the $\lim \sup$ of the sequence of numbers ${\mathbf 1}_{A_n} (\omega)$, that is, $\lim \sup _{n \to \infty } {\mathbf 1}_{A_n }$ is a function on $\Omega$ defined pointwise by $$ (\mathop {\lim \sup }\limits_{n \to \infty } {\mathbf 1}_{A_n } )(\omega ) = \mathop {\lim \sup }\limits_{n \to \infty } {\mathbf 1}_{A_n } (\omega ) \,(\in \{ 0,1 \}). $$

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$\textbf{1}_{A_n}$ is the indicator function.

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If you have a sequence of reals ${a_n}$ then $\limsup a_n=\inf_{k\geq1}{\sup_{m\geq k}{a_m}}$, but I think the problem is that if you have a sequences of sets ${A_n}$ then $\limsup A_n=\cap_{k\geq1}{\cup_{m\geq k}{A_m}}$ and $\limsup a_n=\cap_{k\geq1}{\cup_{m\geq k}{A_m}}$ and $\liminf A_n=\cup_{k\geq1}{\cap_{m\geq k}{A_m}}$, this is naturally if you consider a set $X$ and his power set $2^X$,in the power set you could define a order(not need to be a total order), for any $A,B\in 2^X$, $A\leq B$ if $A\subseteq B$, with this order the union is the sup and the intersection the inf. In real analysis or in calculus you could proof that a sequence of reals numbers converges ${a_n}$ if and only if $\limsup a_n=\liminf a_n$ and $\lim a_n=\liminf a_n$. You could give a notion of convergence of sets with the last proposition, a sequence ${A_n}$ of set converge if $\limsup A_n=\liminf A_n$ and $\lim A_n=\liminf A_n$.

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