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Define the characteristic function of set $A$ to be $${\chi _A}(x) = \left\{ {\begin{array}{*{20}{c}} 1&{x \in A}\\ 0&{x \notin A} \end{array}} \right..$$

For any given collection of sets ${A_n}$, how to prove $$\mathop {\underline {\lim } }\limits_{n \to \infty } {\chi _{{A_n}}}(x) = {\chi _{\mathop {\underline {\lim } }\limits_{n \to \infty } {A_n}}}(x)$$ or in anothor words, $$\mathop {\lim }\limits_{n > 1} \mathop {\inf }\limits_{k \ge n} {\chi _{{A_k}}}(x) = {\chi _{\mathop {\lim }\limits_{n > 1} \bigcap\limits_{k = n}^\infty {{A_k}} }}(x)$$ or in anothor words, $$\mathop {\sup }\limits_{n > 1} \mathop {\inf }\limits_{k \ge n} {\chi _{{A_k}}}(x) = {\chi _{\bigcup\limits_{n = 1}^\infty {\bigcap\limits_{k = n}^\infty {{A_k}} } }}(x)$$

And some intuitive explanation behind this equation?

Asaf Karagila
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zxzx179
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1 Answers1

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I will show that $$ \liminf_{n \to \infty} \mathbf{1}_{A_k}(x) = \mathbf{1}_{\liminf_{n \to \infty} A_n}(x), \qquad (\ast) $$ where $\mathbf{1}_A$ denotes the characteristic function of the set $A$ and $$ \liminf_{n \to \infty} A_n = \bigcup_{n \geq 1} \bigcap_{k \geq n} A_k $$ denotes the limit inferior of the sets $A_1$, $A_2$, ... . We have $x \in \bigcup_{n \geq 1} \bigcap_{k \geq n} A_k$ if and only if for some $n \geq 1$ we have $x \in \bigcap_{k \geq n} A_k$, hence for some $n \geq 1$ we should have $x \in A_k$ for all $k \geq n$. This is to say that $x$ is contained in all $A_k$ from some point onwards. Thus the right hand side of $(\ast)$ is $1$ if and only if $x$ is contained in all but finitely many of the $A_n$. The left hand side is the inferior limit of a sequence of $0$s and $1$s, so it takes the value $1$ if and only if the value $0$ occurs finitely many times (otherwise $0$ is a limit of a subsequence). Therefore, the left hand side is also $1$ if and only if all but finitely many of the $A_n$ contain $x$.

In the same spirit, we have $$ \limsup_{n \to \infty} \mathbf{1}_{A_k}(x) = \mathbf{1}_{\limsup_{n \to \infty} A_n}(x), \qquad (\ast\ast) $$ where $$ \limsup_{n \to \infty} A_n = \bigcap_{n \geq 1} \bigcup_{k \geq n} A_k. $$ Both the left hand side and the right hand side of $(\ast\ast)$ are $1$ if and only if $x$ is contained in infinitely many of the $A_n$.

user133281
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