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Let $(A_n)$ be a seqeunce of sets, I am trying to show that

$$ \limsup_{n \to \infty} \chi_{A_n} = \chi_{\limsup_{n \to \infty}A_n}$$

$$ \liminf_{n \to \infty} \chi_{A_n} = \chi_{\liminf_{n \to \infty}A_n}$$

I am stuck trying to show these identities. I was thinking if let $A = \limsup A_n$, then a good strategy to prove this would be to bound $| \chi_{A_n} - \chi_A | $?? Any ideas would be greatly appreacited. thanks.

Andrés E. Caicedo
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ILoveMath
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1 Answers1

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If $A = \limsup_{n\to \infty} A_n$, then $$\chi_A = \chi_{\cap_{n = 1}^\infty \cup_{k = n}^\infty A_k} = \inf_{n \ge 1} \chi_{\cup_{k = n}^\infty A_k} = \inf_{n\ge 1} \sup_{k \ge n} \chi_{A_k} = \limsup_{n\to \infty} \chi_{A_k}.$$ A similar argument shows $\liminf_{n\to \infty} \chi_{A_n} = \chi_{\liminf A_n}$.

kobe
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