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Prove that $$I_{\limsup A_n} = \limsup I_{A_n}$$

To prove this, we need to prove $I_{\limsup A_n}(\omega) = \limsup I_{A_n}(\omega)$, for each $\omega$ in sample space. But I am getting trouble to understand what $I_{\limsup A_n}(\omega)$ implies.

$I_A(\omega)=1$ if $\omega\in A$ and $0$ otherwise.

Stat_prob_001
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1 Answers1

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Observe that the functions $\limsup I_{A_n}$ and $I_{\limsup A_n}$ only take values in $\{0,1\}$

For an arbitrary $\omega\in\Omega$ the following statements are equivalent:

  • $I_{\limsup A_n}(\omega)=1$
  • $\omega\in \limsup A_n$
  • $\forall n\in\mathbb N\exists k\geq n\;[\omega\in A_k]$
  • $\forall n\in\mathbb N\exists k\geq n\;[I_{A_k}(\omega)=1]$
  • $\limsup I_{A_n}(\omega)=1$

Because the functions mentioned under first bullet and last bullet only take values in $\{0,1\}$ this justifies the conclusion that both functions coincide.

drhab
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