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liminf and limsup with characteristic (indicator) function

Suppose $\{E_k\}_{k\geq 1}$ is a sequence of measurable sets. Then we can define $\limsup_{k\rightarrow\infty}E_k = \bigcap_{k=1}^\infty \bigcup_{j\geq k} E_j$ and $\liminf_{k\rightarrow\infty}E_k=\bigcup_{k=1}^\infty\bigcap_{j\geq k}E_k.$

Let $f_k$ be the sequence of measurable functions defined by $\chi_{E_k}$, where $\chi_{E_k}$ is the characteristic function on the set $E_k$.

I'm asked to show that $\limsup_{k\rightarrow\infty}f_k = \chi_{\limsup_{k\rightarrow\infty} E_k}$ and $\liminf_{k\rightarrow\infty}f_k = \chi_{\liminf_{k\rightarrow\infty}E_k}$

My thoughts thus far: I know that $\limsup f_k=1$ and $\liminf f_k = 0$ For $\limsup$ I have to prove that $\chi_{\limsup E_k}=1$. Is this because $E = \bigcap_{k=1}^\infty\bigcup_{j\geq k}E_k$?

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    Its not a rigorous way. But why don't you see what's happening by taking a monotonic sequence of functions in both cases. It'll help you visualise things easily,I feel.2012-10-22

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