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I was going through Egorov's theorem on wikipedia.

It gives a example why should be $\mu(A)<\infty $. Sequence of real valued indicator function is taken. It claims that the sequence: $ f_n(x)=1_{[n,n+1]}(x) $ converges pointwise for $n\in N$ and $x\in\Re$. I am not able to understand how it converges pointwise to $0$ ?

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  • $\begingroup$ Oh thanks, just didn't strike me soon enough. $\endgroup$
    – Theorem
    May 10, 2012 at 13:35

1 Answer 1

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For the pointwise convergence, you have to fix $x'\in \mathbb R$ and check that a numerical limit exists: in your case that $f_n(x') \to 0$. Clearly, given $x'$ you have that if $n>x'$ then $x'\notin[n,n+1]$ and hence $f_n(x') = 0$ for all $n>x'$, so that for any $x'$ you have $f_n(x')\to 0 $ - which is the pointwise convergence.

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