# Does $\chi_{[n,n+1]}\to 0$ almost everywhere?

Let $f_n=\chi_{[n,n+1]}$ and $f=0$. Is it correct to say $f_n \rightarrow f\$ almost everywhere ?

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Yes, as there is convergence everywhere: for each $x$, $f_n(x)=0$ whenever $n> x$ (in particular, we don't need to specify the measure nor the $\sigma$-algebra).

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$f_n$ converges to $f = 0$ everywhere. You can drop the word 'almost'.

To show this, for any $x$, pick $N \in \mathbb N, N \ge x$. This is possible using the Archimedean property. Now notice that $\forall n > N : f_n(x) = 0$.

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But next you can object: $\int f_n = 1$ for all $n$, but $\int f \ne 1$, so what about the bounced convergence theorem? (Just being argumentative.) –  GEdgar Oct 30 '12 at 23:33
@GEdgar I'm not familiar with the bounced convergence theorem. Do you mean bounded? What's the bound? There isn't an integrable bound. –  Ayman Hourieh Oct 30 '12 at 23:35
Yes, bounded. The bound is 1. Of course $f_n$ is not dominated by an integrable function, so the dominated convergence theorem cannot apply. But (the answer, I guess) the total measure of $[0,\infty)$ is not finite, so neither can the bounded convergence theorem apply. –  GEdgar Oct 30 '12 at 23:43

In fact $f_n \rightarrow f$ everywhere.

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