Let $f_n=\chi_{[n,n+1]}$ and $f=0$. Is it correct to say $f_n \rightarrow f\ $ almost everywhere ?
Does $\chi_{[n,n+1]}\to 0$ almost everywhere?
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real-analysis
measure-theory
convergence-divergence
3 Answers
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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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0Yes, 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. – 2012-10-30
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In fact $f_n \rightarrow f$ everywhere.