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There is the following theorem:

If $(f_n)$ is a sequence in $L^1$ such that $\sum \|f_n\|_1 < \infty$ then

(1) $\sum f_n $ converges almost everywhere (i.e. $\sum f_n(x) = K_x < \infty $)

(2) $\sum f_n \in L^1$

Why does the proof first show (1)? If we just show (2), (1) follows since (2) implies (1).

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    I see. The limit exists if and only if it is in the (range) space. Also, $f_n = n$ is not a Cauchy sequence, so it's not an example.2012-07-04

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Let us recall the following theorem.

Theorem. Let $(X,\| \cdot \|)$ be a normed vector space. The space $X$ is complete (w.r.t. the metric induced by the norm) if and only if every totally convergent series is convergent in norm.

A series $\sum_n x_n$ is totally convergent when $\sum_n \|x_n\|$ converges (as a sum of non-negative real numbers).

Apply this result to $X=L^1$. Your question turns out to be equivalent to (or implied by) the completeness of $L^1$ (w.r.t its natural norm). And how do you prove that $L^1$ is complete? Exactly by choosing any Cauchy sequence and constructing a subsequence that is almost everywhere convergent. This is the reason why (1) is proved before (2).

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    I am not sure if I have understood your question. I think the your question is slightly circular: a necessary condition for a function $f$ to be integrable is that $f$ be a.e. finite. When you try to study the integrability of $\sum_n f_n$, you have first to define this function; if it happened to be a.e. infinite, you'd have to stop.2012-07-05