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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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First, we have to show that the sequence $\{\sum_{k=1}^nf_k(x)\}$ has a limit for almost every $x$ (if it's not the case we can't talk about the sum of the series). – Davide Giraudo Jul 4 '12 at 12:40
@DavideGiraudo I think I confused "limit exists" with "limit is finite". In (1) we don't want to show that the limit is finite we want to show that it exists, i.e. that the sum doesn't oscillate. I didn't think of the case of oscillation. If one ignores that then the sum can only either be finite or infinite and in both cases we have a limit function. – Rudy the Reindeer Jul 4 '12 at 13:04
@DavideGiraudo Do I make any sense? – Rudy the Reindeer Jul 4 '12 at 13:05
@DavideGiraudo But if the limit is finite it does not oscillate so maybe showing that it exists is the same as showing that it's finite. – Rudy the Reindeer Jul 4 '12 at 13:07
In (1), we want the series to be convergent almost everywhere, that is, the sequence $\{\sum_{k=1}^nf_k(x)\}$ converges to a real number for almost every $x$. (in fact writing $K_x<\infty$ is not quite accurate, we should write $K_x\in\Bbb R$ because, in a trivial case $f_n(x):=-n$ we would have $-\infty$). – Davide Giraudo Jul 4 '12 at 13:09
up vote 1 down vote accepted

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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Thank you. In retrospect I think my confusion was about the definition of "converges". I will accept your answer but I have one more question: Can we think of (1) and (2) in my question above as the two properties that an element of $L^1$ needs to satisfy? Even though only (2) explicitly says "$f \in L^1$". In other words: if $L^1$ are functions from $X \to R$ (conv. a.e.) can I think of (1) as showing that $\sum f_n$ is such a function? If the answer is yes then it might make sense to write the proof slightly differently... – Rudy the Reindeer Jul 5 '12 at 5:54
...If the claim is that if $f_n$ is an absolutely converging sequence in $L^1$ then instead of claiming (i) $f = \sum f_n$ converges a.e. (ii)$f$ is in $L^1$ one would claim $f \in L^1$ and then to show this claim one would show (i) $f = \sum f_n$ converges a.e. and (ii) $\int |f| < \infty$. – Rudy the Reindeer Jul 5 '12 at 5:56
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. – Siminore Jul 5 '12 at 7:40

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