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In my notes we prove completeness of $L^p$ by showing that if $\sum\|f_k\|_p < \infty$ then $\sum f_k$ converges in $L^p$. (That's a lemma we prove a bit earlier, namely that $(V, \|\cdot\|)$ is Banach if and only if $\sum \|v_k\|< \infty$ implies $\sum v_k \to v \in V$.)

First we show that if $\sum\|f_k\|_p < \infty$ then $f(x) = \sum f_k(x)$ is in $\mathbb R$ $\mu$ almost everywhere. Once we have that, the notes do the following to show that $\sum_{k=1}^n f_k(x) \to f(x)$ in $\|\cdot\|_p$ (where $g(x) = \sum_{k=1}^\infty |f_k(x)|$):

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My question: Do we really have to apply montone convergence yet again? I would do it like this, but maybe that's wrong so I'd appreciate your correction:

$$ \|f(x) - \sum_{k=1}^n f_k(x)\|_p = \| \sum_{k=1}^\infty f_k(x) - \sum_{k=1}^n f_k(x)\|_p = \|\sum_{k=n+1}^\infty f_k(x)\|_p \stackrel{\Delta-\text{ineq.}}{\leq} \sum_{k=n+1}^\infty \|f_k \|_p $$

Since by assumption $\sum \|f_k \|_p < \infty$we have that $\|f_k\|_p$ must be a null sequence (that is, it tends to $0$) hence for $n$ large enough, $ \sum_{k=n+1}^\infty \|f_k \|_p < \varepsilon$ and we're done.

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    Hm ... as you don't know a priori, that $\sum_{k \ge n+1} f_k$ is a well defined element of $L^p$, you can't apply the triangle inequality to it.2012-07-11
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    @martini But we show that $f(x) = \sum_{k=1}^\infty f_k$ is defined almost everywhere in the previous step of the proof. Does it not follow from that, that $\sum_{k\geq n+1}^\infty f_k$ is defined?2012-07-11
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    ... but you are exchanging pointwise(!) a. e. convergence and integration when you are applying the triangle inequality. You don't know that $\sum_k f_k$ converges in $L^p$! That was what I meant, sry.2012-07-11
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    @Matt: you should be careful not to write $\|f(x)\|_p$ when you mean $\|f\|_p$. This contributes to the confusion between pointwise and $L^p$-convergence.2012-07-11
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    @wildildildlife yes, I think you are right. Thank you. I will delete this (and repost a re-worked version of it).2012-07-11
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    @martini You are right! I was confused. Thank you!2012-07-11

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