In the proof that $L^p(\mu)$ is complete for $p\in[1,\infty]$ (as done in Saxe, Theorem 3.21 or in Folland, Theorem 6.6, the latter of which is outlined here) we make use of the following completeness criterion:
Lemma: Let $(X,\|\cdot\|_X)$ be a normed vector space and consider a sequence $(x_n)_{n\in\mathbb N}$ in $X$. If $\sum_{i=1}^\infty x_i$ converges whenever $\sum_{i=1}^\infty \|x_i\|_X$ converges, then $(X,\|\cdot\|_X)$ is complete.
In order to show completeness, we start with a Cauchy sequence $(f_k)_{k\in\mathbb N}$ in $L^p(\mu)$ and we aim to show, with the help of the above result, that $(f_k)_{k\in\mathbb N}$ converges in $L^p(\mu)$. In particular, we suppose that Cauchy $(f_k)_{k\in\mathbb N}$ is such that $\sum_{i=1}^\infty \|f_k\|_p$ converges. The authors then go to show that the corresponding series converges, which yields the result.
What I don't understand is why is this sufficient to show that any Cauchy sequence of elements converges in $L^p(\mu)$? In considering those Cauchy sequences $(f_k)_{k\in\mathbb N}$ which satisfy that $\sum_{i=1}^\infty \|f_k\|_p$ converges, are we not restricting our attention to a select few Cauchy sequences? Why is focusing on this subset of Cauchy sequences enough?
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I had thought, perhaps, that there was a relation between Cauchy sequences and their corresponding series' converging absolutely, but that has been indicated to not be true here.
Best Answer
Suppose that in some normed space $X$ the following condition holds:
Then, actually, $X$ is complete with respect to the given norm.
Now let's take any Cauchy sequence $\{x_n\}_{n=1}^{\infty}$ in $X$. Our aim is to use the condition above, in order to prove that $\{x_n\}_{n=1}^{\infty}$ is convergent. Note that we are not claiming that $\sum_{n=1}^{\infty}\|x_n\|<\infty$ for the specific Cauchy sequence in hand. Indeed, this may not be the case. But we are claiming that we can use the above condition to prove that $\{x_n\}_{n=1}^{\infty}$ converges. We do this as follows. Pick $k_1$ such that for every $n,m\geq k_1$, we have $\|x_n-x_m\|<\frac{1}{2}$. Next pick $k_2>k_1$ such that for every $n,m\geq k_2$, we have $\|x_n-x_m\|<\frac{1}{4}$. Continue inductively in this way. Put $y_i=x_{k_{i+1}}-x_{k_i}$. The series $\sum_{i=1}^{\infty}y_i$ converges absolutely, because $\|y_i\|<2^{-i}$. Also, $$\sum_{i=1}^my_i=x_{k_2}-x_{k_1}+\cdots + x_{k_{m+1}}-x_{k_m}=x_{k_{m+1}}-x_{k_1}$$ The condition above implies that $\sum_{i=1}^{\infty}y_i$ converges in $X$, hence the sequence $x_{k_{m+1}}$ converges. Being a subsequence of a Cauchy sequence, however, this implies that the original sequence $\{x_n\}_{n=1}^{\infty}$ converges too, because it is well known that if a Cauchy sequence has a convergent subsequence, then it converges as well.