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Theorem (Riesz-Fischer). The vector space $L^1$ is complete in its metric.

Proof. Suppose $\{f_n\}$ is a Cauchy sequence in the norm, so that $\|f_n-f_m\|\to0$ as $n,m \to \infty$. The plan of the proof is to extract a subsequence of $\{f_n\}$ that converges to $f$, both pointwise almost everywhere and in the norm. Under ideal circumstances we would have that the sequence $\{f_n\}$ converges almost everywhere to a limit $f$, and we would then prove that the sequence converges to $f$ also in the norm. Unfortunately, almost every-where convergence does not hold for general Cauchy sequences. The main point, however, is that if the convergence in the norm is rapid enough, then almost everywhere convergence is consequence, and this can be achieved by dealing with an appropriate subsequence of the original sequence.

Indeed, consider a subsequence $\{f_{n_k}\}_{k=1}^\infty$ of $\{f_n\}$ with the following property:

$$\|f_{n_{k+1}}-f_{n_k}\|\le2^{-k}, \quad\quad \text{ for all } k\ge 1.$$

The existence of such a subsequence is guaranteed by the fact that $\|f_n-f_m\|\le\epsilon$ whenever $n,m\ge N(\epsilon)$, so that it suffices to take $n_k= N(2^{-k})$. We now consider the series whose convergence will be seen below, $$\boxed{f(x)=f_{n_1}(x)+\sum_{k=1}^\infty (f_{n_{k+1}}(x)-f_{n_k}(x))}$$ and $$g(x)=\vert f_{n_1}(x)\vert+\sum_{k=1}^\infty \vert f_{n_{k+1}}(x)-f_{n_k}(x)\vert $$ and note that $$\int \vert f_{n_1}\vert+\sum_{k=1}^\infty \int \vert f_{n_{k+1}}-f_{n_k}\vert\le \int \vert f_{n_1}\vert +\sum_{k=1}^\infty 2^{-k}<\infty$$

Is $f(x)$ defined properly? If the series isn't convergent( in extended real numbers), can we still define the limit of the series?

If $\exists x $ such that the series isn't convergent to any real numbers or infinity , the definition is unreasonable , isn't it?

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  • $\begingroup$ The monotone convergence theorem implies that $g$ converges pointwise a.e.. Therefore the series for $f$ converges pointwise a.e. to an $L^1$ function. $\endgroup$ – DisintegratingByParts Apr 11 '18 at 16:47
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The series converges absolutely in the norm of $L^1$ hence it is well defined as an integrable function.

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  • $\begingroup$ Here you use already that $L^1$ is complete! We only know that this is a $L^1$-Cauchy sequence! $\endgroup$ – p4sch Apr 11 '18 at 15:20
  • $\begingroup$ The fact that g is integrable implies that it is finite a.e., which implies f is defined a.e. $\endgroup$ – Sean Apr 11 '18 at 15:29
  • $\begingroup$ The using of the term 'absolutely convergent' is not correct! Absolutely convergence of a series is defined by just satisfying $\sum_{k=1}^\infty \|h_k\| < \infty$. Thus, in incomplete spaces there are absolutely convergent series which are not convergent! In fact, as above argument shows, the absolute convergence of series in normed spaces is equivalent to the completness of the space. You mean that $g(x) = |f_{n_1}(x)| + \sum_{k=1}^\infty |f_{n_{k+1}}(x)-f_{n_k}(x)|$ is integrable and thus $g(x)$ is a.e. finit. Thus, we have pointwise convergence! $\endgroup$ – p4sch Apr 11 '18 at 15:42
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$g(x)$ is integrable, which implys $g(x)<\infty$ almost everywhere.

Note that every term of $g(x)$ is absolutely value of the term of $f(x)$ respectively.

Thus we can conclude that $f(x)$ is absolutely convergent almost everywhere, which implys that $f(x)$ is convergent almost everywhere.

We can define $f(x) = 0$ in the set that the series isn't convergent, which is measure $0$.

The definition of $f(x)$ is skipped in Real Analysis of Stein while it is clear in the book of Rudin.

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