# Completeness of $L^1$

I have written a proof that $L^1$ is complete. Can you read it and tell me if it's right? Thanks.

To show $L^1$ is complete we use the following fact:

Fact: If $f_n$ is a sequence in $L^1$ such that $\sum \|f_n\|_1 < \infty$ then $f = \sum f_n \in L^1$ where $\sum f_n$ is the pointwise limit of $\sum_{n=1}^N f_n(x)$.

Proof: First we show that $\sum f_n$ converges almost everywhere. For this let $Y_n$ be the set of points where $f_n$ does not converge. Then the set of point where $f$ does not converge is $Y = \bigcup Y_n$ which has measure zero. Next we show that $\int |f| < \infty$: Observe that $\sum_{k=1}^n |f_k|$ is a non-decreasing sequence of non-negative measurable functions. Hence we can apply the monotone convergence theorem to get $\int f \leq \int |f| = \int \lim_{n \to \infty} \sum_{k=1}^n |f_k| = \lim_{n \to \infty} \int \sum_{k=1}^n |f_k| = \lim_{n \to \infty} \sum_{k=1}^n \int |f_k| = \sum \|f_n\|_1 < \infty$.

Now we use this to show that $L^1$ is complete: Let $f_n$ be a Cauchy sequence in $L^1$, i.e. for $\varepsilon > 0$ there is $N$ such that for $m,n>N$ we get $\|f_n - f_m\|_1 < \varepsilon$. The idea is to construct a subsequence of $f_n$ that converges to a function in $L^1$. Then $f_n$ will have to converge to the same limit function and hence the claim follows. For this, Let $N_k$ be such that $n,m \geq N_k$ implies $\|f_n - f_m \|_1 < \frac{1}{2^k}$. Then choose $n_k = \max(N_k, n_{k-1}+1)$. Then $\sum \|(f_{n_k} - f_{n_{k+1}})\|_1 \leq 2 < \infty$ and hence by the previously proved fact $\lim_{N\to \infty}\sum_{k=1}^N(f_{n_k} -f_{n_{k+1}}) \in L^1$ (pointwise).

We claim that $\lim_{k \to \infty} f_{n_k} = f_{n_1} - \sum (f_{n_k} -f_{n_{k+1}})$ and is hence in $L^1$ and that convergence is in norm. For this observe that $\sum_{k=1}^N(f_{n_k} -f_{n_{k+1}}) = f_{n_1} - f_{n_{N+1}}$ and hence $f_{n_{N+1}} = f_{n_1} - \sum_{k=1}^N(f_{n_k} -f_{n_{k+1}})$ and hence $\lim_{k \to \infty} f_{n_k}$ is in $L^1$.

We also have convergence in norm: \begin{align} \| f_{n_{N+1}} - f_{n_1} + \sum (f_{n_k} -f_{n_{k+1}}) \|_1 &= \| f_{n_1} - \sum_{k=1}^N(f_{n_k} -f_{n_{k+1}}) -f_{n_1} + \sum (f_{n_k} -f_{n_{k+1}}) \|_1 =\\ &= \| \sum_{k=N+1} (f_{n_k} -f_{n_{k+1}}) \|_1 \leq \sum_{k=N+1} \|(f_{n_k} -f_{n_{k+1}})\|_1 \xrightarrow{N \to \infty} 0 \end{align}

Thank you. I'm still interested in an answer.

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What is $\lim f_n$? – Siminore Jul 5 '12 at 8:08
@Siminore It's the point wise limit of $f_n(x)$. – Rudy the Reindeer Jul 5 '12 at 8:12
@bananalyst Then what is $\sum\lim f_n$? – Alex Becker Jul 5 '12 at 8:13
By the time you have proven your first fact, you are already done actually, since for any normed vector space $(X,|\cdot |)$ we have $X$ complete iff absolutely convergent series converge. – Olivier Bégassat Jul 5 '12 at 8:25
Probably all the stuff about $Y_n$ is a technicality. It is needed when $f_n$ is defined a.e. I think also that "$f_n$ does not converge" is not the correct statement here. – Siminore Jul 5 '12 at 9:23