# $L^1$ is complete in its metric

Theorem: The vector space $L^1$ is complete in its metric. The following proof is from Princeton Lectures in Analysis book $3$ page $70$. Some of my questions about the proof of this theorem are as follows.

1. First assume a Cauchy sequence $(f_n)\in L^1$, then we try to extract a subsequence $\left(f_{n_k}\right)$ of $(f_n)$ which converges to $f$, both point-wise almost everywhere and in the norm. Why do we need to show convergence point-wise almost everywhere? The theorem only says that $L^1$ is complete in its metric, i.e. $L^1$ norm. Right?
2. As an extension to the first question, what is the difference between point-wise convergence and convergence in certain norms? When do I need to show point-wise convergence and when to show convergence in norm and when to show both, please?
3. In the proof, we defined two new functions, and one of the two is $$f := f_{n_1}+\sum_{k=1}^\infty (f_{n_{k+1}}-f_{n_k}).$$ This is, in fact, $$\lim_{k\rightarrow\infty} f_{n_k}=\lim_{n\rightarrow \infty} f_n.$$ This is confusing to me since we are trying to find a limit for $(f_n)$. But in the above definition of $f$ we already assume the existence of $\lim_{n\rightarrow\infty} f_n$. Then what is the point of doing this, please?
4. In addition, the proof states that the series defining $f$ converges almost everywhere. Why is this true? How can I see this, please? Thank you!

Nice question, you have discovered some subtle points there:

1. The point here is that we want to show that some subsequence $(f_{n_k})_k$ converges to some function $f$ which is then a candidate for the limit function in the $L^1$ metric. The problem is more or less, that we only know that $(f_n)_n$ is Cauchy and we have to provide some function $f$ on the measure space to which $(f_n)_n$ could possibly converge. We then show (see below) that this is indeed the case.

Important note: If $(f_n)_n$ also converges pointwise a.e. to some $g$, you see $f=g$. So the proof really tells you something more, namely if $(f_n)_n$ is $L^1$-Cauchy and converges a.e. to $g$, then $g \in L^1$ and $f_n \rightarrow g$ in $L^1$.

2. There is a big difference between convergence in some norm and pointwise convergence. From just pointwise convergence (without further assumptions) you can almost never infere convergence in some norm. As examples note that pointwise convergence does NOT imply uniform convergence (which is convergence in $\Vert \cdot \Vert_{\rm{sup}}$). Also, it does not imply $L^1$ convergence (or $L^p$ convergence, for that matter). As an example consider $f_n = \chi_{[n, n+1]}$ with Lebesgue measure.

Again important note: The proof (with my comment in (1)) will show that under additional assumptions (here: $(f_n)_n$ is $L^1$-Cauchy), pointwise convergence yields convergence in some norm (here: $L^1$-norm).

EDIT 2: Further note: The proof also shows that every $L^1$-convergent sequence $f_n \rightarrow g$ has some subsequence $(f_{n_k})_k$ that converges to some $f$ a.e. and in $L^1$. This then yields $f=g$, i.e. $f_{n_k} \rightarrow g$ a.e.

3. The point here is that the convergence of the series is easier to show than directly showing that the sequence converges. This is done by considering the "absolute value" series $$\sum_{k} |f_{n_{k+1}} - f_{n_k}|.$$ Using the fact that $\Vert f_{n_{k+1}} - f_{n_k}\Vert_1 \leq \frac{1}{2^{k}}$ (I don't have the book, but this is roughly the way the proof normally works), you can derive (using the monotone convergence theorem), that $$\int\sum_{k}\left|f_{n_{k+1}}-f_{n_{k}}\right|\, d\mu=\sum_{k}\int\left|f_{n_{k+1}}-f_{n_{k}}\right|\, d\mu\leq\sum_{k}\frac{1}{2^{k}}<\infty.$$ Now if the integral over some (nonnegative) function is finite, the function itself has to be finite a.e. (why?). This show that the series converges (even absolutely) a.e.

4. See 3.

5. EDIT: Finally, one still has to show that $(f_{n_k})_k$ converges to $f$ in the $L^1$-norm. This can be done by invoking Fatou's Lemma as follows: $$\int\left|f-f_{n_{k}}\right|\, d\mu=\int\liminf_{\ell}\left|f_{n_{\ell}}-f_{n_{k}}\right|\, d\mu\leq\liminf_{\ell}\underbrace{\int\left|f_{n_{\ell}}-f_{n_{k}}\right|\, d\mu}_{<\varepsilon\text{ for }\ell,k\text{ large}}\leq\varepsilon\text{ for }k\text{ large}.$$ Finally, you will need to convince yourself that (because $(f_n)_n$ is Cauchy) the convergence of the subsequence $(f_{n_k})_k$ to $f$ in the $L^1$-Norm suffices for convergence of the whole sequence (why?).

I hope this clarifies your problems.

• Thank you for your answers. Just wondering whether convergence in certain norms implies point-wise convergence please. May 25 '14 at 9:06
• That depends. For the $L^p$ norms, you can show (analogous to the above) that for $f_n \rightarrow f$ there exists a subsequence(!) $(f_{n_k})_k$ that converges to $f$ a.e. In general it is not(!) true that convergence in $L^p$ implies convergence a.e. of the whole(!) sequence. For other norms (like $\Vert \cdot \Vert_{sup}$), it is easy to see that convergence implies pointwise convergence. What other norms are you interested in? May 25 '14 at 9:42
• Thanks. $L^p$ is sufficient for me at the moment. May 25 '14 at 9:54
• Hi! Awesome question and great answer. There is one thing that I am confused about: in the last part of 3., "if the integral over some (nonnegative) function is finite, the function itself has to be finite a.e.. This show that the series converges (even absolutely) a.e." However, I am unfortunately still not convinced why function being finite a.e. shows the series converges a.e. Dec 7 '19 at 4:33
• @James: Since the integral over the function is finite, you have $\sum_k |f_{n_{k+1}} - f_{n_k}|<\infty$ almost everywhere. This means precisely that the series converges absolutely almost everywhere. Dec 7 '19 at 10:46

I was typing this answer when PhoemueX posted his. Since his answer contains sufficient details, I'll just leave this here for completeness concerning the first two points.

For anyone interested, here is a link to the lecture notes mentioned in OP.

1. You do not need to show that there exists a subsequence converging almost everywhere. The trick in this proof is that there exists a subsequence converging almost everywhere. Then, you prove that the convergence is also in $\mathrm L^1$.

2. Convergence almost everywhere means that for a.e. $x$, $f_n(x)\xrightarrow[n\rightarrow\infty]{}f(x)$, and convergence in $\mathrm L^1$-norm means that $\int|f_n-f|\xrightarrow[n\rightarrow\infty]{}0$. One does not imply the other. If you have almost everywhere convergence, then you have have convergence in $\mathrm L^1$ under additional assumptions. Typical assumptions are that the sequence satisfies $|f_n|\le g$, $g\in\mathrm L^1$ (Lebesgue dominated convergence theorem), monotone (monotone convergence theorem), or uniformly integrable. If the sequence converges in $\mathrm L^1$, then you can extract a sequence that converges almost everywhere. Let me again emphasize that here, you indeed want to prove convergence in $\mathrm L^1$. The sketch of the proof is to

• extract a subsequence that converges a.e. to a certain limit $f$;

• prove that the subsequence in fact converges in $\mathrm L^1$ by dominated convergence;

• deduce by the Cauchy property that the whole sequence converges in $\mathrm L^1$ to the limit of the subsequence $f$.

• Quick comment: Boundedness in $L^1$ (if that is to mean $\Vert f_n \Vert_1 \leq C$) will only ensure that the pointwise limit is also $L^1$, it will not(!) give $L^1$ convergence of the sequence. If you mean $|f_n| \leq g$ with $g \in L^1$ then it is of course true. May 25 '14 at 8:09
• @PhoemueX, I meant $|f_n|\le g$, with $g\in\mathrm L^1$, I will clarify this point.
– Ian
May 25 '14 at 8:11