Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question asks us to show that $\Bbb R$ with the following metric is not complete:

Fix a strictly positive function $f \in L^1(\Bbb R)$, and let $d(x,y)=\left|\int_x^y f(t)dt\right|$.

It's easy to see (in line with the answers to the linked question) that any unbounded increasing sequence of real numbers is Cauchy without limit in this metric. What is the completion of this metric space? Is it homeomorphic to any well-known space? If not, what are some interesting properties it has? If the problem is intractable for arbitrary $f$, feel free to choose a suitably interesting $f$ or subclass of such $f$. (In particular, continuous $f$ or smooth $f$.)

share|cite|improve this question
up vote 6 down vote accepted

Put $a = \int_{-\infty}^\infty f(t) \ dt >0$. Define $F : \mathbb{R} \to [0,a]$ by $F(x) = \int_{-\infty}^x f(t) \ dt$. Using that $f$ is strictly positive, it's not hard to see that $F$ has range equal to $(0,a)$. Indeed, $F$ is absolutely continuous, strictly increasing, has $\lim_{x \to -\infty} F(x) = 0$ and $\lim_{x \to \infty} F(x) = a$. Given $x < y$ in $\mathbb{R}$, one has $$ d(x,y) = \int_x^y f(t) \ dt = \int_{-\infty}^y f(t) \ dt - \int_{-\infty}^x f(t) \ dt = F(y) - F(x) = |F(y) - F(x)|$$ which shows $F$ is an isometry from $\mathbb{R}$ with the metric $d$ onto $(0,a) \subset [0,a]$ with the standard metric. Since the completion of $(0,a)$ is isometric to $[0,a]$, the completion of $\mathbb{R}$ in $d$ is isometric to $[0,a]$ as well.

share|cite|improve this answer
Awesome. Thanks, Mike! – Mike Miller May 9 '14 at 5:45
No prob! As an aside, I poked my nose into chat earlier today and now I get all your pings, apparently... That's actually how I found this problem! If I sign out of chat do you think that will stop? – Mike F May 9 '14 at 5:46
I intend to change my name to my full name to avoid exactly that problem; but yes, I think signing out will stop pinging you. – Mike Miller May 9 '14 at 5:50
Haha, great minds think alike, as I am now Mike F – Mike F May 9 '14 at 5:52
I corrected a typo in the definition of $F(x)$. – Pedro Tamaroff May 10 '14 at 1:17

This is $\{-\infty\}\cup\mathbb{R}\cup\{+\infty\}$.

Rough attempt at a proof, perhaps there are mistakes:

Note that every usual Cauchy sequence remains Cauchy when passing to this integral metric.

Suppose $\{c_n\}$ is Cauchy under this metric but is not Cauchy under the usual metric. That is, we can find a fixed positive $\epsilon>0$ such that infinitely far out, we can find successive terms whose distance apart under the usual metric is $\ge \epsilon$. Then by Cauchy-ness under the integral metric, the average value of $f$ on these $\epsilon$-intervals must be going to $0$.

If the intervals are eventually always coming back to some bounded set, we have a contradiction, since for a positive function there is a lower bound on the average value of $f$ over any $\epsilon$-intervals on a compact set (by compactness one can find a single interval where $f$ has the "smallest" average value).

So given any bounded set, eventually all intervals between successive elements fall outside of that bounded set. Well then the terms are diverging in absolute value to $\infty$. But we cannot have arbitrarily large positive and arbitrarily large negative terms, because then the distance between them would be bounded below by, for example $\int_{-1}^1 f(x)dx$. So $c_n$ are diverging to either $+\infty$ or $-\infty$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.