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This is an exercise from "A course in metric geometry" by Burago and Ivanov I am having some troubles with.

Give an example of length structure on the plane for which all continuous curves are admissible, the resulting intrinsic metric is the standard Euclidian one, but lengths of some curves differ from their euclidian lengths.

I have an example for $C^2$ paths but struggling to do this for continuous curves. Thank you.

Vadim
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  • To make it a complete question, you should state the definition of an "admissible curve". – Moishe Kohan Feb 04 '18 at 04:02
  • @MoisheCohen Admissible curves are curves for which we have notion of length. In this question, we need to define different type of length on the class of continuous curves with usual finite Euclidean length, satisfying length structure axioms and above conditions. You can ignore the "admissible curve" by restating that required length structure is defined on continuous paths with finite Euclidean length, it produces Euclidean metric yet some curves have different length compared to standard Euclidean notion of length. – Vadim Feb 04 '18 at 04:46

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A hint: Define the length $L(c)$ of every piecewise-linear path $c$ to be its Euclidean length. Define the length $L(c)$ to be $\infty$ for all other continuous paths $c$. Now, go through the list of axioms of a length structure (see e.g.here) that this is indeed a length structure. Lastly, think of a path $c$ such that $L(c)$ differs from the Euclidean length of $c$.

Edit. Suppose, in addition to the usual axioms of a length space you want the set of paths of finite length to be the same as in the Euclidean case. Here is a modification of my construction above:

Let $A$ be the subset of rectifiable (from Eucldiean viewpoint) continuous curves (maps from finite intervals to $R^2$). Let's define a length structure on $A$ as follows. For each $f:[a,b]\to R^2$ which belongs to $A$, let $S_f\subset [a,b]$ be the union of all open intervals $I\subset [a,b]$ such that $f$ restricted to these intervals is smooth (say, $C^1$). Then the Euclidean length $L(f)$ of $f$ equals $$ L(f)=\int_{S_f}|f'(t)|dt + R_f. $$

Furthermore, let $G_f$ be the union of all open subintervals of $S_f$ such that the restriction of $f$ to each of these subintervals is linear (more precisely, affine). Lastly, let $H_f:= S_f - G_f$. Now, set $$ L'(f):= \int_{G_f}|f'(t)|dt + 2\int_{H_f}|f'(t)|dt + R_f. $$ Hence, for each piecewise-linear path $f$, $L'(f)=L(f)$ but, in general, $L'(f)\ne L(f)$. For each nonrectifiable path $f$ set $L'(f)=\infty$.

I will leave you to check that $L'$ satisfies all the axioms of a length structure and defines the usual (Euclidean) distance on $R^2$ (since we did not change lengths of linear paths).

Moishe Kohan
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    Well, I see. But it seems like cheating. Is there more geometric example, without infinite lengths? Infinite length will definitely answer the question but this answer seems more like finding a loophole in the list of axioms. Thanks for posting, in any case – Vadim Feb 07 '18 at 18:26
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    @Vadim I do not know what you want: There are continuous paths of infinite length even for the standard length structure. You should update your question specifying what extra restrictions besides the axioms you have in mind. For instance, do you want all rectifiable paths (from Euclidean standpoint) to have finite length? That's also easy to accomplish (you can probably do it yourself by thinking about my solution for 5-10 minutes). – Moishe Kohan Feb 07 '18 at 18:34