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I am wondering about the following, possibly naive, question.

Suppose I have a smooth curve, which intersects the horizontal axis in a manner that leads to an accumulation point. More precisely, given a $C^2$ curve $y=f(x)$ such that the set $\{x$ s.t. $f(x)=0\}$ is a set of measure zero (for the Lebesgue measure) which contains an accumulation point, say $x=0$. Does it mean that the $\mathcal{H}^1$ Hausdorff measure of the curve (as a set in $\mathbb{R}^2$) is infinite?

(I hope that it is the case, but I would be very happy with a counter-example as well).

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No, it does not have to be infinite. For example, the graph of $y=x^{3}\sin (1/x)$, $-1\le x\le 1$, is $C^2$ smooth and its length is finite.

Actually, every $C^2$ curve with compact parameter domain has finite length.

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  • $\begingroup$ Thanks! On your example: does any measure type tool distinguish this graph from y=x (due to the accumulation of zeroes), or am I looking in the wrong direction? $\endgroup$ – walt Mar 24 '14 at 16:28
  • $\begingroup$ @walt Hausdorff measure and dimension are too rough for this; you need finer tools. It is true that if $\Gamma$ crosses the line transversally, and $\tilde \Gamma$ is $C^1$-close to $\Gamma$, then transversality holds also for $\tilde \Gamma$, hence the number of intersections is finite. This does not work in $C^0$ norm. $\endgroup$ – user127096 Mar 24 '14 at 16:32
  • $\begingroup$ That's very helpful, thanks again! $\endgroup$ – walt Mar 24 '14 at 16:34

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