Consider the topologist's sine curve:

$$ f(x) = \sin\left(1 \over x \right), x \neq 0 $$

The graph of this function resembles a space-filling curve near $x=0$. It is not a space filling curve, though, because no matter how close we get to the y-axis, there are always points not included in the curve (i.e. $\forall x \neq 0 \exists y \neq \sin(1/x)$). Indeed, "most" points in any given interval are not part of the curve, I think. Is there a meaningful way to express the "area" which this curve fills or appears to fill?

More informally, how much of this image is blue?

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    $\begingroup$ I took a look at that image, and using my mathematician's eyes, I confirmed that yep, there is no blue. $\endgroup$ Mar 11, 2015 at 9:11
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    $\begingroup$ it might be interesting to compare with the en.wikipedia.org/wiki/Osgood_curve which is not space-filling (it is a continuous non self-intersection arc, or a Jordan curve) yet has positive area. $\endgroup$
    – Mirko
    Mar 11, 2015 at 15:04

2 Answers 2


Ross Millikan has observed that the area [= Lebesgue planar measure] of the sine curve is zero. By the same process Ross Millikan used, one can show that the Hausdorff dimension is $1,$ which is a much stronger statement than saying it has planar measure zero. In fact, it has $\sigma$-finite Hausdorff $1$-measure, which also prevents it from having Hausdorff dimension "logarithmically greater than" $1.$ However, for certain more sensitive types of dimension, the sine curve will have dimension greater than $1.$

Azcan/Kocak/Orhun/Üreyen [1] prove that the box-counting dimension [= Minkowski dimension] of the graph of $y = \sin(1/x)$ exists and is equal to $\frac{3}{2}.$ This result is also proved in Goluzina/Lodkin/Makarov/Podkorytov [2] (Problem VIII.5.4b, p. 100; solution on pp. 282-283) and it is stated in Tricot [3] (Section 10.4, pp. 121-122). In fact, Tricot [3] observes more generally that for $0 < \alpha < \beta,$ the graph of $y = x^{\alpha}\cos\left(x^{- \beta}\right)$ (and hence for the corresponding SINE function as well) has a box-dimension that exists and is equal to $2 - \frac{\alpha + 1}{\beta + 1}.$

[1] Hüseyin Azcan, Sahin Kocak, Nevin Orhun, and Mehmet Üreyen, The box-counting dimension of the sine-curve, Mathematica Slovaca 49 (1999), 367-370.

[2] M. G. Goluzina, A. A. Lodkin, B. M. Makarov, and A. N. Podkorytov, Selected Problems in Real Analysis, Translations of Mathematical Monographs #107, American Mathematical Society, 1992, x + 370 pages.

[3] Claude Tricot, Curves and Fractal Dimension, Springer-Verlag, 1995, xiv + 323 pages.


The area of the curve itself is zero. You can divide the interval $(0,1]$ into countably many intervals from one zero crossing to the next, then show that the area of the curve is zero in that interval. As area is countably additive, the total is zero. The curve only appears space filling because we draw it with finite width.

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    $\begingroup$ The closure also has area zero, since it is just the union of the curve and a line segment. $\endgroup$
    – Ian
    Mar 10, 2015 at 20:18
  • $\begingroup$ Then more generally, can you find any (possibly pathological) map $f:\mathbb{R}\to\mathbb{R}$ whose graph $\Gamma = \{ (x,y)\in\mathbb{R}^2 \mid y=f(x) \}$ either has no area (as in non-measurable), or has a strictly positive area? Or will graphs like that always be null sets? $\endgroup$ Mar 11, 2015 at 14:42
  • $\begingroup$ @JeppeStigNielsen: For anything you can write as a function $y=f(x)$ the area will be zero. To get positive area you must have many $y$s that correspond to a single $x$. The classic Peano curve is parameterized as a function of $t$, so is relieved of the requirement to have a single $y$ for each $x$. $\endgroup$ Mar 11, 2015 at 14:45
  • $\begingroup$ That is what I suspected, but I thought I could make you come up with a proof of that. $\endgroup$ Mar 11, 2015 at 14:47
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    $\begingroup$ It looks like the answer in the thread Lebesgue measure of the graph of a function defines an $f$ whose graph is non-measurable. $\endgroup$ Mar 11, 2015 at 16:03

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