"Area" of the topologist's sine curve

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.