Fractal curves area I was reading about the Sierpinski curve on the wiki page and it says that, considering the sequence of Sierpinski curves $S_n$ such that $\lim\limits_{n\rightarrow \infty}S_n$ completely fills the unit square,  the limit of the area enclosed by those curves is of $\frac{5}{12}$. 
I cannot think how it is possible that the limit of the areas enclosed by curves which are converging to a curve which fills the unit square can be different from $1$

It shouldn't be surprising that the area inside the curve is less than $1$.  As the curve wiggles around it leaves lots of area outside.  If you look at the images of the construction on the page you link to there is plenty of area outside the curve.  If the square is $[0,1] \times [0,1]$ the segment $(\frac 12, 0)$ to $(\frac 12, \frac 12)$ including the first point but not the second is outside the curve.  There are lots of branches off it, like $(\frac 14,\frac 14)$ to $(\frac 14, \frac 34)$ as well.  I suspect you can play with the construction to make the area inside be anything you like in the range $(0,1)$.  I don't know if you can get the endpoints.  It is important not to confuse the area inside the curve with the area of the curve.  The area inside is a nice finite number at each stage of the construction.  It looks to me that it approaches the limit from below because at each stage we add more area in squares and trapezoids than we take out in half octagons.  The area of the curve is $0$ at every finite stage, as it is a line of finite length, so the limit of the area of the curve is zero.  On the other hand, the area of the limit curve is $1$ as it covers the whole square.  Taking the limit and computing the area do not commute.
