# Fubini theorem and double integrals

I'm just wondering to know why Fubini theorem does not apply in evaluating the area of snowflakes and how we can evaluated by using double integral as the limit of a sum?

• What do you mean by the "snowflake" shape? – J. M. is a poor mathematician May 10 '11 at 3:41
• Are you referring to the Koch snowflake fractal? Might this have something to do with the boundary of curve having nonzero Lebesgue measure? – Elchanan Solomon May 10 '11 at 6:18
• Maybe you might be interested in this... – draks ... Apr 5 '12 at 18:11

Suppose that $A \subset \mathbb{R}^2$ is a snowflake. Recall that the characteristic function of $A$ is $$\chi_A(x,y) = \begin{cases} 1, & (x,y) \in A,\\ 0 & (x,y) \notin A\end{cases}.$$ Remember that the area of $A,$ which we call $\mu(A),$ equals $$\mu(A) = \iint_{\mathbb{R}^2} \chi_A(x,y)\,dx\,dy.$$
For any sufficiently nice function $f:\mathbb{R}^2 \to \mathbb{R},$ let us define $f^{\uparrow}:\mathbb{R}\to\mathbb{R}$ by $f^{\uparrow}(x) = \int_{\mathbb{R}} f(x,y) dy.$ Similarly, let us define $f^{\rightarrow}(y) = \int_{\mathbb{R}} f(x,y) dx.$ Then Fubini's theorem tells us that $$\int_\mathbb{R} f^{\uparrow}(x) dx = \int_{\mathbb{R}} f^{\rightarrow}(y) dy = \iint_{\mathbb{R}^2} f(x,y) dx\,dy.$$
If we wanted to apply this to a formula for $\mu(A),$ we would have to calculate one of the functions $f^{\uparrow}$ or $f^{\rightarrow},$ and then integrate it. But if $A$ is a snowflake, $\chi_A$ is very complicated, and $f^{\uparrow}$ is complicated, and $f^{\rightarrow}$ is very complicated, so their integrals are very difficult to evaluate directly.