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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?

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What do you mean by the "snowflake" shape? – J. M. 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? – Isaac 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.

Therefore, I would not say that Fubini's theorem does not "apply;" instead, I would say that it is not very useful in this situation.

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