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?
Fubini theorem and double integrals
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0Maybe you might be interested in [this](http://math.stackexchange.com/q/104210/19341)... – 2012-04-05
1 Answers
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.