fubini's theorem, multivariable integration Evaluate the integral
$$ \int_{[0,1]^n} \max{(x_1,x_2,x_3, \cdots,x_n)} \,dx_1dx_2\cdots dx_n $$
Any hints comments are appreciated. 
Thanks 

The set of points
$$
H = \{(x_1,\ldots,x_n): x_i=x_j=\max(x_1,\ldots,x_n) \text{ for some } i\not=j\}
$$
lies inside the union of hyperplanes $\bigcup_{i\not=j}\{(x_1,\ldots,x_n): x_i=x_j\}$. Hence it has zero measure. Therefore
\begin{align}
&\int_{[0,1]^n} \max{(x_1,\cdots,x_n)} \,dx_1dx_2\cdots dx_n\\
=&n \int_{[0,1]^n} x_1{\bf 1}(x_2,\cdots,x_n< x_1)\,dx_1dx_2\cdots dx_n\tag{a}\\
=&n \int_0^1x_1\int_0^1 {\bf 1}(x_2<x_1)\int_0^1 {\bf 1}(x_3<x_1)\cdots\int_0^1 {\bf 1}(x_n<x_1)\,dx_n\cdots dx_2\,dx_1\tag{b}\\
=&n \int_0^1x_1\int_0^{x_1} \int_0^{x_1} \cdots\int_0^{x_1} \,dx_n\cdots dx_2\,dx_1\\
=&n \int_0^1x_1^n\,dx_1,
\end{align}
where ${\bf 1}(\cdot)$ denotes the indicator function, equality $(a)$ is due to symmetry and the fact that $H$ has measure zero, and $(b)$ is due to Fubini's theorem.
