For the volume of the n-dimensional unit ball $B_1(0) \subset \mathbb R_n$ one can write $$ |B_1(0)| = \int_{B_1} 1 dx $$
in this document the author is using induction to obtain the volume but there is still one step which is confusing me:
For $n>2$ let $x \in B_1(0)$ then we can write $x=(x',x'')$ with $x'=(x_1,x_2)$ and $x''=(x_3, \dots, x_n)$ such that $$ x' \in D_1= \{ x \in \mathbb R ^2 : |x|<1 \} $$ and $$ x'' \in (B_1)_{x'} = \{x'' \in \mathbb R ^ {n-2} : (x',x'') \in B_1(0) \} $$ Now the author is appealing Fubini's theorem: $$ \int_{B_1} 1 dx = \int_{D_1} \int_{(B_1)_{x'}} dx''dx' $$ For me this makes somehow geometrically sense (like e.g. the coarea formula for integrals) but a rigorous argument is missing.
Thanks to the tips of Andreas Blass I understand the application of Fubini now. Here a quite detailed argument:
$$ \int_{B_1} 1(x) dx = \int_{\mathbb{R}^n} 1_{B_1}(x) dx = \int_{\mathbb{R}^{n-1}}\int_{\mathbb{R}^2} 1_{B_1}(x_1, \dots,x_n)dx_3\dots dx_ndx_1 dx_2 $$ $$ =\int_{\mathbb{R}^{n-1}}\int_{\mathbb{R}^2} 1_{ \{ x_3^2+\dots + x_n^2<1-x_1^2-x_2^2 \}}(x_1, \dots,x_n)dx_3\dots dx_ndx_1 dx_2 $$ $$ =\int_{\mathbb{R}^{n-1}} 1_{\{ x_1^2+x_2^2<1 \}} \int_{\mathbb{R}^2} 1_{ \{ x_3^2+\dots + x_n^2<1-x_1^2-x_2^2 \}}(x_1, \dots,x_n)dx_3\dots dx_ndx_1 dx_2 $$ $$ =\int_{D_1}\int_{(B_1)_{x'}} 1 dx''dx' $$