# Using Fubini to obtain the volume of the $n$-dimensional unit ball

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.

• I don't see why Fubini's theorem isn't a rigorous argument here. – Andreas Blass Mar 21 '18 at 23:57
• I know that Fubini is valid for some product space. But what es the exact product space here. The expression $D_1 \times (B_1)_{x'}$ is dependent on x' ... do we use $B_1 = D_1 \times (B_1)_{x'}$? – Muzi Mar 22 '18 at 19:26
• Use the product space $\mathbb R^2\times\mathbb R^{n-2}$ and integrate the characteristic function of the ball. – Andreas Blass Mar 22 '18 at 19:29

$$\int_{B_1} 1(x) dx = \int_{\mathbb{R}^n} 1_{B_1}(x) dx = \int_{\mathbb{R}^{n-1}}\int_{\mathbb{R}^2} 1_{B_1}(x_1, \dots,x_n)dx_3\dots dx_ndx_1 dx_2$$ $$=\int_{\mathbb{R}^{n-1}}\int_{\mathbb{R}^2} 1_{ \{ x_3^2+\dots + x_n^2<1-x_1^2-x_2^2 \}}(x_1, \dots,x_n)dx_3\dots dx_ndx_1 dx_2$$ $$=\int_{\mathbb{R}^{n-1}} 1_{\{ x_1^2+x_2^2<1 \}} \int_{\mathbb{R}^2} 1_{ \{ x_3^2+\dots + x_n^2<1-x_1^2-x_2^2 \}}(x_1, \dots,x_n)dx_3\dots dx_ndx_1 dx_2$$ $$=\int_{D_1}\int_{(B_1)_{x'}} 1 dx''dx'$$