0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ I don't see why Fubini's theorem isn't a rigorous argument here. $\endgroup$ – Andreas Blass Mar 21 '18 at 23:57
  • $\begingroup$ 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 $B_1 = D_1 \times (B_1)_{x'}$? $\endgroup$ – Muzi Mar 22 '18 at 19:26
  • 1
    $\begingroup$ Use the product space $\mathbb R^2\times\mathbb R^{n-2}$ and integrate the characteristic function of the ball. $\endgroup$ – Andreas Blass Mar 22 '18 at 19:29
1
$\begingroup$

Thanks to the tips of Andreas Blass I understand the application of Fubini now. Here a quite detailed argument:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.