oleifere45

2022-06-22

I am trying to find $n$-dimensional Lebesgue measure of $A$. Let $A\subset {\mathbb{R}}^{n}$ be a set:
$A=\left\{\left({x}_{1},{x}_{2},\dots ,{x}_{n}\right)\in {\mathbb{R}}^{n}:\left(\sum _{i=1}^{n}|{x}_{i}|\right)\le 1\right\}$
1. Exact same problem. The same problem appears here:
"Let n $\ge$ 1 and n is an integer. Use Fubini's theorem to calculate the n-th Lebesgue measure of this set:
$\left\{\left({x}_{1},{x}_{2},\dots ,{x}_{n}\right)\in {\mathbb{R}}^{n}:{x}_{i}\ge 0,i=1,2,\dots ,\sum _{i=1}^{n}{x}_{i}\le 1\right\}$"
1.1 For $n=1$, we have $0\le \lambda \left(A\right)\le 1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\lambda \left(A\right)\le 1$. It seems to be obvious to me.
1.2 For $n=2$, we have $0\le \lambda \left(A\right)\le 2\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\lambda \left(A\right)\le 2$. Sqaure with $a=\sqrt{2}$
2. Similar problem. I have noticed similar problem, where A:

gaiageoucm5p

Here is a probabilistic intepretation. Consider independent random variables $\left({u}_{k}{\right)}_{k\in \left\{1,2,...n\right\}}$ s.t. ${u}_{k}:\mathrm{\Omega }\to \left[-1,1\right]$ IID uniform. We have
${F}_{|{u}_{1}|}\left(z\right)=P\left(|{u}_{1}|\le z\right)=P\left(\left\{{u}_{1}\in \left[0,z\right]\right\}\cup \left\{{u}_{1}\in \left[-z,0\right]\right\}\right)=2\frac{z}{2}=z,\phantom{\rule{thinmathspace}{0ex}}z\in \left[0,1\right]$
and by independence
$\begin{array}{rl}{P}_{n}=P\left(|{u}_{1}|+|{u}_{2}|+...+|{u}_{n}|\le 1\right)& ={\int }_{\left[0,1{\right]}^{n}}{\mathbf{1}}_{\left\{x:\sum {x}_{k}\le 1\right\}}\left(x\right)dx=\\ & ={\int }_{\left[0,1\right]}{\int }_{\left[0,{x}_{1}\right]}{\int }_{\left[0,{x}_{2}\right]}\left(...\right){\int }_{\left[0,{x}_{n-1}\right]}d{x}_{n}\left(...\right)d{x}_{3}d{x}_{2}d{x}_{1}\end{array}$
So
$\begin{array}{rl}n=1& \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{P}_{1}=1\\ n=2& \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{P}_{2}={2}^{-1}\\ n=3& \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{P}_{3}=\left(3\cdot 2{\right)}^{-1}\\ & \left(...\right)\\ n=k& \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{P}_{k}=\left(k!{\right)}^{-1}\end{array}$
Now to find the volume ${V}_{n}$ of $A$ (i.e. $\lambda \left(A\right)\right)$ as we wanted, we solve the proportion
${P}_{n}:1={V}_{n}:{2}^{n}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{V}_{n}={2}^{n}{P}_{n}$
where $\lambda \left(\left[-1,1{\right]}^{n}\right)={2}^{n}$.

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