Find $n$-dimensional measure of set $A$

Problem. Find $n$-dimensional Lebesgue measure of $A$. Let $A \subset \mathbb{R}^{n} $ be a set:

$$ A = \left\{ (x_1, x_2, \ldots, x_n) \in \mathbb{R}^{n}: \left( \sum_{i=1}^n |x_{i}| \right) \leq 1 \right\} $$

• Exact same problem. The same problem appears [2].

  • For $n = 1$, we have $0 \leq \lambda(A) \leq 1 \implies \lambda(A) \leq 1$. It seems to be obvious to me.

  • For $n = 2$, we have $0 \leq \lambda(A) \leq 2 \implies \lambda(A) \leq 2$.
    Sqaure with $a = \sqrt{2}$.

• Similar problem. I have noticed similar problem [1], where $A$: $$ A = \left\{ \left( x_1, x_2, \ldots, x_{n} \right) \in \mathbb{R}^n : \sum_{i=1}^n |x_{i}| : \forall |x_{i}| \leq 1\ \right\}. $$

Solution 1:

Here is a probabilistic intepretation. Consider independent random variables $(u_k)_{k\in \{1,2,...n\}}$ s.t. $u_k:\Omega \to [-1,1]$ IID uniform. We have $$F_{|u_1|}(z)=P(|u_1|\leq z)=P(\{u_1 \in [0,z]\}\cup\{u_1 \in [-z,0]\})=2\frac{z}{2}=z,\,z \in [0,1]$$ and by independence $$\begin{aligned}P_n=P(|u_1|+|u_2|+...+|u_n|\leq 1)&=\int_{[0,1]^n}\mathbf{1}_{\{x:\sum x_k \leq 1\}}(x)dx=\\ &=\int_{[0,1]}\int_{[0,x_1]}\int_{[0,x_2]}(...)\int_{[0,x_{n-1}]}dx_n(...)dx_3dx_2dx_1\end{aligned}$$ So $$\begin{aligned}n=1&\implies P_1=1\\ n=2&\implies P_2=2^{-1}\\ n=3&\implies P_3=(3\cdot 2)^{-1}\\ &(...)\\ n=k&\implies P_k=(k!)^{-1}\end{aligned}$$ Now to find the volume $V_n$ of $A$ (i.e. $\lambda(A))$ as we wanted, we solve the proportion $$P_n:1=V_n:2^n\implies V_n=2^nP_n$$ where $\lambda([-1,1]^n)=2^n$.