I am trying to find n-dimensional Lebesgue measure of A. Let A ⊂ R n be a set: A = { ( x 1 , x 2 , … , x n ) ∈ R n : ( ∑ i = 1 n | x i | ) ≤ 1 } 1. Exact same problem. The same problem appears here:"Let n ≥ 1 and n is an integer. Use Fubini's theorem to calculate the n-th Lebesgue measure of this set: { ( x 1 , x 2 , … , x n ) ∈ R n : x i ≥ 0 , i = 1 , 2 , … , ∑ i = 1 n x i ≤ 1 } "1.1 For n = 1, we have 0 ≤ λ ( A ) ≤ 1 ⟹ λ ( A ) ≤ 1. It seems to be obvious to me.1.2 For n = 2, we have 0 ≤ λ ( A ) ≤ 2 ⟹ λ ( A ) ≤ 2. Sqaure with a = 2 2. Similar problem. I have noticed similar problem, where A: A = { ( x 1 , x 2 , … , x n ) ∈ R n : ∑ i = 1 n | x i | : ∀ | x i | ≤ 1 } .

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I am trying to find   n-dimensional Lebesgue measure of   A. Let   A  ⊂            R              n       be a set:  A  =      {    (          x      1        ,          x      2        ,    …    ,          x      n        )    ∈                  R                    n              :          (              ∑                  i          =          1                n                    |                    x                  i                            |            )        ≤    1    }  1. Exact same problem. The same problem appears here:&quot;Let n   ≥ 1 and n is an integer. Use Fubini&#039;s theorem to calculate the n-th Lebesgue measure of this set:      {    (          x      1        ,          x      2        ,    …    ,          x      n        )    ∈                  R            n        :          x      i        ≥    0    ,    i    =    1    ,    2    ,    …    ,          ∑              i        =        1            n              x      i        ≤    1    }  &quot;1.1 For   n  =  1, we have   0  ≤  λ  (  A  )  ≤  1    ⟹    λ  (  A  )  ≤  1. It seems to be obvious to me.1.2 For   n  =  2, we have   0  ≤  λ  (  A  )  ≤  2    ⟹    λ  (  A  )  ≤  2. Sqaure with   a  =      2  2. Similar problem. I have noticed similar problem, where A:  A  =      {          (              x        1            ,              x        2            ,      …      ,              x                  n                    )        ∈                  R            n        :          ∑              i        =        1            n              |              x              i                    |        :    ∀          |              x              i                    |        ≤    1         }    .

gaiageoucm5p · Accepted Answer

Here is a probabilistic intepretation. Consider independent random variables   (      u    k        )          k      ∈      {      1      ,      2      ,      .      .      .      n      }       s.t.       u    k    :  Ω  →  [  −  1  ,  1  ] IID uniform. We have      F                  |                    u        1                    |              (  z  )  =  P  (      |        u    1        |    ≤  z  )  =  P  (  {      u    1    ∈  [  0  ,  z  ]  }  ∪  {      u    1    ∈  [  −  z  ,  0  ]  }  )  =  2      z    2    =  z  ,    z  ∈  [  0  ,  1  ]and by independence                              P          n                =        P        (                  |                          u          1                          |                +                  |                          u          2                          |                +        .        .        .        +                  |                          u          n                          |                ≤        1        )                            =                  ∫                      [            0            ,            1                          ]              n                                                            1                                {            x            :            ∑                          x              k                        ≤            1            }                          (        x        )        d        x        =                                          =                  ∫                      [            0            ,            1            ]                                    ∫                      [            0            ,                          x              1                        ]                                    ∫                      [            0            ,                          x              2                        ]                          (        .        .        .        )                  ∫                      [            0            ,                          x                              n                −                1                                      ]                          d                  x          n                (        .        .        .        )        d                  x          3                d                  x          2                d                  x          1                    So                    n        =        1                                    ⟹                          P          1                =        1                            n        =        2                                    ⟹                          P          2                =                  2                      −            1                                              n        =        3                                    ⟹                          P          3                =        (        3        ⋅        2                  )                      −            1                                                            (        .        .        .        )                            n        =        k                                    ⟹                          P          k                =        (        k        !                  )                      −            1                              Now to find the volume       V    n   of   A (i.e.   λ  (  A  )  ) as we wanted, we solve the proportion      P    n    :  1  =      V    n    :      2    n      ⟹        V    n    =      2    n        P    n  where   λ  (  [  −  1  ,  1      ]    n    )  =      2    n  .

I am trying to find n -dimensional Lebesgue measure of A . Let A ⊂

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