I know that the n-dimensional Lebesgue-Measure λ n is defined at least on all Borel Sets B n = B ( R n ). Let's assume A = ( a , b ) ⊆ R n .(i) My first question is whether it is possible to define λ n ( A )? I suppose that this leads to λ n ( A ) = 0?(ii) Is it allowed to use the 1-dimensional Lebesgue-Measure here or can we do no other than using λ n since the space is R n ?May anybody have an answer/explanation to this?

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I know that the n-dimensional Lebesgue-Measure       λ    n   is defined at least on all Borel Sets             B        n    =      B    (            R        n    ). Let&#039;s assume   A  =  (  a  ,  b  )  ⊆            R        n  .(i) My first question is whether it is possible to define       λ    n    (  A  )? I suppose that this leads to       λ    n    (  A  )  =  0?(ii) Is it allowed to use the 1-dimensional Lebesgue-Measure here or can we do no other than using       λ    n   since the space is             R        n  ?May anybody have an answer/explanation to this?

odvucimo1pp17 · Accepted Answer

Firstly,   (  a  ,  b  ) is not a subset of             R        n   for   n  ≥  2. Perhaps you want to embed   (  a  ,  b  )  ⊂      R   in             R        n   through a homeomorphism. In particular, you can consider the set       A    ′    =  (  a  ,  b  )  ×                              {          0          }          ×          …          ×          {          0          }                ⏟                    n      −      1       times        ⊂            R        n  .   A and       A    ′   are homeomorphic via the obvious map   f  :  A  →      A    ′    ,  t  ↦  (  t  ,  0  ,  …  ,  0  ).1. Instead of       λ    n    (  A  ), it makes sense to find       λ    n    (      A    ′    ). Indeed,       λ    n    (      A    ′    )  =  0.2. No, you cannot use the one-dimensional Lebesgue measure here, since your set is a subset of             R        n   (and not       R  ).

I know that the n-dimensional Lebesgue-Measure λ n </msup> is d

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