Lebesgue measure using special polygonsI want to prove that the following set satisfies m ( G ) = 1. G = { ( x , y ) ∈ R 2 : x &gt; 0 , 0 &lt; y &lt; e − x }I have to prove this without using the Lebesgue integral.This set is open, so its measure is defined as it follows: m ( G ) = sup { m ( P ) : P ⊂ G }, where P is a special polygon.We can construct a sequence of special polygons P n ⊂ G: P n = ⋃ j = 1 n [ 0 , l o g ( n + 1 j ) ] × [ j − 1 n + 1 , j n + 1 ] .The measure of one of these special polygons is m ( P n ) = ∑ j = 1 n l o g ( n + 1 j ) 1 n + 1 .As the sequence of measures { m ( P n ) } is increasing, the supremum of this sequence is its limit: sup ( { m ( P n ) } ) = lim n → ∞ ∑ j = 1 n l o g ( n + 1 j ) 1 n + 1 = lim n → ∞ log ⁡ ( ( n + 1 ) n n ! ) 1 n + 1 .I know that this limit is 1, so the logarithm argument must tend to e, but I don't know how to prove this analytically. I've tried working on the final expression to get the definition of e, but I haven't been able to do so. Could yo help me?

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Lebesgue measure using special polygonsI want to prove that the following set satisfies   m  (  G  )  =  1.  G  =  {  (  x  ,  y  )  ∈            R        2    :  x  &amp;gt;  0  ,  0  &amp;lt;  y  &amp;lt;      e          −      x        }I have to prove this without using the Lebesgue integral.This set is open, so its measure is defined as it follows:  m  (  G  )  =  sup  {  m  (  P  )  :  P  ⊂  G  }, where P is a special polygon.We can construct a sequence of special polygons       P    n    ⊂  G:      P    n    =      ⋃          j      =      1        n        [    0    ,    l    o    g          (                                    n            +            1                    j                    )        ]    ×      [                  j        −        1                    n        +        1              ,          j              n        +        1              ]    .The measure of one of these special polygons is   m  (      P    n    )  =      ∑          j      =      1        n    l  o  g      (                            n          +          1                j              )        1          n      +      1        .As the sequence of measures   {  m  (      P    n    )  } is increasing, the supremum of this sequence is its limit:  sup  (  {  m  (      P    n    )  }  )  =      lim          n      →      ∞            ∑          j      =      1        n    l  o  g      (                            n          +          1                j              )        1          n      +      1        =      lim          n      →      ∞        log  ⁡            (                        (          n          +          1                      )            n                                    n          !                    )                      1                  n          +          1                      .I know that this limit is 1, so the logarithm argument must tend to e, but I don&#039;t know how to prove this analytically. I&#039;ve tried working on the final expression to get the definition of e, but I haven&#039;t been able to do so. Could yo help me?

Uriel Whitehead · Accepted Answer

Step 1I would use different approximating sets. Adding to G the set   0  ≤  y  ≤  1 will not affect the calculation. Then, the union,       B    n  , over   1  ≤  i  ≤      n    2   of the disjoint sets       A          n      i        =      [                  i        −        1            n        ,          i      n        )    ×  [  0  ,      e          −      i              /            n        ] is closed, contained in G and has measure       1    n        ∑          i      =      1                      n        2                  e          −      i              /            n        =      1    n        ∑          i      =      0                      n        2            −      1            e          −      (      i      +      1      )              /            n        =            e              −        1                  /                n              n        ∑          i      =      0                      n        2            −      1        (      e          −      1              /            n            )    i    =            e              −        1                  /                n              n    ⋅            1      −              e                  −          n                            1      −              e                  −          1                      /                    n                      →  1 as   n  →  ∞  ,, as you can check.Step 2To conclude, note that each       B    n   is measurable, that       B    n    ⊆      B          n      +      1       and that       ⋃          n      ∈              N                  B    n    =  G, from which it follows that G is measurable and   m  (  G  )  =  m      (          ⋃              n        ∈                  N                            B      n        )    =      lim          n      →      ∞        m  (      B    n    )  =      lim          n      →      ∞            m    ∗    (      B    n    )  =  1.

I want to prove that the following set satisfies m(G)=1. G={(x,y) in mathbb{R}^2:x>0,0<y<e^{-x}}

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