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}}

ghettoking6q 2022-08-12 Answered
Lebesgue measure using special polygons
I want to prove that the following set satisfies m ( G ) = 1.
G = { ( x , y ) R 2 : x > 0 , 0 < y < 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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Answers (1)

Uriel Whitehead
Answered 2022-08-13 Author has 8 answers
Step 1
I 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 2
To 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.

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