 # 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 $m\left(G\right)=1$.
$G=\left\{\left(x,y\right)\in {\mathbb{R}}^{2}:x>0,0
I have to prove this without using the Lebesgue integral.
This set is open, so its measure is defined as it follows:
$m\left(G\right)=sup\left\{m\left(P\right):P\subset G\right\}$, where P is a special polygon.
We can construct a sequence of special polygons ${P}_{n}\subset G$:
${P}_{n}=\bigcup _{j=1}^{n}\left[0,log\left(\frac{n+1}{j}\right)\right]×\left[\frac{j-1}{n+1},\frac{j}{n+1}\right].$
The measure of one of these special polygons is $m\left({P}_{n}\right)=\sum _{j=1}^{n}log\left(\frac{n+1}{j}\right)\frac{1}{n+1}.$
As the sequence of measures $\left\{m\left({P}_{n}\right)\right\}$ is increasing, the supremum of this sequence is its limit:
$sup\left(\left\{m\left({P}_{n}\right)\right\}\right)=\underset{n\to \mathrm{\infty }}{lim}\sum _{j=1}^{n}log\left(\frac{n+1}{j}\right)\frac{1}{n+1}=\underset{n\to \mathrm{\infty }}{lim}\mathrm{log}{\left(\frac{\left(n+1{\right)}^{n}}{n!}\right)}^{\frac{1}{n+1}}.$
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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Step 1
I would use different approximating sets. Adding to G the set $0\le y\le 1$ will not affect the calculation. Then, the union, ${B}_{n}$, over $1\le i\le {n}^{2}$ of the disjoint sets ${A}_{ni}=\left[\frac{i-1}{n},\frac{i}{n}\right)×\left[0,{e}^{-i/n}\right]$ is closed, contained in G and has measure $\frac{1}{n}\sum _{i=1}^{{n}^{2}}{e}^{-i/n}=\frac{1}{n}\sum _{i=0}^{{n}^{2}-1}{e}^{-\left(i+1\right)/n}=\frac{{e}^{-1/n}}{n}\sum _{i=0}^{{n}^{2}-1}\left({e}^{-1/n}{\right)}^{i}=\frac{{e}^{-1/n}}{n}\cdot \frac{1-{e}^{-n}}{1-{e}^{-1/n}}\to 1$ as $n\to \mathrm{\infty },$, as you can check.
Step 2
To conclude, note that each ${B}_{n}$ is measurable, that ${B}_{n}\subseteq {B}_{n+1}$ and that $\bigcup _{n\in \mathbb{N}}{B}_{n}=G$, from which it follows that G is measurable and $m\left(G\right)=m\left(\bigcup _{n\in \mathbb{N}}{B}_{n}\right)=\underset{n\to \mathrm{\infty }}{lim}m\left({B}_{n}\right)=\underset{n\to \mathrm{\infty }}{lim}{m}_{\ast }\left({B}_{n}\right)=1.$

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