I want to prove that the following set satisfies $m(G)=1$. $$G=\{(x,y)\in \mathbb{R}^2 : x>0, 0<y<e^{-x}\}$$
(This exercise is proposed on Lebesgue Integration on Euclidean Space by Frank Jones, page 30).
I have to prove this without using the Lebesgue integral.
This set is open, so its measure is defined as it follows: $m(G)=\sup\{m(P) : P\subset G\}$, where $P$ is a special polygon.
We can construct a sequence of special polygons $P_n\subset G$: $$P_n=\bigcup_{j=1}^n\left[0,log\left({\frac {n+1}j}\right)\right]\times \left[\frac{j-1}{n+1},\frac{j}{n+1}\right].$$
The measure of one of these special polygons is $$m(P_n)=\sum_{j=1}^n log\left({\frac {n+1}j}\right)\frac{1}{n+1}.$$
As the sequence of measures $\{m(P_n)\}$ is increasing, the supremum of this sequence is its limit: $$\sup(\{m(P_n)\})=\lim_{n\to\infty}\sum_{j=1}^n log\left({\frac {n+1}j}\right)\frac{1}{n+1}=\lim_{n\to\infty}\log \left(\frac{(n+1)^n}{n!}\right)^{\frac 1 {n+1}}.$$
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?
Thanks in advance!
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 $A_{ni}=\left[\frac{i-1}{n},\frac{i}{n}\right)\times [0,e^{-i/n}]$ is closed, contained in $G$ and has measure $\frac{1}{n}\sum^{n^2}_{i=1}e^{-i/n}=\frac{1}{n}\sum^{n^2-1}_{i=0}e^{-(i+1)/n}=\frac{e^{-1/n}}{n}\sum^{n^2-1}_{i=0}(e^{-1/n})^i=\frac{e^{-1/n}}{n}\cdot \frac{1-e^{-n}}{1-e^{-1/n}}\to 1$ as $n\to \infty, $ as you can check.
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 $m(G)=m\left(\bigcup_{n\in \mathbb N}B_n\right)=\underset{n\to\infty}\lim m(B_n)=\underset{n\to\infty}\lim m_*(B_n)=1.$