This is a problem from Royden's Real Analysis. Fix $a,b\in\mathbb{R}$, and let $\Delta:=\left\{(x,y)\in\mathbb{R}^2\Big|x\in[0,a];y\in\left[0,\frac{b}{a}x\right]\right\}$, so that $\Delta$ is a closed triangle with vertices $(0,0)$, $(a,0)$, and $(a,b)$. I "know" that $m_{2}(\Delta)=\frac{ab}{2}$, since the Lebesgue measure in $\mathbb{R}^2$ is simply the area. I want to prove this using rectangles and applying the continuity of the Lebesgue measure, so I need to use rectangles to construct either (a) a countable ascending chain of measurable sets whose union is $\Delta$, or (b) a countable descending chain of measurable sets of finite measure whose intersection is $\Delta$. I haven't been able to do either.
[Math] An application of continuity of Lebesgue measure — the area of a triangle in $\mathbb{R}^2$
measure-theoryreal-analysis
Best Answer
I've found an alternative solution to the problem:
Let $\Delta':=\left\{\left(x,y\right)\in\mathbb{R}^2\Big|x\in\left[0,1\right];y\in\left[0,x\right]\right\}$. Then $\Delta$ is an inhomogeneous dilation of $\Delta'$; namely, $\Delta=\varphi\Delta'$, where $\varphi:(x,y)\mapsto(ax,by)$.
Hence it is sufficient to show that $m_{2}\left(\Delta'\right)=\frac{1}{2}$. We shall construct a decreasing sequence of measurable sets of finite measure whose intersection is $\Delta$ as follows: We start with $E_1:=[0,1]\times[0,1]$. We then break the unit square into four subsquares, each of measure $\frac{1}{4}$; $E_2$ is the almost disjoint union of those three such subsquares which intersect $\mathrm{Int}\left(\Delta'\right)$ nontrivially, viz., $E_2:=\left(\left[0,\frac{1}{2}\right]\times\left[0,\frac{1}{2}\right]\right)\cup\left(\left[\frac{1}{2},1\right]\times\left[0,\frac{1}{2}\right]\right)\cup\left(\left[0,\frac{1}{2}\right]\times\left[\frac{1}{2},1\right]\right)$. We repeat this process inductively, so that $\forall n\in\mathbb{N}$, $E_n$ is the almost disjoint union of those subsquares of measure $\frac{1}{2^{2n-2}}$ which have nonempty intersection with $\mathrm{Int}\left(\Delta'\right)$. Each $E_n$ is measurable, as it is a finite union of measurable sets, and, moreover, each $E_n$ is of finite measure. By construction, $E_n\supseteq E_{n+1}\hspace{3pt}\forall n\in\mathbb{N}$, and $m_{2}\left(E_n\right)=\frac{2^{2n-3}+2^{n-1}}{2^{2n-2}}=\frac{\left(2^{n-1}\right)\left(2^{n-2}+1\right)}{\left(2^{n-1}\right)^2}=\frac{2^{n-2}+1}{2^{n-1}}=\frac{2^{n-2}}{2^{n-1}}+\frac{1}{2^{n-1}}=\frac{1}{2}+\frac{1}{2^{n-1}}$. Observe that $\Delta'=\bigcap_{n\in\mathbb{N}}{E_n}$. Then $m_{2}\left(\Delta'\right)=m_{2}\left(\bigcap_{n\in\mathbb{N}}{E_n}\right)$. By continuity of measure, $m_{2}\left(\bigcap_{n\in\mathbb{N}}{E_n}\right)=\lim_{n\rightarrow\infty}{m_{2}\left(E_n\right)}=\lim_{n\rightarrow\infty}{\left(\frac{1}{2}+\frac{1}{2^{n-1}}\right)}=\frac{1}{2}+0=\frac{1}{2}=m_{2}\left(\Delta'\right)$.