An application of continuity of Lebesgue measure — the area of a triangle in $\mathbb{R}^2$ 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.
 A: 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, $m_{2}\left(\bigcap_{n\in\mathbb{N}}{E_n}\right)=\lim_{n\rightarrow\infty}{m_{2}\left(E_n\right)}=\lim_{n\rightarrow\infty}{\left(\frac{1}{2}+\frac{1}{2^{n-1}}\right)}=\frac{1}{2}+0=\frac{1}{2}=m_{2}\left(\Delta'\right)$.
A: Consider the unit square with corners at $(0,0),\ (1,0),\ (1,1),\ (0,1)$ and the closed triangle $x \ge y$ lying below the "main diagonal" from $(0,0)$ to $(1,1)$. If this can be shown to be partitioned into pairwise disjoint squares whose area sums to $1/2$, then by applying the map $(x,y) \to (ax,by)$ one gets rectangles covering all the area (but not all the points, of course) of your triangle $\Delta.$
The idea of a construction is based on a "cantor set" idea of succesively removing middle ninths of a square, at each stage leaving $(8/9)^{th}$ of the area of that square, so that at the next iteration one has left $(8/9)^2$ of the area, and so on. Since $(8/9)^n \to 0$ as $n \to \infty$ the remaining area after all the middle squares are removed is $0$. One has definitely not removed all the points in the given square, in the same way as the one-dimensional cantor set has measure zero, so that one has removed a total length of 1 from the unit interval, while the (uncountable, nowhere dense) cantor set remains not removed.
At step one of the iteration we cut the unit square into 9 subsquares (like a tic-tac-toe board). We then try to remove the middle square, but cannot do this at step one since this middle square does not lie on or below the main diagonal. So nothing is removed at step one, all that is done is the partitioning of the unit square into the nine subsquares. 
Note that three of the nine subsquares lie completely in $\Delta$, so that on succesively removing middle ninths from them, the area of these three squares will have been exhausted. What is left is now three subtriangles, which are the lower halves of the three squares from step one which lie along the main diagonal. 
At step 2 each of these three squares along the main diagonal will be again subdivided into nine squares each, and the process iterated. 
A few things to note: Since at any stage we're removing a middle subsquare of a given square at a specific stage, no two of the removed squares share any boundary points. Also it seems in this way we will end up removing all the area, i.e. that the total area of the removed squares is $1/2$. There are some details in showing the total removed area is indeed $1/2$, and I'll try to fill that in; however I think this idea works to give your countable ascending chain of sets whose area approaches $1/2$ from below. However this construction does not partition the triangle into pairwise disjoint closed squares, which is impossible anyway. It just uses up all the area.
