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, $m_{2}\left(\bigcap_{n\in\mathbb{N}}{E_n}\right)=\lim_{n\rightarrow\infty}{m_{2}\left(E_n\right)}=\lim_{n\rightarrow\infty}{\left(\frac{1}{2}+\frac{1}{2^{n-1}}\right)}=\frac{1}{2}+0=\frac{1}{2}=m_{2}\left(\Delta'\right)$.
We shall be using the following result:
Lemma. if $E$ is Lebesgue measurable and $m(E)>0$, then for every $\alpha\in (0, 1)$, there is an interval $I$ such that $m(E \cap I) > \alpha \, m(I)$.
Proof. It is a consequence of the Lebesgue Density Theorem, according to which, if $m(E)>0$, then for almost every point $x \in E$
$$
\lim_{\varepsilon\to 0}
\frac{m\big(E \cap (x-\varepsilon,x+\varepsilon)\big)}{2\varepsilon} = 1.
$$
Hence for such an $x$ we can choose an $\varepsilon$, such that
$$
\frac{m\big(E \cap (x-\varepsilon,x+\varepsilon)\big)}{2\varepsilon} > \alpha
$$
or equivalently
$$
m\big(E \cap (x-\varepsilon,x+\varepsilon)\big)> \alpha\,m\big((x-\varepsilon,x+\varepsilon)\big)\big). \tag*{$\Box$}
$$
In our case, as $m(E)>0$, let an interval $I=[a,b]$, such that $m(E\cap I)>\frac{4}{5}m(I)$, and set $t=m(I)/5>0$.
Clearly, writing $E_t=E+t$,
$$
m\big(E_t\cap [a+t,b]\big)=m\big(E\cap [a,b-t]\big)
\ge m(E\cap I)-m\big([b-t,b]\big)\ge \tfrac{3}{5}m(I),
$$
and similarly
$$
m\big(E\cap [a+t,b-t]\big),\,m\big(E_t\cap [a+t,b-t]\big)\ge\frac{2}{5}m(I),
$$
while $m\big([a+t,b-t]\big)=\frac{3}{5}m(I)$. Therefore
$$
m(E\cap E_t)\ge m\big(E\cap E_t\cap [a+t,b-t]\big)\ge\frac{1}{5}m(I).
$$
Best Answer
My understanding:
If a function $f$ is continuous and $\lim_{n\to \infty}x_n=x$, then $f(x)=f(\lim_{n\to \infty}x_n)=\lim_{n\to \infty}f(x_n)$. That is, you can take the limit out. Similarly, $\lim_{n\to \infty}\cup_{k=1}^nA_k=\cup_{k=1}^{\infty}A_k$, the "continuity" property of measure implies $$ m(\cup_{k=1}^{\infty}A_k)=m(\lim_{n\to \infty}\cup_{k=1}^nA_k)=\lim_{n\to \infty}m(\cup_{k=1}^nA_k)=\lim_{n\to \infty}A_n $$
Note $\{A_n\}_{n=1}^{\infty}$ is an ascending collection of measurable sets, so $A_n=\cup_{k=1}^nA_k$.