I'm reading Apostol's Calculus. And I did the $(a)$ item in the following exercise:
As follows:
- Def: A set consisting of a single point is $P=\{(x,y)|x=0,y=0 \}$.
I don't need to specify a formula to construct all the points in the plane because of $(4)$, other points in the plane are going to be invariant under congruence.
- Due to $(5)$, $P$ is in $\mathcal{M}$ and $a(P)=0\cdot 0=0$, which is legitimate due to $(1)$, since $a(P)\geq 0$.
$$ \text{Reference for the axioms:}$$
By axiom 5, the rectangle having all four vertices at the point $(0,0)$ is measurable and has area $hk=(0-0)(0-0)=0$.
That was your argument. However, someone may object that this is not a true "rectangle" for axiom 5. Since axiom 5 is not clear on what a "rectangle" is, this objection may be valid.
For another proof that avoids this ambiguity, you could use limits. By axiom 3, using any measurable set $U$ we have
$$a(\emptyset)=a(U-U)=a(U)-a(U)=0$$
Consider the sets $S=\emptyset$, $Q=\{(0,0)\},$ and $T$ the rectangle with the vertices $(\pm\delta,\pm\delta)$ for some $\delta>0$. By axiom 5, $a(T)=4\delta^2$. We see that $S\subseteq Q\subseteq T$. The only number $c$ satisfying
$$0=a(\emptyset)=a(S)\le c\le a(T)= 4\delta^2$$
for all $\delta>0$ is $c=0$. Thus, by axiom 6, $Q$ is measurable and $a(Q)=0$.