I'm trying to prove that the Lebesgue measure of a straight line in $\mathbb{R}^2$ is zero. I've found certain posts here on Stackexchange concerning this, but I have a slightly different solution which I would like to discuss.
Let $L \subset \mathbb{R}^2$ be a straight line, and let $(p_n, q_n)_{n=1}^\infty$ be an enumeration of the countable set $L \cap \mathbb{Q}^2$. Fix $\epsilon > 0$ and define
$$ A_n = \left\{ (x,y)\in \mathbb{R}^2 \,\mid\, -\frac{\sqrt{\epsilon}}{2\sqrt{2^n}} \leq x-p_n < \frac{\sqrt{\epsilon}}{2\sqrt{2^n}}, \;
-\frac{\sqrt{\epsilon}}{2\sqrt{2^n}} \leq y-q_n < \frac{\sqrt{\epsilon}}{2\sqrt{2^n}} \right\}. $$
Since $L \cap \mathbb{Q}^2$ is dense in $L$, it follows that
$$ L \subset \bigcup_{n=1}^\infty A_n, $$
and so
$$ \lambda(L) \leq \sum_{n=1}^\infty \lambda(A_n) = \sum_{n=1}^\infty \text{vol}(A_n) = \sum_{n=1}^\infty \epsilon/2^n = \epsilon. $$
But $\epsilon$ was arbitrary, so $\lambda(L) = 0$.
Is this argument fine?
Best Answer
The outline and overall plan of your argument are fine: cover $L$ with countably many sets of small total area, and so on.
But your execution is too complicated and (as the comments indicate) has technical flaws.
My recommendation would be: come up with a proof that the $x$-axis has measure $0$, and then shift and rotate your construction to take care of arbitrary $L$.
Exactly how you do this might depend on what kinds of sets you know how to measure, and what you already know about 2 dimensional measure. You might for instance cover $L$ with disks, the $n$th disk with radius $\epsilon/n$ so the sum of their diameters is big enough to cover all of $L$, but yet with finite small total area.