Prove that set $A=\{(x,y)\in \mathbb{R}; \left|x\right| + \left|y\right|=q; q \in \mathbb{Q}\}$ has measure $0$

Solution 1:

It suffices to show that $A_q := \{ (x,y)\in\mathbb R^2 : |x| + |y| = q\}$ is of measure $0$ for all $q\in\mathbb Q$, because then we know by the subadditivity of the Lebesgue-measure $\lambda$ and the countability of $\mathbb Q$, that $$ \lambda(A) = \lambda\left( \bigcup_{q\in\mathbb Q} A_q\right) \leq \sum_{q\in\mathbb Q} \lambda(A_q) = 0. $$

Now, as you probably have already noticed: $$ A_q = \{(x, q-x): x\in[0,q] \} \cup \{ (x, x-q) : x\in[0,q]\} \cup \{ (x, x + q) \mid x\in[-q,0] \} \cup \{ (x, q-x) : x\in[-q,0] \} $$

Now, $A_q$ is the union of finitely many graphs of continuous functions defined on a bounded interval. Each of them is a Lebesgue-null set which concludes the proof of the claim.

Alternatively you could also manually approximate each of the four subsets of $A_q$ with small squares $K_j$ as follows:

We'll do the proof for an interval $[a,b]$ where $a,b\in\{0, q, -q\}$ and a function $f$ with $f'\equiv c$ with $c\in\{-1, 1\}$. For $n\in \mathbb N$, we split up $[a,b]$ in equisized subintervals $I_1,\dots, I_n$ of length $\frac{b-a}{n}$. We define the square $K_i = I_i \times [\min_{x\in I_i}f(x), \max_{x\in I_i}f(x)]$ (which is a square as $f'\equiv c$ with $c\in\{-1,1\}$). Then we have $A_q\subseteq \bigcup_{i=1}^n K_i$ with $\lambda(\bigcup_{i=1}^n K_i) = \sum_{i=1}^n \lambda(K_i) = n\cdot (\frac{b-a}{n})^2 = \frac{b-a}{n}$. As this tends to $0$ for $n\rightarrow\infty$, this concludes the proof.