Vitali set proof: Why is $1 \leq \sum_{k \in \mathbb{N}}m(\mathcal{N}_k) \leq 3$ impossible?

Question

So I am reading the construction of non-measurable set $\mathcal{N}$. This is done so by considering an enumeration $\{r_k\}$ of $\mathbb{Q} \cap [-1,1]$ I am stuck on seeing the obviousness in why (by monotonicity) The following is a contradiction. So we claim and show the following Inclusion

$$[0,1] \subset \bigcup_{k} \mathcal{N_k} \subset [-1,2]$$

Where $\mathcal{N}_k= \mathcal{N}+r_k$

I get that by monotonicity, we have

$$1 \leq \sum_{k \in \mathbb{N}}m(\mathcal{N}_k) \leq 3$$

But I do not see the contradiction as to why since neither $m(\mathcal{N})=0$ nor $m(\mathcal{N})>0$ we are done by contradiction, can't the measure be strictly between 1 and 3? so 2?

Answer 1

Since $m(\mathcal N_k)=m(\mathcal N)$ for each $k$, both cases lead to a contradiction:

• If $m(\mathcal N)=0$, then the infinite sum equals $0$ as well, yielding $1\leq 0\leq 3$, which is absurd.
• If $m(\mathcal N)>0$, the infinite sum equals infinity, which gives $1\leq\infty\leq 3$, again a contradiction.