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> **Proposition.** $\lambda [0,1] = 0$
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*Proof.* Let $\varepsilon>0$ be arbitrary. We will prove that $\lambda[0,1] <\varepsilon$. Let $q : \mathbb{N} \rightarrow \mathbb{R}$ denote an injection with image equal to $\mathbb{Q} \cap [0,1]$. Let $p : \mathbb{N} \rightarrow \mathbb{R}\_{>0}$ denote a sequence with $\sum\_{i \in \mathbb{N}} p\_i < \varepsilon$. Then $$\lambda [0,1] = \lambda \bigcup\_{i \in \mathbb{N}} (q\_i+[-p\_i/2,p\_i/2]) \leq \sum\_{i \in \mathbb{N}} \lambda (q\_i+[-p\_i/2,p\_i/2]) = \sum\_{i\in \mathbb{N}} p\_i < \varepsilon$$

So $\lambda[0,1] < \varepsilon$. Since this is true for all $\varepsilon>0$, we deduce that $$\lambda [0,1] = 0.$$

> 
> **Question.** What gives?
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