# Different proofs that the outer measure of a singleton is equal to zero

Let $\lambda^*$ be the outer measure defined as $$\lambda^*(E)=\inf_{E\space\subset\space\cup_{n=0}^\infty\space I_n}\space\sum_{n\space\in\space\mathbb N}\space l(I_n)\qquad\forall\space E\in\mathscr P(\mathbb R)$$ where $I_n=(a_n,b_n)$ and $l(I_n)=b_n-a_n$.

Let $x \in \mathbb R$. I want to prove that $$\lambda^*(\{x\})=0\qquad\qquad(1)$$ I can easily prove that the outer measure of an interval is its length using Heine-Borel theorem at some point of the proof, so$$\lambda^*((a,b))=b-a\quad\quad(2)$$ Using this fact I can prove $(1)$ by showing that $\{x\}\subset I$ such that the measure of $I$ is arbitrarely small and using monotonicity (which I've already proved). If till now I am correct, my answer is: can I prove $(1)$ without using $(2)$? Maybe using only monotonicity.

Comment (you can skip): since I think the fact that signletons have zero measure is quite foundamental I would like to prove it using as less other results as possible.

• Given $\epsilon>0$ it really isn't hard to find an interval containing $\{x\}$ of length $\epsilon$. Or indeed even a sequence of intervals of total length $\epsilon$... Commented Oct 27, 2017 at 15:39

I'm a bit confused by your question. As far as I understand what you're asking, yes you can prove that $\lambda^*(\{x\}) = 0$ without knowledge of the outer measure of intervals. Simply note that $$\{x\}\subset \left(x-\frac{1}{n},x+\frac{1}{n}\right)$$ for all $n \in \mathbb{N}$ and hence $$\lambda^*(\{x\}) \le \inf_{n \in \mathbb{N}}l\left(x-\frac{1}{n},x+\frac{1}{n}\right)=\inf_{n \in \mathbb{N}} \frac{2}{n}=0.$$