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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.

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  • $\begingroup$ 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 $\epsilon$... $\endgroup$ Commented Oct 27, 2017 at 15:39

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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 $$\lambda^*(\{x\}) \le \inf_{n \in \mathbb{N}}l\left(x-\frac{1}{n},x+\frac{1}{n}\right)=\inf_{n \in \mathbb{N}} \frac{2}{n}=0.$$

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    $\begingroup$ Oh, of course, that is really trivial! I 've been confused by the definition of outer measure and didn't recall that it already contains the notion of lenght and there was no need to use the length of the intervals. My bad, thank you $\endgroup$
    – bdbit
    Commented Oct 27, 2017 at 15:49

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