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How to prove that if $A \subset \mathbb{R}$ is null set, then for every $\varepsilon > 0$ there exists a countable collection of open intervals $(A_{n})_{n=1}^{\infty}$ such that $A \subseteq \bigcup\limits_{n=1}^{\infty} A_{n}$ and $\sum\limits_{n=1}^{\infty} \ell(A_{n}) < \varepsilon$.

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    $\begingroup$ That's one possible definition. If you don't tell us your definition, we can't help you. $\endgroup$ – PhoemueX Apr 2 '16 at 17:54
  • $\begingroup$ What would happen if this were not the case? Try proof by contradiction. $\endgroup$ – siegehalver Apr 2 '16 at 17:57
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The out measure is defined by the infimum of the sums of open coverings. If this infimum is 0, for each $\epsilon>0$, must exist a covering by open sets such that the sum is $<\epsilon$. (If not, the infimum will be $>\epsilon $, an absurd

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