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In the context of Lebesgue measure on $\mathbb{R}^n$, is null set the same thing as a set of measure zero?

I understand that null set implies measure zero, not sure about the other direction.

Update: By null set I mean https://en.m.wikipedia.org/wiki/Null_set a set that can be covered by a countable union of intervals(balls) of arbitrarily small total length.

To be precise: Is a set with measure zero coverable by countable union of balls of arbitrarily small total length?

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    $\begingroup$ What is your definition of a null set? Usually the term null set means a set of measure zero. $\endgroup$ – carmichael561 May 11 '16 at 4:43
  • $\begingroup$ I have never seen a distinction. $\endgroup$ – copper.hat May 11 '16 at 5:02
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    $\begingroup$ For signed measures they are different. $\endgroup$ – Rick Sanchez May 11 '16 at 5:12
  • $\begingroup$ @carmichael561 By null set I mean it can be covered by countable union of balls of arbitrary small total measure. $\endgroup$ – yoyostein May 11 '16 at 5:19
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    $\begingroup$ @yoyostein so by definition you give it means Lebesgue measure of the set is 0. $\endgroup$ – user175968 May 11 '16 at 5:21
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Is a set with measure zero coverable by countable union of balls of arbitrarily small total length?

Yes. This follows from the fact that the Lebesgue measure is constructed using a generator of the Borel $\sigma$-algebra (say the set of open intervals in $\mathbb{R}$) using Caratheodory's Extension Theorem.

In particular, if we denote $\mu((a,b)) = b-a$ for any open interval $(a,b)$, we can define for any $A\subset\mathbb{R}$

$$\lambda^*(A) =\inf\left\{\sum_{n=1}^\infty\mu(A_n): \text{each $A_n$ an open interval, $A \subset \bigcup_{n=1}^\infty A_n, $}\right\}.$$

This $\lambda^*$ is the Lebesgue outer measure. If $\lambda$ denotes the Lebesgue measure then any measurable set satisfies $\lambda(A) = \lambda^*(A)$. The result then follows by the definition of the infimum.

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