Simple question about Lebesgue and Borel sigma algebra

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If N is of null Lebesgue measure. Can we find a Borel set B of null measure such that N is entirely contained in B?
 
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it's the other way around. we want B borel such that N is contained in B
 
quasar987 said:
it's the other way around. we want B borel such that N is contained in B
Oops!

This is also true. It follows from the fact that the Lebesgue measure is regular.