Can null Lebesgue measurable sets be contained in null Borel sets?

[quasar987]

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?

 

[morphism]

The empty set!

 

[quasar987]

it's the other way around. we want B borel such that N is contained in B

 

[quasar987]

bump!

 

[morphism]

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.