# Non-measurable subsets of a null set

Given the Lebesgue measure on $\mathbb{R}$, I am wondering if a non-measurable set can always be included in a null set? More precisely, let $A$ be a set in the Borel σ-algebra $\mathcal{B}(\mathbb{R})$, to which we assign a strictly positive Lebesgue measure. Given a theorem in measure theory, it contains non-measurable sets.

Can we say, according to some theorem (please state the theorem) that all the non-measurable sets contained in the set $A$ are also contained in some measurable subset $B$ of $A$ to which we can assign the value zero? I know that this is possible for null sets by defining the negligeable sets and eventually the completion of the measurable space.

• No, because if $A$ is non-measurable, then $A^c$ is non-measurable too – Tryss Apr 29 '15 at 8:58
• A is in fact measurable as a borel set. – ivo Apr 29 '15 at 9:30
• My $A$ is not the Borel set $A$, I should have called this set $C$. Sorry for the confusion – Tryss Apr 29 '15 at 12:46

Every subset $A$ of a null set $N$ is itself null. This can be shown directly by observing that every covering of $N$ also covers $A$.