is it possible to find a measurable subset $A$ of a non-measurable set $B$ such that $A$ can be assigned the measure zero or more particularly a non zero value ( in the case of the Lebesgue measure then a strictly positive value ) ? Can you give an example ?
I am also looking for an explicit example of a non-measurable set contained in a null set given a certain measure.
Thanks.
I am not sure what you are hoping for by "explicit" non-measurable set. The construction (?) of such a set requires the use of the Axiom of choice. I realize that that sentence is paradoxical, but so be it.
Regardless, given a non-measurable set $C$ disjoint from $A$, then $B= A\cup C$ should work. Try showing that this $B$ is, in fact, non-measurable given that $C$ is.
The existence of nonmeasurable sets is standard in any Real analysis book. For example, let $C$ be a set of coset representatives of the quotient $\mathbb{R}/\mathbb{Q}$. One can show this is not measurable. (Try it!)