every subset of a measurable set is measurable Is it true that every subset of a measurable set is measurable? for any measure. So if A is a measurable set then, B as a subset of A must be measurable wrt the same measure.
 A: No. This hold only for sets of measure $0$ assuming the measure is complete. The whole space itself is always measurable, so it would mean every set is measurable in that space whis not true for $\mathbb R$.
A: The answer to that question depends on a choice of the consistent system of axioms in which we prefer to work.
For example, if we consider the question asking  whether every subset of the real axis $R$ is Lebesgue measurable, then we will get the following different  pictures:
Theorem 1 (Vitali(1903)). (ZFC)  There exists a subset of the real axis $R$ which is not measurable in the Lebesgue sense.
Theorem 2 (Mycielski-Swierczkowski(1964)). (ZF+DC+AD) Every subset of the real axis $R$ is measurable in the Lebesgue sense.  
A: Certainly not. The set $[0,1]$ is Lebesgue measurable, but (if we assume the Axiom of Choice) it has non-measurable subsets. 
If $A$ is a set of Lebesgue measure $0$, then any subset of $A$ is Lebesgue measurable.
A: Say $(X,\mathfrak S, \mu)$ is a measure space. Measurability of any subset of a measurable set is not a property of the measure space, but rather of the measurable space $(X,\mathfrak S)$ associated to it, thus, in order to show it is false that every subset of a measurable set is measurable, it is in fact sufficient to show that $\mathfrak S$ is not closed wrt $\subset$. It should be therefore noticed that measurability is independent of the measure.
There are trivial counterexamples, e.g. any set $X$ of more than one point endowed with the trivial $\sigma$-algebra $\mathfrak S\doteq\{\varnothing, X\}$. The singleton constituted by any point $\{x\}$ in $X$ does not belong to $\mathfrak S$, making it a nonmeasurable set.
Many non-trivial counterexamples are also known for relatively well-behaved measurable spaces, such as the Vitali set, which rely on the Axiom of Choice.
As noted above, it is true that, given $(X,\mathfrak S,\mu)$ a measure space and $A\subset X$ any measurable set of measure $0$, every subset $B\subset A$ is measurable if and only if $(X,\mathfrak S,\mu)$ is a complete measure space (this may in fact be taken as a definition of complete measure space, as e.g. here).
