The classic example of a non-measurable set is described by [wikipedia](http://en.wikipedia.org/wiki/Vitali_set#Construction_and_proof). However, this particular construction is reliant on the axiom of choice; in order to choose representatives of $\mathbb{R} /\mathbb{Q}$.


"Since each element intersects [0,1], we can use the axiom of choice to choose a set containing exactly one representative out of each element of R / Q."



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> Is it possible to construct a non-measurable set (in $\mathbb{R}$ for example) without requiring the A.o.C.?
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Does constructing non-measurable sets require the axiom of choice?
In the 1960's, Bob Solovay constructed a model of ZF + the axiom of dependent choice (DC) + "all sets of reals are Lebesgue measurable." DC is a weak form of choice, sufficient for developing the "non-pathological" parts of real analysis, for example the countable additivity of Lebesgue measure (which is not provable in ZF alone). Solovay's construction begins by assuming that there is a model of ZFC in which there is an inaccessible cardinal. Later, Saharon Shelah showed that the inaccessible cardinal is really needed for this result.


