What will happen when measuring unmeasurable object?

There is a set called Vitali Set which is not Lebesgue measurable.

Analogously, there also exists a Vitali set $Y$ in $\mathbb R^3$ which is a subset of $[0,1]^3$ and $|Y\cap q|=1$ for all $q\in \mathbb R^3/\mathbb Q^3$. However, I'm curious about if it fulfilled a kind of isotropic uniform medium, let this isotropic uniform medium has density $\rho$, and put it on a electronic scale to weigh, what reading can we get? Note that $m_Y=\rho V_Y$ but $V_Y$ seems to be undefined... So it seems we cannot get any real reading. But on the other hand, since we are using a electronic scale, it also seems we must get a reading...A paradox?

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The existence of Vitali sets relies on the axiom of choice, which makes the question a duplicate of this Phys.SE post. Also related: physics.stackexchange.com/q/20370/2451 – Qmechanic Dec 12 '12 at 16:20

 If an unmeasurable object existed, which law would it go against? – Popopo Dec 12 '12 at 17:41 It could not be made of matter since matter is made of atoms and particles and it could not be a field since fields are continuous. So if you discovered such an object it would be a new form of substance which would not fit into any law of physics that we currently understand. So in a sense such an object would violate all of our laws. – FrankH Dec 12 '12 at 19:43 Well, in newtonian mechanics objects are systems of mass points. So my question is about a Vitali set fulfilled with mass points. It sounds possible to exist. – Popopo Dec 13 '12 at 3:42 But it is a FINITE number of points, immeasurable sets have an uncountably INFINITE number of points. So they are not comparable. – FrankH Dec 13 '12 at 4:14 What about a mass point system contains countable infinite many mass points? It sounds strange that a mass point system can contains either finite or $\beth_1$ mass points but not to $\aleph_0$ which is actually between them. – Popopo Dec 13 '12 at 5:46