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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?

Popopo
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1 Answers1

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There's no paradox. We are on physics stackexchange, not mathematics stack exchange. Non-measurable sets are purely mathematical concepts that cannot be physically instantiated. Any medium in our universe is either made out of particles that are discrete or fields which, as far as we know, can be modeled as being continuous in our 4 dimensional space-time. There is no way to construct a non-measurable set in this universe.

FrankH
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