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Nicolo is asking about functions where the inverse image of a Lebesgue measurable set is Lebesgue measurable. This is stronger than the usual definition of measurability where it is required only the …

There is a very naïve argument for this. As Henri says, it reduces to a zero set of a polynomial $f$. Write $$f(z_1,\dotsc,z_n)=\sum_{j=0}^d g_j(z_1,\dotsc,z_{n-1})z_n^j$$ where the polynomial $g_d$ i …

Every discrete space is a metric space. If we consider a measurable cardinal $\kappa$ as a discrete space, then it has an ultrafilter $\mathcal{F}$ in which the intersection of fewer than $\kappa$ ele …

It follows from Vitali's covering theorem but not in an entirely trivial fashion. We can reduce to the case where $E$ is open of finite measure. The set of all open balls contained in $E$ is then a Vi …

Let $B_p$ denote the 1-ball with centre 0 with respect to the $l^p$ norm. For any $p$ and $q$ there is a number $N$ such that $B_p$ is covered by $N$ translates of $B_q$. Then any $\epsilon$-ball in t …