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EDIT: This answer is wrong. I am not deleting it at the request of the OP. See the counterexample at the bottom. Here is a terrible mathematician answer that you didn't know that you didn't care for …

In the early 20th century there was a lot of fuss over the axiom of choice implying that there are Lebesgue non-measurable sets of reals. In his book about The Axiom of Choice, Gregory Moore points to …

No, you can't have that. It is consistent that the real numbers are a countable union of countable sets, in which case you immediately have that there is no nontrivial measure which is countably addit …

This is not really a problem. If $\kappa$ is a measurable cardinal then $V_\kappa$, or the set of sets which are hereditarily have size smaller than $\kappa$, is a model of $\sf ZFC$. This means that …

Some remarks: First of all it is known since 1906 that the axiom of choice implies the existence of non-measurable sets. What Solovay have shown is that it is consistent (relative to the existence of …

Suppose that for all $\alpha<\kappa$ we have that $A_\alpha\subseteq\kappa$. We define the diagonal intersection to be $$\bigtriangleup_{\alpha<\kappa}A_\alpha = \left\lbrace\xi<\kappa\ \middle|\ \xi\ …

The existence of a non-measurable set is completely too weak to prove the axiom of choice. It is consistent with ZF (without large cardinals at all) that the real numbers are a countable union of co …

Suppose $M$ is a model of ZFC, and $\mu^M$ is the Lebesgue measure on $\mathbb R^M$ such that $\mu^M(\mathbb R^M)=1$. It is known that if $r$ is a Cohen real over $M$ and $N=M[r]$ then $\mu^N(\mathbb …

The metric function is continuous, therefore measurable. One can show that measurable functions can be built with a hierarchy similar to the Borel hierarchy. You start with continuous functions (into …