Measurability of $\frak c$ Can the continuum cardinality $\frak c$ be a measurable cardinal ?
Why not, what is the simplest reason besides that ZFC proves existence of $\frak c$ but not existence of a measurable cardinal?
 A: Measurable cardinals are all strong limits, and the continuum cannot be a strong limit since $\aleph_0$ is less than it and $2^{\aleph_0}$ is equal to it.
On the other hand, it is consistent that the continuum is a real-valued measurable cardinal. (While real-values measurability implies certain modest largeness properties, it is much less than for measurables, which are truly large cardinals. For instance, real-valued measurables need to be as big as a weakly inaccessible, but there’s nothing stopping the continuum from being weakly inaccessible.)
As a side note, the fact that ZFC proves the existence of the continuum and not the existence of measurables has nothing to do with it, since both properties could be consistent and consistently coincide. For instance ZFC doesn’t prove the existences of weak inaccessibles but it is consistent that the continuum is weakly inaccessible. (Assuming weak inaccessibles are consistent.)
A: The continuum cannot be a measurable cardinal. Indeed, measurable cardinals are always strong limit cardinals: if $\kappa$ is measurable and $\lambda<\kappa$ then $2^\lambda<\kappa$. (And much, much, much more is true; measurables are extremely large.)
In the specific case of $\mathfrak{c}$, there's a quick way to see this: topology. Suppose $U$ is a countably complete ultrafilter on $\mathbb{R}$. Then for each $i$ there is a (non-unique) closed interval of size $2^{-i}$ with rational endpoints which is in $U$. 


*

*Why? Well, the set of all such rationals is countable and covers $\mathbb{R}$; so if each such interval were not in $U$, the set of their complements would be a countable set of elements of $U$ whose intersection (being empty) is not in $U$, contradicting countable closure. 


But taking the intersection of these closed intervals, we get that $U$ contains a singleton: since $U$ is countably closed, the intersection of countably many elements of $U$ is again in $U$.
So every countably complete ultrafilter on $\mathbb{R}$ is principal, and so $\vert\mathbb{R}\vert=\mathfrak{c}$ is not measurable.
