A space is realcompact if its a closed subspace of an arbitrary product of real lines, with product topology.
A cardinal $\kappa$ is called measurable if there exists a (countably additive) $\{0, 1\}$-valued measure $\mu:\kappa\to \{0, 1\}$ with $\mu(\kappa) = 1$ and $\mu(\{x\}) = 0$ for $x\in \kappa$. This is not the standard meaning of what set theorists call a measurable cardinal. Apparently the proper terminology is "$\sigma$-measurable cardinal".
Its known that a discrete space is realcompact if and only if it has non-measurable cardinality. The proof of this result basically follows trivially from definitions. Its also known that a metrizable space of non-measurable cardinality must be realcompact.
For discussion on above see Rings of continuous functions by Gillman and Jerison.
In the same book, they never question if realcompactness is equivalent to non-measurable cardinality also for metric spaces. Is that true?
Also see this post for some previous discussion of whetever every metrizable space is realcompact.
Best Answer
Yes, its true.
See this great answer by KP Hart
I present a slightly different proof below.
Note the following result from the article The sup = max problem for the extent and the Lindelöf degree of generalized metric spaces, II by Hirata:
Theorem. (corollary 2.3) Let $X$ be a semi-stratifiable space with $e(X) = \kappa$, where $\text{cf}(\kappa) > \omega$. Assume that $\tau^\omega < \kappa$ for each $\tau < \kappa$. Then $X$ has a closed discrete subset of size $\kappa$.
Here $e(X) = \sup\{|D| : D\subseteq X, D\text{ is closed and discrete}\}+\omega$ is the extent of $X$.
Suppose that $X$ is a metrizable space (in particular, semi-stratifiable) with $|X|$ a measurable cardinal. Since $X$ is metrizable, $e(X) = w(X)$ and $|X|\leq w(X)^\omega$ so that $|X|\leq e(X)^\omega$, see Handbook of set-theoretic topology by Kunen and Vaughan. From $|X|\leq e(X)^\omega$ it follows that $e(X)$ is a measurable cardinal. If $e(X) = \kappa$ is the least measurable cardinal then $\kappa$ is strongly inaccessible, so conditions of Hirata's theorem are satisfied and there exists a closed discrete subset $D\subseteq X$ with $|D|$ measurable. Since closed subspace of realcompact space is realcompact, $X$ is not realcompact. If $\kappa$ is not the least measurable cardinal then again there exists a closed discrete set $D\subseteq X$ with $|D|$ measurable, which shows that $X$ is not realcompact.
Thus we established that a metrizable space $X$ is realcompact iff $|X|$ is non-measurable.