Here is some elaboration on Joel David Hamkins's answer.

A (two-valued) measurable cardinal is an uncountable cardinal $\kappa$ such that there is a continuous $({<\kappa})$-additive $\{0,1\}$-valued probability measure $\mu$ defined on all subsets of $\kappa$. To say that $\mu$ is $({<\kappa})$-additive means that if $(X_i)_{i \in I}$ is a family of pairwise disjoint subsets of $\kappa$ with $|I|<\kappa$, then $\mu\left(\bigcup_{i \in I} X_i\right) = \sum_{i\in I} \mu(X_i)$. Since $\mu$ can only take values $0$ and $1$, this is equivalent to saying that (a) at most one of the $X_i$ can have measure $1$, and (b) if they all have measure $0$ then so does $\bigcup_{i \in I} X_i$. (Some people say $\kappa$-additive instead of $({<\kappa})$-additive, but I prefer to use $\kappa$-additive to mean the above for families with index set of size *equal* to $\kappa$.)

The existence of a continuous countably additive $\{0,1\}$-valued probability measure $\mu$ defined on all subsets of a set $S$ implies the existence of a measurable cardinal. Indeed, I claim that if $\kappa \geq \aleph_1$ is the *smallest* cardinal such that $\mu$ is not $\kappa$-additive, then $\kappa$ is a measurable cardinal. To see this, let $(X_i)_{i<\kappa}$ be a family of pairwise disjoint measure $0$ sets such that $\bigcup_{i<\kappa} X_i$ has measure $1$ (i.e. the family contradicts $\kappa$-additivity in the only possible way). Defining $\bar{\mu}(I) = \mu\left(\bigcup_{i \in I} X_i\right)$ for every $I \subseteq \kappa$, we obtain a $({<\kappa})$-additive $\{0,1\}$-valued probability measure $\bar\mu$ defined on all subsets of $\kappa$.

So the existence of a continuous countably additive $\{0,1\}$-valued measure defined on all subsets of a set $S$ is exactly equivalent to the existence of a measurable cardinal. However, since a probability measure is also allowed to take values strictly between $0$ and $1$, this is not quite equivalent to the statement you asked about.

By analogy with the above, a real-valued measurable cardinal is an uncountable cardinal $\kappa$ such that there is a continuous $({<\kappa})$-additive probability measure defined on all subsets of $\kappa$. The existence of a real-valued measurable cardinal is equivalent to your statement by a variation of the trick used above.

In 1930, Ulam (Zur Masstheorie in der allgemeinen Mengenlehre, Fund. Math. 16) showed that if $\kappa$ is real-valued measurable then $\kappa \geq 2^{\aleph_0}$, and that if $\kappa > 2^{\aleph_0}$ is real-valued measurable then $\kappa$ is in fact measurable (with a possibly different measure). Ulam also showed that successor cardinals like $\aleph_1$ cannot be real-valued measurable.

In the 1960's, Solovay (MR290961) finally resolved the boundary case. He showed that if $\kappa = 2^{\aleph_0}$ is real-valued measurable then there is an inner model (namely $L[I]$ where $I$ is the ideal of null sets) wherein $\kappa$ is still real-valued measurable and GCH holds, therefore $\kappa$ is measurable in that inner model by Ulam's earlier results. While this doesn't mean that the existence of a real-valued measurable cardinal and the existence of a measurable cardinal are equivalent, it shows that the two statements are equiconsistent over ZFC.

Using forcing (and another result of Ulam), Solovay also showed that if there is a model with a measurable cardinal then there is a model in which the Lebesgue measure on $[0,1]$ can be extended to a probability measure defined on all subsets of $[0,1]$.