If one forces with measure algebra, then every formula $\phi(\tau\_{1},\tau\_{2},...,\tau\_{k})$ where $\tau\_{i}$'s are names, has a truth value, the measure of which is a real number. It is just a curiosity that is the "set"(may I say set?) of measures of all truth values equals to $[0,1]$? If fixed a name for a random real, letting $\phi$ run over, is the set of measures of all truth values (seems countable) dense in $[0,1]$ ?



distibution of truth values of all formulas on [0,1]
To answer your latter question, if by a name for a random real you mean the canonical name for the real determined by the generic filter over the measure algebra, which is known as a *random* real, then the answer is yes. This is true because for any notion of forcing $\mathbb{B}$, then the boolean value $[\![\check b\in\dot G]\!]$ is precisely $b$ for every $b\in \mathbb{G}$. So in the case of random real forcing, if $\dot r$ is the canonical name for the random real, then by using an interval $I$ of measure $q$, say, the boolean value of $[\![\dot r\in \check I]\!]$ will be precisely $I$, which has measure $q$, and these are dense in the unit interval. (And this was essentially the point of Goldstern's comment.)


But if you by a name for "a random real" you instead mean any old real, perhaps from the ground model, then it is no longer true that the boolean values of statements will have measures dense in the unit interval. The reason is that the measure algebra forcing is almost homogeneous, and therefore statements $\varphi(\check t\_0,\ldots,\check t\_n)$ in the forcing language involving only check names $\check t\_i$ will always have boolean value either $0$ or $1$. 


