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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 $[0,1]$ ?

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    $\begingroup$ I am not sure this is a "research level" question. The truth value of a formula is not a number, but a measurable set (or an equivalence class of measurable sets). Perhaps you meant to ask about the measure of this set? In that case, I don't see the point of this question, but the answer seems easy -- just compute the truth value of $\dot r \le p/q$, where $\dot r$ is the canonical name of the random real. $\endgroup$
    – Goldstern
    Commented Dec 31, 2011 at 10:07

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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 $1$.

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  • $\begingroup$ Since this thread got needlessly bumped, I took the liberty of fixing the MathJax excessive `` (and turn them into a Boolean norm symbol, or so). $\endgroup$
    – Asaf Karagila
    Commented Jan 29, 2014 at 22:05

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