I have a feeling that uncountable products in $MBool$ do not exist in general. Suppose One thing is for example sure: if $B$ is MBool$ has uncountable products, then the 4-element Boolean algebra where we define forgetful functor $\mu(x) = 1/2$ for either of MBool \to Bool$ cannot preserve them (whence the elements $x$ strictly between forgetful functor couldn't have a left adjoint.)

Indeed, if $0$ and 2$ is the 2-element Boolean algebra with its unique measure $1$. \mu$. If the uncountable product $(B, (2, \mu)^{\omega_1}$ existed in $MBool$, where $\omega_1$ is the first uncountable ordinal, then and if this product is calculated as it seems would be in $Bool$ (which is the set-theoretic infinite product), then we could construct an uncountable strictly decreasing chain of maps $f_i: B \to B^{\omega_1}$, uniquely determined by the equations $(\pi_j \circ f_i)(x) = x$ for $j \leq i$ and $(\pi_j \circ f_i)(x) = 1$ for $j > i$ (here $\pi_j$ is projection onto the $j^{th}$ component). The elements $f_i(x) \in (B, \mu)^{\omega_1}$ are ordered in the manner suggested below: uncountable tuples)

$$(1, 1, 1, \ldots) > (x, 0, 1, 1, \ldots) > (x, x0, 0, 1, \ldots) > \ldots$$

$$\mu(1, 1, 1, \ldots) > \mu(x, mu(0, 1, 1, \ldots) > \mu(x, xmu(0, 0, 1, \ldots) > \ldots$$

by the axioms on metric Boolean algebras; this . This is impossible by countable cofinality considerations.

There

Regardless of whether there are infinite products in $MBool$, there is a trivial reason why the forgetful functor $U: MBool \to Bool$ cannot have a left adjoint $F$. F$ under the axioms given for metric Boolean algebras. Suppose WLOG that $\mu(1) = 1$ where the $1$ on the left is the top element of the putative free metric Boolean algebra $FB$. Then define a different metric $\mu'$ on $UFB$ by $\mu'(b) = r\mu(b)$ where $r > 1$, giving a different metric Boolean algebra $B'$. Then there is no nonexpansive metric Boolean algebra map $FB \to B'$ which extends the Boolean algebra embedding $i: B \to UB'$ along $i: B \to UFB$.

One might think this trivial objection could be remedied by adding an extra axiom like $\mu(1) = 1$ (so we are working with probability measures in a sense), but it seems highly doubtful even in that case that a left adjoint out of $Bool$ would exist. The lack of certain limits in As noted above, it could only exist if $MBool$ lacked many limits like uncountable products, and this would rule out a straightforward application of an adjoint functor theorem.

I have a feeling that uncountable products in $MBool$ do not exist in general. Suppose for example $B$ is the 4-element Boolean algebra where we define $\mu(x) = 1/2$ for either of the elements $x$ strictly between $0$ and $1$. If the uncountable product $(B, \mu)^{\omega_1}$ existed in $MBool$, where $\omega_1$ is the first uncountable ordinal, then it seems we could construct an uncountable chain of maps $f_i: B \to B^{\omega_1}$, uniquely determined by the equations $(\pi_j \circ f_i)(x) = x$ for $j \leq i$ and $(\pi_j \circ f_i)(x) = 1$ for $j > i$ (here $\pi_j$ is projection onto the $j^{th}$ component). The elements $f_i(x) \in (B, \mu)^{\omega_1}$ are ordered in the manner suggested below:

$$(1, 1, 1, \ldots) > (x, 1, 1, \ldots) > (x, x, 1, \ldots) > \ldots$$

and there would have to be a corresponding uncountable strictly decreasing chain of positive real numbers

$$\mu(1, 1, 1, \ldots) > \mu(x, 1, 1, \ldots) > \mu(x, x, 1, \ldots) > \ldots$$

by the axioms on metric Boolean algebras; this is impossible by countable cofinality considerations.

There is a trivial reason why the forgetful functor $U: MBool \to Bool$ cannot have a left adjoint $F$. Suppose WLOG that $\mu(1) = 1$ where the $1$ on the left is the top element of the putative free metric Boolean algebra $FB$. Then define a different metric $\mu'$ on $UFB$ by $\mu'(b) = r\mu(b)$ where $r > 1$, giving a different metric Boolean algebra $B'$. Then there is no nonexpansive metric Boolean algebra map $FB \to B'$ which extends the Boolean algebra embedding $i: B \to UB'$ along $i: B \to UFB$.

One might think this trivial objection could be remedied by adding an extra axiom like $\mu(1) = 1$ (so we are working with probability measures in a sense), but it seems highly doubtful even in that case that a left adjoint out of $Bool$ would exist. The lack of certain limits in $MBool$ would rule out a straightforward application of an adjoint functor theorem.

I am not sure about a left adjoint out of $Met$.