Let $(X_\alpha)_{\alpha \in A}$ be a family of boolean random variables $X_\alpha: \Omega \to \{0,1\}$ on a probability space $\Omega = (\Omega, {\mathcal F}, {\mathbf P})$. Let ${\mathcal S}$ be a family of boolean sentences that each involve finitely many of the $X_\alpha$. Suppose that each sentence $S \in {\mathcal S}$ is almost surely satisfied by the $(X_\alpha)_{\alpha \in A}$. Can one then "repair" the random variables by locating further random variables $(\tilde X_\alpha)_{\alpha \in A}$ with each $\tilde X_\alpha$ almost surely equal to $X_\alpha$, such that the $\tilde X_\alpha$ surely satisfy all the sentences $S \in {\mathcal S}$?
If $|A| \leq \aleph_0$ (that is to say there are at most countably many random variables) then the task is easy, for then the set of sentences $S$ is also at most countable, and (because the countable union of null events is null) there is a single null event $N$ outside of which the $X_\alpha$ already surely satisfy all the sentences $S$. In particular there is a deterministic choice $X_\alpha^0 \in \{0,1\}$ of boolean inputs that satisfy all the sentences, and if one sets $\tilde X_\alpha$ to equal $X_\alpha$ outside of $N$ and $X_\alpha^0$ in $N$, we obtain the claim.
If $|A| \leq \aleph_1$ (that is to say $A$ has at most the cardinality of the first uncountable ordinal) and $\Omega$ is complete, then a slight variant of the above argument also works. We may well order $A$ so that every element $\alpha$ has at most countably many predecessors. We then use transfinite induction to recursively select $\tilde X_\alpha$ almost surely equal to $X_\alpha$, with the property that for all (not just almost all) sample points $\omega \in \Omega$, the tuple $(\tilde X_\beta(\omega))_{\beta \leq \alpha}$ may be extended to a tuple $(x_\beta)_{\beta \in A}$ solving all the sentences $S \in {\mathcal S}$. Indeed, if such variables $\tilde X_\beta$ have already been constructed for all $\beta < \alpha$, then the random variable $X_\alpha$ will already have this property outside of a null set $N_\alpha$ (here we use the fact that the set of tuples in the metrisable space $\{0,1\}^{\{ \beta: \beta \leq \alpha\}}$ that can be extended is the continuous image of a compact set and is thus closed and measurable). For each $\omega \in N_\alpha$, there exists at least one choice of $\tilde X_\alpha(\omega)$ that will obey the required extension property, thanks to the compactness theorem; using the axiom of choice to arbitrarily define $\tilde X_\alpha$ on this null set, we obtain a $\tilde X_\alpha$ with the required properties (it is measurable because $\Omega$ is assumed complete), and then the entire tuple $(\tilde X_\alpha)_{\alpha \in A}$ will surely satisfy all the sentences $S \in {\mathcal S}$. [It may be possible to drop the completeness hypothesis here by appealing to a measurable selection theorem; I have not thought about this carefully.]
Another illustrative case where the answer is affirmative is if $A$ is arbitrary and ${\mathcal S}$ is just the collection of equality sentences $X_\alpha = X_\beta$ for $\alpha,\beta \in A$. Thus we have $X_\alpha=X_\beta$ almost surely for each $\alpha,\beta$, and we wish to modify each $X_\alpha$ on a null set to create new random variables $\tilde X_\alpha$ such that $\tilde X_\alpha = \tilde X_\beta$. Note that for each $\omega \in \Omega$ it is not necessarily the case (even after deleting a null set) that all the $X_\alpha(\omega)$ are equal to each other (e.g., suppose $A=\Omega=[0,1]$ and $X_\alpha(\omega) = 1_{\alpha=\omega}$), but nevertheless the problem is easily solved in this case by arbitrarily selecting one element $\alpha_0$ of $A$ and defining $\tilde X_\alpha := X_{\alpha_0}$.
However, I do not have a good intuition as to whether the answer to this question is affirmative in general, even if one assumes good properties on the probability space $\Omega$ (e.g., that it is a standard probability space). The appearance of the cardinal $\aleph_1$ hints that perhaps the answer is sensitive to undecidable axioms in set theory.
(For my ultimate application I would eventually like to replace the boolean space $\{0,1\}$ with the interval $[0,1]$ or other Polish spaces, and the sentences $S$ with closed conditions involving finitely many or countably many of the variables at a time, but the Boolean case already seems nontrivial and captures much of the essence of the problem.)
EDIT: The following "near-counterexample" may also be suggestive. Set $\Omega = [0,1]$, let $A = 2^{[0,1]}$ be the power set of $\Omega$, and let $\mathcal{S}$ be the set of sentences $X_\alpha = X_\beta$ where $\alpha,\beta \subset [0,1]$ differ by at most one point. If one sets $X_\alpha(\omega) := 1_{\omega \in \alpha}$, then one morally has that the $X_\alpha$ almost surely satisfy all the sentences in $S$, but that there is no way to repair the $X_\alpha$ to random variables $\tilde X_\alpha$ that surely satisfy the equations as this would force $\tilde X_{[0,1]} = \tilde X_\emptyset$ while $X_{[0,1]}=1$ and $X_\emptyset = 0$. However this is not actually a counterexample because most of the $X_\alpha$ are non-measurable. (Removed due to errors)
Best Answer
In Terry's answer, he shows that his original question reduces to the question of whether, given a $\sigma$-algebra $\mathcal F$ on some set $X$ and a measure $\mu$ on $(X,\mathcal F)$, there is a ``splitting'' of the quotient algebra $\mathcal F / \mathcal N$, where $\mathcal N$ denotes the ideal of $\mu$-null sets. In this context, a splitting is a Boolean homomorphism $\Phi: \mathcal F / \mathcal N \rightarrow \mathcal F$ such that $\Phi([A]) \in [A]$ for all $A \in \mathcal F$. (Some authors call this a lifting instead of a splitting.) When some such $\Phi$ exists, let us say that $(X,\mathcal F,\mu)$ has a splitting.
I did some digging on this question this afternoon, and found two very good sources of information: David Fremlin's article in the Handbook of Boolean Algebras (available here) and a survey paper by Maxim Burke entitled "Liftings for noncomplete probability spaces" (available here). I'll summarize some of what I found below to supplement what Terry mentions in his answer. He mentions already that it is independent of ZFC whether $([0,1],\text{Borel},\text{Lebesgue})$ has a splitting:
Also mentioned already is the fact that if we expand the $\sigma$-algebra in question from the Borel sets to all Lebesgue-measurable sets, then the situation is more straightforward:
Now on to some not-yet-mentioned results. First, it's worth pointing out that one can obtain splittings with nice extra properties.
Once again, shrinking our $\sigma$-algebra from all $\mu$-measurable sets to only the Borel sets causes problems.
Thus, interestingly, Shelah's consistency result becomes a theorem of $\mathsf{ZFC}$ if we insist on the splitting being translation-invariant (with respect to mod-$1$ addition). More generally:
What stood out to me most in Fremlin and Burke's articles is how many questions seem to be wide open.
If yes, this would give a consistent positive answer to Terry's original question.
(Carlson showed that it is consistent to have $2^{\aleph_0} = \aleph_2$ and for $([0,1],\text{Borel},\text{Lebesgue})$ to have a splitting. Specifically, he showed that this holds whenever one adds precisely $\aleph_2$ Cohen reals to a model of $\mathsf{CH}$.)