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Essential undecidability of PA says every complete extension of PA includes a non r.e. set of new "axioms," all undecidable in PA.  In that sense lots of sentences of PA are undecidable in PA.  And it is trivial to design a measure on, say sentences of PA with Godel number below $n$, where the set of decidable sentences has measure 1 (for all $n$ above the Godel number of the first decidable sentence, which would be 0=0 for the usual Godel numberings).  Just take the counting measure while restricting the count to decidable sentences.  Similar cheap tricks will give other uninformative results. 

But is there a useful, nontrivial way to define the density of decidable sentences in PA? 

The answer seems to be the state of the art.  It has nothing special to do with PA of course.  It holds for any incomplete theory, just using any theorem in place of $0=0$ and any refutable sentence in place of $0=1$. Chaitin's "heuristic principle" uses complexity to suggest that in fact almost all true sentences are independent in any recursive extension of PA, as cited in discussion of How many of the true sentences are provable? But people do not seem persuaded by this.

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Asymptotic density seems a very natural measure. The density of a set of sentences is the limit as $n\to\infty$ of the proportion of those sentences of length at most $n$ amongst all sentences of length at most $n$.

(One should use a formalism that has only finitely many sentences of a given length—for example, one can use variable symbols $x$, $x'$, $x''$, and so on, where the prime counts as another symbol, instead of indices; or else just stratify the sentences as a union of finite sets, such as by Gödel codes.)

Having density 1 means that as one considers longer and longer sentences, the proportion of them in the set goes to 100%. Having density 0 means that as one considers longer and longer sentences, the proportion of them in the set goes to 0.

We similarly get a notion of lower density and upper density by using $\liminf$ and $\limsup$.

Using this measure, it turns out that the set of theorems, the set of refutable sentences, and the set of independent sentences (over some fixed theory) all have a lower density bounded away from zero and an upper density bounded away from 1.

For example, every sentence of the form $\varphi\vee( 1=1)$ is a theorem, and this is a positive density set. Every sentence of the form $\varphi\wedge (1=0)$ is refutable, and this is a positive density set. And every sentence of the form $(\varphi\vee (1=1))\wedge\psi$ is independent, where $\psi$ is some fixed independent sentence, and this also is a positive density set.

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    $\begingroup$ (copy of my Oct 2015 answer to the MSE question Proportion of true statements that are provable) -- Marek Zaionc has studied this issue for the last several years in the case of various systems of propositional logic. See also the StackExchange question Are there more true statements than false ones?. $\endgroup$ Commented Jul 14, 2023 at 20:44
  • $\begingroup$ But also, David's link goes back to 2009. It seems this question has been considered many times. $\endgroup$ Commented Jul 14, 2023 at 20:46
  • $\begingroup$ It is not obvious to me that sentences of the form $\varphi\vee( 1=1)$ are a positive density set. How can I see that? $\endgroup$ Commented Jul 14, 2023 at 21:38
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    $\begingroup$ It is easier to see in the space of all strings of a specific length, not just wffs. For example, there are $26^n$ strings of length $n$ from an alphabet of size 26. If one fixes 6 of the characters to have specific values, there are $26^{n-6}$ strings in agreement, which makes proportion $1/26^6$, which is positive. In first-order logic, we don't allow all strings of characters as sentences, but the rate of growth is still basically exponential, and so fixing some characters to have specific values will still have about the corresponding fraction of the whole space. $\endgroup$ Commented Jul 14, 2023 at 21:44
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    $\begingroup$ I agree with that inclination, but the issue is that if there are infinitely many variables, you'd have infinitely many formulas of that size, and this prevents the asymptotic density from working. One needs to stratify the collection of all sentences by a tower of finite families. It seems that there are many ways to do so, such as the methods we've been discussing. $\endgroup$ Commented Jul 15, 2023 at 12:55

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