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Unlike the collection $L$ of Lebesgue measurable sets, the collection $J$ of Jordan measurable sets do not form a Sigma algebra. (A set is Jordan measurable if and only if its characteristic function is Riemann integrable, and the characteristic function of a singleton is Riemann integrable, but the characteristic function of the rational numbers is not.) But we can still talk about $\sigma(J)$, the Sigma algebra generated by $J$.

Now in this 1993 journal paper, K.G. Johnson shows that the Borel Sigma algebra is a proper subset of $\sigma(J)$ which is a proper subset of the Lebesgue Sigma algebra, and that a set is in $\sigma(J)$ if and only if it can be written as a union of a Borel set and a subset of a measure $0$ $F_\sigma$ set. But Johnsons says there are two questions he doesn’t know the answer to:

Where does $\sigma(J)$ stand relative to the analytic sets ... and relative to the universally measurable sets[?]

My question is, have either of these questions been answered in the two decades since this paper was published? Is it known whether there are analytic sets which are not in $\sigma(J)$ or vice versa? Is it known whether there are universally measurable sets which are not in $\sigma(J)$ or vice versa? Or are these still open problems?

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Here is an answer to your question.

Assume $V=L$, there is a null $\Pi^1_1$-set which is not in $\sigma(J)$.

Proof: Let $A=\{x\mid x\in L_{\omega_1^x}\}$ be the $\Pi^1_1$-null set. Now suppose that $A=B\cup C$ where $B$ is a null Borel set and $C$ is a subset of a $\Sigma^0_2(x)$ null set $F$ for some $x$. Since $B$ is Borel, there must be some countable $\alpha$ so that for any $z$ with $z\in A$ and $\omega_1^z>\alpha$, $z$ does not belong to $B$. Fix this $\alpha$ and let $A_{\alpha}=\{z\mid z\in A\wedge \omega_1^z>\alpha\}$. Then $A_{\alpha}\subseteq F$. By a result in Kjos-Hanssen, Bjørn; Nies, André; Stephan, Frank; Yu, Liang, Higher Kurtz randomness, Ann. Pure Appl. Logic 161, No. 10, 1280-1290 (2010). ZBL1223.03025, there is a real $z\in A_{\alpha}$ which is $x$-Kurtz random and so $z\not\in F$, a contradiction.

There are lots of higher recursion theory facts hidden in the proof. You are welcome to ask any questions concerning these facts.

This method may be pushed further to have a $ZFC$-proof by defining $A=\{x\oplus y\mid y\in L_{\omega_1^{x\oplus y}}\ [x]\}$.

And clearly there is a Jordan measurable set which is not $\mathbf{\Sigma}^1_1$.

OK, the question was already answered in Mauldin, R. Daniel(1-NTXS) Analytic non-Borel sets modulo null sets. (English summary) Set theory (Boise, ID, 1992–1994), 69–70, Contemp. Math., 192, Amer. Math. Soc., Providence, RI, 1996.

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Regarding the collections of analytic sets and Jordan measurable sets, a simple cardinality argument shows there exist Jordan measurable sets that are not analytic. Any subset of the Cantor middle thirds set is Jordan measurable (indeed, any such subset has Jordan measure zero), and thus there are $2^c$ many Jordan measurable sets. This result was essentially proved by Philip E. B. Jourdain around 1903. However, there are only $c$ many analytic sets, a fact that follows easily from each of several characterizations of analytic sets.

喻 良 has mentioned a paper by Mauldin in which results stronger in certain ways are obtained. Slightly more information about this paper and a related paper by Mauldin can be found by searching for the phrase "In [28] Mauldin proves" in my 30 April 2000 sci.math essay on $F_{\sigma}$ Lebesgue measure zero sets (this may also be of interest).

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  • $\begingroup$ The nontrivial part is whether there is an analytic set which is not in $\sigma(J)$. $\endgroup$ – 喻 良 Nov 24 '18 at 8:45

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