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We know that the consistency of ZFC+"Exists an inaccessible cardinal" implies the consistency of ZF+DC+"All sets are Lebesgue measurable"; and DC proves the existence of non-Borel sets.

J. Truss proved that repeating Solovay's construction by collapsing any limit cardinal to be $\aleph_1$ we obtain a model of ZF+"All sets are Lebesgue measurable", and in that model DC holds if and only if we collapsed an inaccessible. If we collapsed a singular cardinal then the resulting model has the property that all sets are Borel.

If we assume ZF+"All sets are Lebesgue measurable"+"There exists a non-Borel set", can we conclude that there is an inner model with an inaccessible cardinal?

Some clarifications:

  1. When I say Borel sets, I mean elements of the $\sigma$-algebra generated by the open sets.
  2. When I say Lebesgue sets, I mean elements of the $\sigma$-algebra generated by completing the Borel $\sigma$-algebra with respect to the null ideal.

  3. As the Borel measure may fail to be $\sigma$-additive, we can as the following to complement the above question:

    Assume that ZF+"All sets are Lebesgue measurable"+"The Borel measure is $\sigma$-additive", can we conclude that there is an inner model with an inaccessible cardinal?

    Now, taking the Borel and Lebesgue sets as defined above makes more sense.

  4. The above leads to the next question:

    If the Borel measure is not $\sigma$-additive, can we represent $\mathbb R$ as a countable union of null sets?

    (It is tempting to say immediately yes, but remember that countable unions of countable sets need not be countable anymore.)

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    $\begingroup$ Could you say what you mean exactly by "all sets are Lebesgue measurable" in a context without DC? After all, the theory of Lebesgue measure in ZF can seem to break down (imagine that the reals are a countable union of countable sets). $\endgroup$
    – Joel David Hamkins
    Commented Nov 1, 2012 at 11:19
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    $\begingroup$ While you're at it, could you say what you mean exactly by "Borel set"? I assume that you mean a set in the smallest $\sigma$-algebra containing the open sets, but, if you talk with enough set theorists, you might mean a set with a Borel code (i.e., a real that encodes how to build the set from open sets by iterated complementation and countable union). The two meanings can differ in the absence of choice. $\endgroup$
    – Andreas Blass
    Commented Nov 1, 2012 at 12:00
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    $\begingroup$ There is always a set without a Borel code: the set of all Borel codes $b$ which are not elements of the set coded by $b$. $\endgroup$
    – Joel David Hamkins
    Commented Nov 1, 2012 at 12:48
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    $\begingroup$ Incidentally, Asaf, I think you mean to raise the question, rather than beg it. begthequestion.info $\endgroup$
    – Joel David Hamkins
    Commented Nov 1, 2012 at 13:30
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    $\begingroup$ In view of Joel’s example, the definition is still unclear to me. The set $I$ of all sets that can be covered for every $\epsilon>0$ by a countable sequence of open intervals whose lengths sum to $\epsilon$ is an ideal, but not ZF-provably a $\sigma$-ideal. Is the “null ideal” $I$, the $\sigma$-ideal generated by $I$, or something else? $\endgroup$
    – Emil Jeřábek
    Commented Nov 1, 2012 at 14:01

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