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Asaf Karagila
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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.

It raises the question, ifIf 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.

  1. 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.)

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.

It raises the question, 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.

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.

  1. 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.)

All this begging and whatnot...; deleted 4 characters in body
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Asaf Karagila
  • 39.8k
  • 8
  • 135
  • 283

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.

It begsraises the question, 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 an elementelements of the $\sigma$-algebra generated by the open sets.
  2. When I say Lebesgue sets, I mean an elementelements of the $\sigma$-algebra generated by completing the Borel $\sigma$-algebra with respect to the null ideal.

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.

It begs the question, 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 an element of the $\sigma$-algebra generated by the open sets.
  2. When I say Lebesgue sets, I mean an element of the $\sigma$-algebra generated by completing the Borel $\sigma$-algebra with respect to the null ideal.

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.

It raises the question, 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.
added 287 characters in body
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Asaf Karagila
  • 39.8k
  • 8
  • 135
  • 283

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.

It begs the question, 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 an element of the $\sigma$-algebra generated by the open sets.
  2. When I say Lebesgue sets, I mean an element of the $\sigma$-algebra generated by completing the Borel $\sigma$-algebra with respect to the null ideal.

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.

It begs the question, 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?

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.

It begs the question, 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 an element of the $\sigma$-algebra generated by the open sets.
  2. When I say Lebesgue sets, I mean an element of the $\sigma$-algebra generated by completing the Borel $\sigma$-algebra with respect to the null ideal.
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Asaf Karagila
  • 39.8k
  • 8
  • 135
  • 283
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