Return to Question

The Solovay model shows that ZF (plus an inaccessible) is consistent with every Lesbesgue measurable set being measurable. How far can we go in that direction? We can always have non-measurable sets by making the $\sigma$-algebra small, so here are two ways of making them big (either seems interesting to me):

1. For every set $X$ there exists a non-trivial measure on the entire power set of $X$. (Here non-trivial means that points have measure zero.)

2. For every set $X$, the completion of every probability measure on a set $X$ contains the entire power set of $X$.

Are there counterexamples to either of these already in ZF? If not, what is their consistency strength? Are either compatible with other axioms set theorists find attractive, such as dependent choice or the axiom of determinacy?

Edit James Hanson points out I need to exclude countable sets.

The Solovay model shows that ZF (plus an inaccessible) is consistent with every subset of $\mathbb{R}$ being measurable. How far can we go in that direction? We can always have non-measurable sets by making the $\sigma$-algebra small, so here are two ways of making them big (either seems interesting to me):

1. For every set $X$ there exists a non-trivial measure on the entire power set of $X$. (Here non-trivial means that points have measure zero.)

2. For every set $X$, the completion of every probability measure on a set $X$ contains the entire power set of $X$.

Are there counterexamples to either of these already in ZF? If not, what is their consistency strength? Are either compatible with other axioms set theorists find attractive, such as dependent choice or the axiom of determinacy?

Edit James Hanson points out I need to exclude countable sets.

The Solovay model shows that ZF is consistent with every Lesbesgue measurable set being measurable. How far can we go in that direction? We can always have non-measurable sets by making the $\sigma$-algebra small, so here are two ways of making them big (either seems interesting to me):

1. For every set $X$ there exists a non-trivial measure on the entire power set of $X$. (Here non-trivial means that points have measure zero.)

2. For every set $X$, the completion of every probability measure on a set $X$ contains the entire power set of $X$.

Are there counterexamples to either of these already in ZF? If not, what is their consistency strength? Are either compatible with other axioms set theorists find attractive, such as dependent choice or the axiom of determinacy?

Edit James Hanson points out I need to exclude countable sets.

The Solovay model shows that ZF (plus an inaccessible) is consistent with every Lesbesgue measurable set being measurable. How far can we go in that direction? We can always have non-measurable sets by making the $\sigma$-algebra small, so here are two ways of making them big (either seems interesting to me):

1. For every set $X$ there exists a non-trivial measure on the entire power set of $X$. (Here non-trivial means that points have measure zero.)

2. For every set $X$, the completion of every probability measure on a set $X$ contains the entire power set of $X$.

Are there counterexamples to either of these already in ZF? If not, what is their consistency strength? Are either compatible with other axioms set theorists find attractive, such as dependent choice or the axiom of determinacy?

Edit James Hanson points out I need to exclude countable sets.

The Solovay model shows that ZF is consistent with every Lesbesgue measurable set being measurable. How far can we go in that direction? We can always have non-measurable sets by making the $\sigma$-algebra small, so here are two ways of making them big (either seems interesting to me):

1. For every set $X$ there exists a non-trivial measure on the entire power set of $X$. (Here non-trivial means that points have measure zero.)

2. For every set $X$, the completion of every probability measure on a set $X$ contains the entire power set of $X$.

Are there counterexamples to either of these already in ZF? If not, what is their consistency strength? Are either compatible with other axioms set theorists find attractive, such as dependent choice or the axiom of determinacy?

The Solovay model shows that ZF is consistent with every Lesbesgue measurable set being measurable. How far can we go in that direction? We can always have non-measurable sets by making the $\sigma$-algebra small, so here are two ways of making them big (either seems interesting to me):

1. For every set $X$ there exists a non-trivial measure on the entire power set of $X$. (Here non-trivial means that points have measure zero.)

2. For every set $X$, the completion of every probability measure on a set $X$ contains the entire power set of $X$.

Are there counterexamples to either of these already in ZF? If not, what is their consistency strength? Are either compatible with other axioms set theorists find attractive, such as dependent choice or the axiom of determinacy?

Edit James Hanson points out I need to exclude countable sets.