A very quick way to prove a set is measurable.

Asked Dec 12 '14 at 00:03

Active Dec 12 '14 at 00:11

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All examples of non-measurable subset of $\mathbb{R}$ (in the Lebesgue sense) seem to need the axiom of choice in some way or the other. Hence, can we say:

The set $A\subseteq \mathbb{R}$ is measurable because the axiom of choice was not used to define it.

Is that a valid argument? Or does there exist non-measurable sets that do not require the axiom of choice?

asked Dec 12 '14 at 00:03

Sylvain

1

I suspect it might be hard to prove that you never used choice in your definition. Also, you can use choice to define measurable sets. – Jonny Dec 12 '14 at 00:07
You don't need the full strength of the axiom of choice. You could limit the choice axiom to collections that have at most $c$ sets. – Matt Samuel Dec 12 '14 at 00:09
@yyyyyyy I disagree that this is a duplicate. I read the nice answer by AsafKaragila but I am still puzzled about my question of whether this can be used as a proof that a set is measurable. – Sylvain Dec 12 '14 at 00:21
See also Axiom of choice and non measurable set – augurar Dec 12 '14 at 00:27

A very quick way to prove a set is measurable.

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