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If $(X,\mathcal {B},\mu)$ and $(X,\mathcal{B}'\mu')$ are measure spaces on the same set $X$, we say that the latter is a refinement of the former if $\mathcal{B}\subset \mathcal{B}'$ and $\mu'\downharpoonright_\mathcal{B}=\mu$.

Any measure space admits a refinement to a complete measure space via completion, and the resulting measure space is the coarsest refinement of the original measure space to a complete measure space. This implies that it is possible for a complete measure space to admit a further refinement, and such an example can be found in here.

Now the Lebesgue measure space $(\mathbb{R}^d,\mathcal{L}(\mathbb{R}^d),m_d)$ is complete; but unlike the above example above, many subsets of $\mathbb{R}^d$ are Lebesgue measurable and I was wondering if it admits a further refinement. Does it?

Ken
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  • This is an interesting question and I hope you get a good answer. Nonmeasurable sets generally require the axiom of choice and are not constructible from unions and intersections of intervals. I find it unlikely that an interesting refinement exists but I don't know. – jdods Jul 16 '19 at 02:20
  • Instead of "refinement", the more usual word for this would be an extension of Lebesgue measure. I think there have been some relevant previous questions either here or on MathOverflow. – Nate Eldredge Jul 16 '19 at 04:46
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    @NateEldredge I did some research by with the word "extension" and found out that my question was a duplicate. Thanks! – Ken Jul 16 '19 at 05:23
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    Duplicate of Extension of the Lebesgue measurable sets – Ken Jul 16 '19 at 05:24

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