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If $(X,\mathcal{A},\mu)$ is a measure space, $(X,\mathcal{B},\overline{\mu})$ is complete, $\mathcal{B}\supset\mathcal{A}$, and $\overline{\mu}(A)=\mu(A)$ for every $A\in\mathcal{A}$, is $(X,\mathcal{B},\overline{\mu})$ necessarily the completion of $(X,\mathcal{A},\mu)$?

My definition of completion is:

The completion of $\mathcal{A}$ is the smallest $\sigma$-algebra $\mathcal{B}$ containing $\mathcal{A}$ such that $(X,\mathcal{B},\mu)$ is complete.

It seems like the answer to my question is no, because it isn't clear that $\mathcal{B}$ is necessarily the smallest $\sigma$-algebra satisfying the required properties. But I have been looking at the Completion Theorem where they seem to assume that the answer to my question is yes. How can I see that the smallest $\sigma$-algebra satisfying the required properties is the only $\sigma$-algebra satisfying the required properties?

A. Goodier
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3 Answers3

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As stated the result is surely false. If $(X,\mathcal A, \mu)$ is already complete you are asserting that we cannot extend the measure to a larger sigma algebra. Take $\mathcal A$ to be trivial sigma algebra (consisting of just the empty set and the whole space) to get obvious counter-examples.

Kavi Rama Murthy
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The completion theorem is first an existence theorem. The proof gives an explicit construction of the completion.

If you look into the proof, then you see that the $\sigma$-algebra $\Sigma^*$ constructed there is the smallest possible. It contains the original $\mathcal A$ and all subsets of sets of zero measure.

Completion of measure space is usually defined without the requirement of smallest possible $\sigma$-algebra. The completion theorem gives you this smallest completion for free.

daw
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In general, the extension you are looking for is not unique. Let $X=[0,1]$, for example, and consider the trivial $\sigma$-algebra $\mathcal A=\{\emptyset, X\}$. Let $\mu(\emptyset)=0$ and $\mu(X)=1$.

This is complete, but there are a lot of different ways to extend it to a complete measure on a finer $\sigma$-algebra. For example, $\mathcal B$ can be all subsets of $X$, and the measure can be any Dirac measure.

Or the Lebesgue measure is also good.

A. Pongrácz
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