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I'm reviewing measure theory. In several texts, e.g. A Course in Real Analysis by McDonald and Weiss, complete measure spaces are measure spaces for which all subsets of sets of measure zero also have measure zero. Incomplete measure spaces may have non-measureable subsets of zero measure sets.

What I don't understand is why we can't just define that subset as non existent. First off, I'm failing to think of an example of an incomplete measure. We already have a measure of a Sigma algebra so, intuitively, any subset of a zero measure set worth considering could be expressed as the complement of an infinite union. Furthermore, why is it subsets of zero measure sets we care about? Is there some reason we shouldn't consider that finite nonzero measure sets may have nonmeasurable subsets?

For instance, take the Sigma algebra of all nonsingular open subsets of the real line. Define their complement to be the interior of the usual real (closed) complement (now open). By Sigma additivity and a complement I could consider the shrinking interval (-1/n, 1/n) and see that the measure goes to zero but the limit of the set goes to a nonmeasurable singleton.

AdamO
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2 Answers2

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You are forgetting that $\sigma$-Algebras are only closed under countable unions.

A well-known example of a non-complete measure is the borel measure on $\mathbb{R}$. The corresponding $\sigma$-Algebra $\mathcal{B}$ $\subset \mathcal{P}(\mathbb{R})$ is the smallest $\sigma$-Algebra that includes all the intervals $(-\infty,a]$. That $\sigma$-Algebra contains all the open sets, all the closed sets, all the singleton sets, and much more. But why should it contain all the subsets of every set of measure zero? See here for a construction of a set that is lebesgue measurable, but not borel measurable.

But you're right of course that we can always easily complete a given measure - we can simply add all the subsets of sets of measure zero to our $\sigma$-Algebra, and define their measure to be zero. It's relatively straight-forward to show that if we add those sets, plus all other sets that are required to make the resulting collection of sets a $\sigma$-Algebra, we can indeed define a measure on the new $\sigma$-Algebra that extends the old one.

So if we have an arbitrary measure, we can always choose to look at it's completion instead, and in that sense we don't, indeed, need to bother with incomplete measures.

But if we do that, we're changing the $\sigma$-Algebra that we're working in, and possbily make it much more complex. So if we're relying on the particular construction of the $\sigma$-Algebra, then switching to it's closure might not be feasable.

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fgp
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  • That makes a lot of sense. When we have continuous mappings with measure zero discontinuity sets, we would like to know that there are good regularity properties as we invert that image. For instance inverting x^2 of 1 gives {-1, 1}. Hence I would like to know both - 1 and 1 have measure zero and are measurable. – AdamO May 08 '14 at 14:22
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A simple example of an incomplete measure is the following:

Let $\nu$, $\lambda$ each be the Lebesgue Measure on $\mathbb{R}$, and for a rectangle $I_1\times I_2 \subset \mathbb{R}\times\mathbb{R}$, define $\mu\, : \, \mathcal{P}(\mathbb{R}\times\mathbb{R}) \to \mathbb{R}_{>0}$ by $\mu(I_1\times I_2) = \nu(A)\lambda(B)$

Now, consider the rectangle $Q = [0,1] \times [0,1]$, and the subset

$\Omega = \{(x,y) \in Q \, | \, x=0,\, 0\leq y\leq 1 \} \subset Q$.

Then surely, $\mu(\Omega) = 0$. Now, there exists a nonmeasurable subset $A\subset [0,1]$ (for instance, the Vitali set). Let $A_1 = \{0\} \times A \subset \Omega \subset Q$. Then (denoting by $\mu^*$ the completion of $\mu$), we have that $\mu^*(A_1) = 0$, but $A_1$ is not $\mu$-measurable.

While any subset of a zero measure set can be expressed as the complement of an infinite union, this doesn't help to classify, complete, or otherwise resolve the issue we have above with the set $A_1$, because the infinite union need not be countable (take, for instance, the Cantor Set). Measure is countably additive, and the expression of a subset as a complement of an infinite union does not guarantee that the set has measure $0$, so I'm not sure what this gives you, in general.

One of the things that we must resolve in measure theory is eliminating the possibility of nonmeasurable subsets of sets of measure zero, which we accomplish by completing the measure. We can't simply ignore the existence of these sets by saying they don't exist (surely I have given an example above). So we have to address these in a way that would remain logically consistent with the monotonicity of measure, and other properties of measure we have. Namely, we would want the subset of any null set to be measurable (and null), which is another way of saying the measure should be complete.

We don't necessarily need to be overly concerned about nonmeasurable subsets of finite measure sets, because even when we complete the measure, we may still have nonmeasurable sets. Lebesgue measure on $\mathbb{R}$ is complete, but the Vitali set $A$ is nonmeasurable, and can be scaled down or up to be a subset of a set of arbitrary measure, so we can't guarantee this.

What we can do, in general, is pass to the completion of our $\sigma$-Algebra (guaranteeing all subsets of null sets are measurable), extending the measure to be complete, in which case we need not concern ourselves too much with incomplete measures, but as the previous answer noted, this may not be feasible, or may make the $\sigma$-Algebra far more complicated than the original $\sigma$-Algebra you were working with.

Nicholas Stull
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