A metric space X is called complete if every Cauchy sequence of points in X has a limit that is also in X. It's perfectly clear to me.

A measure space $(X, \chi, \mu)$ is complete if the $\sigma$-algebra contains all subsets of sets of measure zero. That is, $(X, \chi, \mu)$ is complete if $N \in \chi$, $\mu (N) = 0$ and $A \subseteq N$ imply $A \in \chi$. Technically, I could understand the definition, but can't get the logic behind it.


1) Why do we care only about subsets of sets of measure zero to determine completeness?

2) How does the completeness of measure spaces relate to a completeness of metric spaces?

3) Could you suggest a concrete elementary example of a measure space (preferably, with simple sets) that isn't initially complete and then is completed?

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    $\begingroup$ 2) It's unrelated. 3)$\mathbb{R}$ with the Borel sigma-algebra and the Lebesgue measure on it. $\endgroup$ – Jon-S Oct 29 '16 at 20:00
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    $\begingroup$ There is a somewhat tenuous link between this notion and completeness for metric spaces, but it's probably not the best way to think about completeness. See my response in this thread: math.stackexchange.com/questions/926435/complete-measure-space/… $\endgroup$ – Josh Keneda Oct 30 '16 at 0:00

In some proofs in measure theory, one often encounters the following: we have a set $E$ of measure $0$ and a subset $A\subseteq E$, and we want to show that $A$ has measure $0$. However, if the measure space is not complete, this will not always be the case, since $A$ might not be measurable. This is fixed by considering the completion of the measure space.

Moreover, if a measure space is complete, then the following property holds: if $A,B$ are measurable with $A\subseteq B$, $\mu(A)=\mu(B)$ and $A\subseteq C\subseteq B$, then $C$ is measurable, with $\mu(C)=\mu(A)=\mu(B)$. That's why we only care for sets of measure zero, since it implies that "completeness" extends to sets of measure different than zero.

Edit: Suppose that $A\subseteq B$, $\mu(A)=\mu(B)<\infty$ and the space is complete. Suppose now that $A\subseteq C\subseteq B$. Then $C\setminus A\subseteq B\setminus A$, and the last set has measure zero, so by completeness $C\setminus A$ is measurable, with $\mu(C\setminus A)=0$. From this, we obtain that $C$ is measurable, with measure $$\mu(C)\leq\mu(C\setminus A)+\mu(A)=\mu(A).$$ But, $A\subseteq C$, so $\mu(A)\leq \mu(C)$, therefore $\mu(C)=\mu(A)$.

This fact explains my last comment before the edit, and could be a definition for "completeness for sets of positive measure". Therefore, this shows that if we complete the measure to include all sets that should have measure zero, then we include all sets that should have any given positive measure.

  • $\begingroup$ detnvvp, thank you for the reply! Could you please explain how exactly completeness of sets of measure zero implies "completeness" of sets of measure different than zero. $\endgroup$ – Konstantin Oct 29 '16 at 21:19
  • $\begingroup$ Please check the edit. $\endgroup$ – detnvvp Oct 29 '16 at 23:42

3) Take the sample space to be $\Omega=\{1,2,3\}$, the $\sigma$-algebra to be $\mathcal F=\{\emptyset,\Omega,\{1,2\},\{3\}\}$, and let $P$ be the probability measure on $(\Omega,\mathcal F)$ such that $P(\{3\})=1$. Then $P(\{1,2\})=0$ and $\{1\}$ is a non-$\mathcal F$-measurable subset of $\{1,2\}$. The probability space $(\Omega,\mathcal F,P)$ is not complete.

  • $\begingroup$ This helps me understand the concept of completion, but do you have another example that does not involve a probability measure ? The Borel $\sigma$-algebra on $\mathbb{R}^n$ is not complete, and its completion with respect to the Lebesgue measure is the $\sigma$-algebra of Lebesgue measurable sets. Could you give a concrete example (for example on the real line $\mathbb{R}$) of the sets/elements that were added to make the completion ? $\endgroup$ – matthieu Feb 28 '18 at 10:00

1) There's only one possible way to assign subsets of a set of measure zero a measure: Its measure is zero. This is not possible for other sets.

2) Unrelated (as far as I know at least)

3) Either the Borel sigma-algebra with Lebesgue measure or - if you want a very trivial example - the trivial measure on a sigma-algebra which is not the power set.


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