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Give an example of a measure which is not complete ? A measure is complete if its domain contains the null sets.

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See the "motivation" and "examples" sections here. –  David Mitra Jul 3 '12 at 16:41
@ArturoMagidin I wonder if this is the same user. –  Rudy the Reindeer Jul 3 '12 at 17:12
@MattN. No way for me to know; you could flag and ask a moderator, though. –  Arturo Magidin Jul 3 '12 at 17:13
@ArturoMagidin I'm not sure whether that's the right thing to do. So I won't. –  Rudy the Reindeer Jul 3 '12 at 17:41

3 Answers 3

up vote 8 down vote accepted

The canonical example is the Borel measure on the Borel $\sigma$ algebra (the $\sigma$ algebra generated by the open intervals) on $\mathbb{R}.$ This example is often used as a motivation for the construction of the Lebesgue measure.

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For a really trivial example: let $X$ be any set with at least two points, take the trivial $\sigma$-algebra $\mathcal{F} = \{X, \emptyset\}$, and define $\mu$ on $\mathcal{F}$ by $\mu(X) = \mu(\emptyset) = 0$.

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Here's one:

Define $\mu (A) = 0$ (the zero measure) for all $A$ in the sigma algebra. Then pick any set $B$ that contains a set that is not in the sigma algebra.

And here's another one, taken from "A Course in Real Analysis" by McDonald/Weiss, page 250:

Let $(\mathbb R, \Sigma, \lambda)$ denote $\mathbb R$ with the Lebesgue measure. Then this space is complete. But the product space $(\mathbb R^2, \Sigma^2, \lambda \times \lambda)$ isn't. To see this, pick any non-Lebesgue measurable set $N$ and let $A := \{0\} \times N $ and $B:=\{0\} \times \mathbb R$. Then $B$ has measure zero and $A \subset B$. But $A$ is not measurable in the product measure.

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