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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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Welcome to math.SE. Since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, statements phrased in the imperative ("Give", "Prove") when asking for help are considered rude by many here (even with a question mark added) – Arturo Magidin Jul 3 '12 at 16:41
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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