Give an example of a measure which is not complete ? A measure is complete if its domain contains the null sets.
Give an example of a measure which is not complete
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0See the "motivation" and "examples" sections [here](http://en.wikipedia.org/wiki/Complete_measure). – 2012-07-03
3 Answers
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.
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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0Thanks Nate! Also, could you help me this question from Bogachev's book? I guess that you are familiar with his book. https://math.stackexchange.com/questions/2998689/the-proof-of-caratheodorys-criterion-from-bogachevs-book – 2018-11-14
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.