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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](http://en.wikipedia.org/wiki/Complete_measure).2012-07-03

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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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    Dear Nate! Let me ask you question on your example: I am working on measure theory from Bogachev's book "measure theory". And he gave ecxactly the same example. I am not able to understand why this measure is not complete. I was trying in this way: let $E$ be subset of $X=[0,1]$ with $E\neq \varnothing$ and $E\neq X$. Let $\mathcal{A}_{\mu}$ be the set of $\mu$-measurable sets (definition from Bogachev's). I am trying to show that $E\notin \mathcal{A}_{\mu}$.2018-11-14
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    Suppose $E\in \mathcal{A}_{\mu}$ then $\forall \varepsilon>0$ $\exists A_{ \varepsilon}\in \mathcal{A}$ with $\mu^*(A\triangle A_{\varepsilon})<\varepsilon.$ Unfortunately I am not able to get contradiction. Would be very grateful for help!2018-11-14
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    @RFZ: I think you have got some definitions mixed up. Following Bogachev, if you have a measure $\mu$ defined on a $\sigma$-algebra $\mathcal{A}$, you may form a new $\sigma$-algebra $\mathcal{A}_\mu$. We would say that $\mu$ is complete only if $\mathcal{A}_\mu = \mathcal{A}$. For this example, with $\mathcal{A} = \{X,\emptyset\}$ and $\mu = 0$, we in fact have $\mathcal{A}_\mu = 2^X$, i.e. every set is $\mu$-measurable. But not every set was in $\mathcal{A}$, so $\mu$ is not complete.2018-11-14
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    Thanks 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-book2018-11-14
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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.