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I'm playing around with a collection of subsets of $\mathbb{R}^n$ let's call them $X_i$. What I want to know is, is the following condition sufficient for some $X_i$ being measurable?

Almost all elements of $X_i$ are surrounded by a neighbourhood contained in $X_i$

This is motivated by the idea that non measurable sets are "fractally wiggly"

Assuming standard topology and measure on $\mathbb{R}^n$ but I'm also interested in minimal assumptions regarding topology and measure that keep this a sufficient condition for being measurable.

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up vote 4 down vote accepted

I believe this is sufficient, though not necessary. The set of $x\in X_i$ that are contained in a neighborhood in $X_i$ is the interior of $X_i$, which is open. If $X_i$ satisfies this condition, then it's an open set union a set of measure zero, both of which are measurable. This will still be true for any measure defined at least on the Borel $\sigma$-algebra for your space.

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Right, and to see that it's not necessary one could take for example any set of positive measure and remove a countable dense subset, or less trivial examples like fat Cantor sets. – Jonas Meyer Oct 13 '10 at 22:01
Good point. The irrationals are a good example. – Paul VanKoughnett Oct 14 '10 at 3:34

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