Show whether E is a Measurable Set

Let $(E_k)_{k=1}^{\infty}$ be a sequence of measurable sets in $\mathbb{R}^n$ . Let $E$ be a subset of $\mathbb{R}^n$ so that $x \in E$ if and only if $x$ belongs to exactly 2 of the sets $E_k$ . Determine whether E is measurable. If it is, show it. If not, give a counterexample.

I suppose it is measurable? No? Can't think of any counterexample.

Thanks in advance.

Re: Show whether E is a Measurable Set

for any two positive integers $1\leq i <j$ consider the set $A_{i,j}=E_i\bigcap E_j\bigcap (\bigcap_{k\ne i,j}\mathbb{R}^n\setminus E_k)$ . Such sets are measurable and $E$ is a union of such sets

Re: Show whether E is a Measurable Set

Quote:

Originally Posted by Markeur

Let $(E_k)_{k=1}^{\infty}$ be a sequence of measurable sets in $\mathbb{R}^n$ . Let $E$ be a subset of $\mathbb{R}^n$ so that $x \in E$ if and only if $x$ belongs to exactly 2 of the sets $E_k$ . Determine whether E is measurable. If it is, show it. If not, give a counterexample.

I suppose it is measurable? No? Can't think of any counterexample.

Thanks in advance.

In my opinion any question about "something" should start off with a definition of "something" and a demonstration that the definition is satis fied. So here is my little contribution as a matter of principle from http://en.wikipedia.org/wiki/Measure_(mathematics)

σ-algebra, meaning that unions, intersections and complements of sequences of measurable subsets are measurable.

Let Σ be a σ-algebra over a set X. A function μ from Σ to the extended real number line is called a measure if it satisfies the following properties:

Non-negativity:
http://upload.wikimedia.org/wikipedi...359301d7ec.png for all http://upload.wikimedia.org/wikipedi...0fce1e8199.png

Countable additivity (or σ-additivity): For all countable collections http://upload.wikimedia.org/wikipedi...42c71bc19f.png of pairwise disjoint sets in Σ:
http://upload.wikimedia.org/wikipedi...46cdd70cb1.png

Null empty set:
http://upload.wikimedia.org/wikipedi...9076c0df86.png
I would find it interesting to see how this would be done in the case of the previous post.