show a set is lebesgue measurable

Question

For lebesgue measure, is it true that the union/intersection of measurable sets is also measurable (finite or infinite unions or intersections)? But it's not true for subsets? (i.e.,a subset of a measurable set is not necessarily measurable.) And also, if A is measurable, -A is also a measurable set? So in general, how do I prove if a set is measurable or not? What should I show? For example, if I know that A is measurable, how do I show that E={$\sqrt x$: x$\in$ A} is also measurable?

Thank you.

Answer 1

The set of measurable sets form a $\sigma - algebra$. So they are closed under countable unions and intersections.

Subset of a measurable set need not be measurable in general.See the http://en.wikipedia.org/wiki/Vitali_set. But if the measurable set has measure zero and if the space is complete then all its subsets are measurable.

If $E$ is a measurable set then $E+t$ is measurable as well as $cE$ is measurable. Here $E+t$ is translation and the other operation is dilation.

A set is measurable if you can find an open set $O$ such that $m^*(O-E)<\epsilon ,\forall \epsilon>0$. Here in the particular example it is easy to find that open set with the desired property if given an open set for the original set.

Answer 2

A large collection of Lebesgue measurable sets are the Borel sets. These are countable intersections, unions, and complements of open sets.

Zero measure sets and their subsets are also useful as a test for measurability.

A powerful, but not always necessary, is the characterization: $E$ is measurable if and only it for every $\epsilon>0$ there is an open set $U\supset E$ such that $\mu^*(U\setminus E)<\epsilon$, where $\mu^*$ is the exterior measure.

It is not true in general that the inverse image of a measurable set, by a continuous function, is measurable. For example: Consider the function $g(x)=x+C(x)$, where $C(x)$ is the Devil's staircase. This function is strictly increasing and maps the Cantor set (a set of measure zero) to a set of positive measure. Let $f(x)=g^{-1}(x)$. Then we can pick a non-meaurable set $A$ in $\text{Im}(g)$. This will be sent by $f$ to a subset of the Cantor set. Since the Cantor set is of measure zero, and the Lebesgue measure is complete, then $B:=f(A)$ is also of measure zero and in particular measurable. Therefore $B$ is measurable, but $f^{-1}(B)=A$ is not.

We need the functions to be more continuous then, to preserve sets of measure zero, to send sets of small outer measure to sets of small outer measure. The right notion is absolutely continuous.

We can check that $\sqrt{x}$ is absolutely continuous. For this we might use that $$|\sqrt{y}-\sqrt{x}|=\frac{|y-x|}{|\sqrt{y}+\sqrt{x}|}.$$ This allows us to make sums $\sum |\sqrt{y_i}-\sqrt{x_i}|$ arbitrarily small when $\sum |y_i-x_i|$ is small.