# College Math Teaching

## January 25, 2011

### Some Sets Which Are Lebesgue Measurable

We’ve yet to show some Lebesgue measurable sets; we’ll do that now.
Note: by subaddivity of Lebesgue outer measure, we know that for all sets $T$, we have:
$m^*(T) \leq m^*(T \cap E) + m^*(T \cap C(E))$ for all measurable sets $E$; again here we use $C( )$ to denote the complement of a set.
So to show that a set is measurable, we need only show $m^*(T) \geq m^*(T \cap E) + m^*(T \cap C(E))$ for the set in question.

$\mathbb{R}$ and $\emptyset$ are measurable.

Proof: $m^*(T) = m^*(T \cap R) + m^*(T\cap \emptyset) =m^*(T)$

So now we know that the set of measurable sets isn’t empty. 🙂
Actually we can do better than that.

Proposition If a set has Lebesgue outer measure zero, then it is measurable.
Proof: Let $E$ have measure zero; that is, $m^*(E) = 0$. Let $T$ be arbitrary:
$m^*(T \cap E) + m^*(T \cap C(E))$. But $T \cap E \subset E$ and therefore $m^*(E) \geq m^*(T \cap E)$ which implies that $m^*(T \cap E) = 0$ But $T \cap E \subset T$ which implies that $m^*(T) \geq m^*(T\cap E) = m^*(T \cap E) + m^*(T \cap C(E))$ which is what we had to show.

Now have from the definition: $m^*({x}) = 0$ (one point sets have measure zero). Thus by countable additivity of measurable sets, any countable set (e. g., the rationals) has measure zero.

What about sets (other than $\mathbb{R}$) that have positive measure?
Here is how we are going to proceed: we will show that sets of the form $(a, \infty)$ are measurable. Then that will mean that sets like $[a,\infty)$ are measurable which will then imply that open intervals $(a,b)$ are measurable. But then, because of the fact that the measurable sets form a sigma algebra (closed with respect to countable unions, intersections, complements), we will get, free of charge, all of the topologically open sets to be measurable (remember that the reals are a second countable topological space), all closed sets, all countable intersections of open sets (called G-delta sets) and all countable unions of closed sets (the F-sigma sets). Note: the smallest sigma-algebra that contains the open intervals is called the Borel Sets.

It is true that not all measurable sets are Borel sets, but that is a topic for another day.

Showing the intervals $(a, \infty)$ are measurable.
Let $T$ be arbitrary and let $T_{1} = T \cap (a, \infty)$ and let $T_{2} = T \cap (-\infty, a]$. With no loss of generality we can assume that $m^*(T)$ is finite, otherwise there is nothing to show.
So, given any $\epsilon > 0$ we can find a countable collection of intervals $I_{i}$ that cover $T$ such that $m^*(T) + \epsilon \geq \sum_{i=1}^{\infty} l(I_{i}) =\sum_{i=1}^{\infty} m^*(I_{i})$ (note that $l(I_{i})$ denotes the length of the interval, which is its outer measure).
Let $T_{1} = T \cap (a,\infty)$ and $T_{2} = T \cap (-\infty, a]$, $I_{1,i} = I_{i} \cap (a,\infty)$ and $I_{2,i} = I_{i} \cap (-\infty, a]$
Now $m^*(T_1) \leq \sum_{i=1}^{\infty} l(I_{1,i} = \sum_{i=1}^{\infty} m^*(I_{1,i})$ because $T_{1} \subset \cup_{i=1}^{\infty} I_{1,i}$ and
$m^*(T_2) \leq \sum_{i=1}^{\infty} l(I_{2,i}) = \sum_{i=1}^{\infty} m^*(I_{2,i})$ because $T_{2} \subset \cup_{i=1}^{\infty} I_{2,i}$

So $m^*(T_{1}) + m^*(T_{2}) \leq \sum_{i=1}^{\infty} (m^*(I_{1,i}) + m^*(I_{2,i})) = \sum_{i=1}^{\infty} (m^*(I_{i})) \leq m^*(T) + \epsilon$
Since this is true for all $\epsilon$ it follows that $m^*(T_{1}) + m^*(T_{2}) \leq m*^(T)$ which is what we had to show.

So now that we have a feel for what sorts of sets are measurable (at least the Borel sets, and a bit more than that), we are ready to get back to Lebesgue integration. We’ve defined the Lebesgue integral for bounded functions over a closed interval; we can now move to unbounded functions and to some promised convergence theorems.

Note from my past
I’d imagine that most of us have written moronic things on examinations from time to time. I still remember writing the following on an analysis exam: “$E$ has measure zero and is therefore countable”….to which my professor replied: “Nonsense…whatever happened to the Cantor set?”.

I’ll have to deal with the Cantor set in it’s own post; well show that we can construct a Cantor set with zero measure and one of any given finite measure. This isn’t just a “cool thing”; it is also an essential part of some interesting counterexamples.