# College Math Teaching

## January 24, 2011

### Lebesgue Measure: Outer Measure, Measure, Why the Caratheodory Criterion works

Filed under: advanced mathematics, analysis, Lebesgue Integration, Measure Theory — collegemathteaching @ 9:10 pm

In this post, I will distinguish between Lebesgue outer measure (denoted by $m^{*}(E)$) and Lebesgue measure (denoted by $m(E)$ ). Recall that Lebesgue outer measure of a set $E$ is $inf \sum_{i=1}^l(I_{i})$ where $\cup_{i=1} I_{i}$ is a covering of $E$ by open intervals $I_{i} = (a_{i},b_{i})$ and $l(I_{i}) = b_{i} - a_{i}$. Lebesgue outer measure is defined for all subsets of the real line $E$ but in a previous post we showed an example of a subset for which the combination of translation invariance and countable additivity do NOT hold for outer measure.

We need to restrict to subsets of the real line for which translation invariance and countable additivity holds; when we use Lebesgue outer measure on these sets we call it Lebesgue measure.

So, what we are going to show is this: if we restrict our sets to those sets $E$ for which:
$m^{*}(T) = m^{*}(T \cap E) + m^*(T\cap\tilde{E})$ (where $\tilde{E}$ is the set complement of $E$) for ALL subsets $T$, then Lebesgue outer measure IS countably additive with respect to those subsets. Lebesgue outer measure applied to these sets is called Lebesgue measure.

Note: If $m^{*}(T) = m^{*}(T \cap E) + m^*(T\cap\tilde{E})$ is true we say that “$E$ splits $T$ and sometimes refer to $T$ as a “test set”.

So, why is this condition the one that we want? Well, well prove the following results assuming that the given sets $E_{i}$ splits all test sets $T$.

1. If $E_{1}$ and $E_{2}$ are measurable sets (e. g., splits all sets $T$) then $E_{1} \cap E_{2}$ is also measurable (splits all sets) and $E_{1} \cup E_{2}$ is also measurable.
Proof: first recall that $m^{*}(T) \leq m^{*}(T\cap E) + m^{*}(T\cap\tilde{E})$ by countable subadditivity of Lebesgue outer measure. We need to show the other inequality.

First recall that $\tilde{E_{1}} \cap\tilde{E_{2}} = \tilde{(E_{1}\cup E_{2})}$ by DeMorgan’s laws. (the latter is supposed to be the compliment of the union of the set $E_{1}\cup E_{2}$.

Now because $E_{2}$ is measurable, (1) $m^*(T\cap\tilde{E_{1}})=m^*(T\cap\tilde{E_{1}}\cap E_{2})+m^*(T\cap\tilde{E_{1}}\cap\tilde{E_{2}})$

Now recall that, from basic set theory, $E_{1}\cup E_{2} = E_{1}\cup(E_{2}\cap\tilde{E_{1}})$
So from countable subadditivity of outer measure:
(2) $m^*(T\cap E_{1}\cup E_{2}) \leq m^*(T\cap E_{1})) + m^*(T\cap (E_{2} \cap\tilde{E_{1}}))$

So now attempt to compute: (3)
$m^*(T\cap (E_{1} \cup E_{2})) + m^*(T \cap\tilde{E_{1} \cup E_{2}}) \leq m^*(T\cap E_{1}) + m^*(T\cap (E_{2} \cap\tilde{E_{1}}) +m^*(T \cap\tilde{E_{1}} \cap\tilde{E_{2}} )$

But $E_{2}$ is measurable and therefore splits $T \cap\tilde{E_{1}}$ and so the last two terms can be combined to $m^*(T \cap\tilde{E_{1}})$ so (3) becomes $m^*(T \cap E_{1}) + m^*(T \cap\tilde{E_{1}}) = m^*(T)$ which is the right hand of inequality (2).

So, it follows by a routine induction argument than a finite union of measurable sets is measurable.
Now, what about the intersection? If $E_{1}$ and $E_{2}$ are measurable, so are their complements (and vica-versa; the definition is symmetric). Now recall that $E_{1} \cap E_{2}$ = $C({(\tilde{E_{1}} \cap\tilde{E_{2}})})$ (note: the outer “$C()$” denotes set complement as I couldn’t get the LaTex command for the outer “tilde” to work) and the result follows.

2. Now we show finite additivity of disjoint measurable sets $E_{i}$:
We need to show that $m^*(T \cap \cup_{i=1}^{n} E_{i}) \geq \sum_{i=1}^{n} m^*(T \cap E_{i})$
Clearly, the statement is true for $n = 1$. Assume that the statement is true for all integers up to $n - 1$.
Now by disjointness, $T \cap (\cup_{i=1}^{n} E_{i}) \cup E_{n} = T\cup E_{n}$ and $T \cap \cup_{i=1}^{n} \cap\tilde{E_{n}} = T\cap(\cup_{i=1}^{n-1} E_{i}$.
Now $E_{n}$ splits $T \cap (\cup_{i=1}^{n}) E_{i}$ therefore
$m^*( T \cap (\cup_{i=1}^{n}) E_{i} )= m^*(T \cap E_{n}) +m^*(T \cap (\cup_{i=1}^{n-1} E_{i}) =$
$m^*(T \cap E_{n})+ \sum_{i=1}^{n-1} m^*(T \cap E_{i})$

3. We now need to show that the countable union of measurable sets is measurable.
First note that if $E = \cap_{i=1}^{\infty}E_{i}$ we can assume that the $E_{i}$ are disjoint. Here is why: Let $F_{1} = E_{1}$, $F_{2} = E_{2} \cap\tilde{E_{1}}$, $F_{3} = E_{3} \cap\tilde{E_{1}\cup E_{2}}$ and so on. Then $cup_{i=1}^{\infty} E_{i} = cup_{i=1}^{\infty} F_{i}$ and the $F_{i}$ are mutually disjoint. So we can assume with no loss of generality that the $E_{i}$ have this property.

Note: I am getting tired of the “tilde” notation and so will be using the $C()$ notation to denote the set complement.

Now let $G_{n} = \cup_{i=1}^{n}E_{i}$. Then $G_{n}$ is measurable and $C(G_{n}) \supset C(E)$. Then:
$m^*(T) = m^*(T \cap G_{n})+m^*(T\cap C(G_{n}) \geq m^*(T \cap G_{n})+ m^*(T\cap C(E))$
By finite additivity $m^*(T \cap G_{n}) = \sum_{i=1}^{n} m^*(T \cap E_{i})$ Hence we can substitute into the right hand side of the inequality to obtain:
$m^*(T) \geq \sum_{i=1}^{n} (m^*(T \cap E_{i})+m^*(T \cap C(E)))$
This is true for all values of $n$
This means $m^*(T) \geq \sum_{i=1}^{\infty} (m^*(T \cap E_{i})+m^*(T \cap C(E))) \geq m^*(T \cap E) + m^*(T\cap C(E))$ ; the latter inequality following from countable subaddivity.

4. Now we can show finite additivity for for disjoint measurable sets in Lebesgue measure (no longer outer measure) by replacing $T$ with $\mathbb{R}$.

5. We now consider a countably infinite collection of disjoint measurable sets. We must show that $m(\cup_{i=1}^{\infty} E_{i}) \geq \sum_{i=1}^{\infty} m(E_{i})$ This follows from the fact that $\cup_{i=1}^{\infty} E_{i} \supset \cup_{i=1}^{k} E_{i}$ for all $k$. This means that $m(\cup_{i=1}^{\infty} E_{i} ) \geq m(\cup_{i=1}^{k} E_{i} )$ for all $k$. Hence the infinite sum is either bounded by $m(\cup_{i=1}^{\infty} E_{i} )$ and therefore converges or is infinite as is $m(\cup_{i=1}^{\infty} E_{i} )$

All of this shows that restricting the sets we consider to those who obey the Caratheodory criterion with respect to Lebesgue outer measure gives us the properties that we want.

Of course, we’ve yet to classify WHICH sets are measurable or which sets form important subclasses of measurable sets.
Here is an outline of what we are going to do: we’ll expand on lemmas which state things like “doing operations X, Y and Z to measurable sets yields a measurable set”; we’ve done this with respect to finite unions, finite intersections, and now countable unions. We’ll then see that such sets form a type of “set algebra” and what is known as a “sigma-algebra”.
That is for another day (soon) š

Update Since we are close to closing the deal as far as the “set algebra” and “sigma-algebra” result, I’ll go ahead and finish in this post. First I’d like to recall something from basic set theory (I’ll use $C(A)$ to denote the complement of the set $A$ ):
$C(B) \cup C(A) = C(A \cap B)$. I’ll leave the formal proof of this to the reader, but the basic idea is that if something isn’t an element of both $A$ and $B$, then it must be in the complement of one or the other set. This leads to the following result for sets $E_{i}$:
$\cup_{i=1}^{\infty} C(E_{i}) = C(\cap_{i=1}^{\infty} E_{i})$ Again, I won’t prove this, but the idea is as follows: if something is an element of the left hand side, then it must be in the complement of at least one of the $E_{i}$ which means it can’t be in ALL of them and therefore can’t be in the intersection.

What does this have to do with our measurable sets? We’ve just seen that the countable union of measurable sets is measurable and by symmetry, the complement of a measurable set is measurable. Hence the countable intersection of measurable sets is measurable.

Set algebras An algebra of sets is a collection of sets which is closed with respect to finite unions and complements; hence it is immediate that the set of measurable sets forms a set algebra.

Sigma algebras A set algebra is called a sigma-algebra if it is closed with respect to countable unions as well (and by DeMorgan’s laws: closed with respect to countable intersections as well). So we’ve just shown that the collection of measurable sets forms a sigma-algebra.

What we haven’t shown is a single measurable set as yet! THAT is what we are going to do next.