Measure Theory/Countable Additivity

Properties of Length-measure

We can finally achieve what has been elusive for so long: a measure of sets of real numbers, which is countably additive.

In this lesson, we prove that the $\lambda$ which we just constructed is countably additive.

Theorem: Countable Additivity $\lambda$ is countably additive.

Pairwise Additive

As a warmup to countable additivity, let's prove the easier claim of pairwise additivity. Let $E,F\in {\mathcal {M}}$ be two measurable sets which are disjoint, $E\cap F=\emptyset$.

We would like to show that $\lambda (E\sqcup F)=\lambda (E)+\lambda (F)$.

Exercise 1. Apply Measurability Prove the desired result by applying the measurability of F to $E\sqcup F$. Don't forget to use the fact that E and F are disjoint.

Exercise 2. Generalize Notice that the proof did not actually require E to also be measurable. Therefore state a generalization of the above result.

Exercise 3. Finite Additivity Prove by induction that $\lambda$ is therefore finitely additive. As part of the exercise, state what "finitely additive" should mean.

Exercise 4. Countable Additivity Let $E_{1},E_{2},\dots \in {\mathcal {M}}$ be any countable collection of disjoint measurable sets. We would like to show $\lambda \left(\bigsqcup _{i=1}^{\infty }E_{i}\right)=\sum _{i=1}^{\infty }\lambda (E_{i})$ To do so, state the result which you just proved for finite additivity. Then apply monotonicity and then take the limit as $n\to \infty$. Finally, use the above to prove countable additivity.

Exercise 5. Use Countable Additivity Find $\lambda ((0,1)\sqcup (2,3))$ and then find $\lambda \left(\bigsqcup _{i=1}(i,i+1)\right)$. Also find $\lambda \left(\bigcup _{k=1}^{\infty }(k,k+1/2^{k})\right)$.

Excision

We have additivity results, and one would hope that we have something like results which resemble subtraction.

Definition: excision

Let $E,F\in {\mathcal {M}}$ be any two measurable sets and $E\subseteq F$. Then the property that $\lambda (F\smallsetminus E)=\lambda (F)-\lambda (E)$ is called excision.

Exercise 6. Prove Excision. Prove that $\lambda$ satisfies excision.

Continuity of Measure

Recall that, roughly stated, if a function f is continuous then $\lim _{x\to x_{0}}f(x)=f\left(\lim _{x\to x_{0}}x\right)=f(x)$. Effectively, continuity of f means that the limit passes into the function.

There is a similar property for length measure. If $E_{1},E_{2},\dots$ is an ascending sequence of measurable sets (i.e. $E_{i}\subseteq E_{i+1}$ for $1\leq i$) then

$\lim _{n\to \infty }\lambda \left(\bigcup _{i=1}^{n}E_{i}\right)=\lambda \left(\lim _{n\to \infty }\bigcup _{i=1}^{n}E_{i}\right)$

One small problem with the statement above is that the expression $\lim _{n\to \infty }\bigcup _{i=1}^{n}E_{n}$ is ... not even defined, actually.

But of course it makes good sense to identify this as $\bigcup _{i=1}^{\infty }E_{i}$.

(Also note that $\bigcup _{i=1}^{n}E_{i}$ is superfluous because the sequence is ascending. This is just the same thing as $E_{n}$.)

Definition: continuity of measure

Let $E_{1},E_{2},\dots \in {\mathcal {M}}$ be an ascending sequence of measurable sets. The property that

$\lim _{n\to \infty }\lambda (E_{n})=\lambda \left(\bigcup _{i=1}^{\infty }E_{i}\right)$

is called the upward continuity of measure.

A sequence of sets $E_{1},E_{2},\dots \subseteq {\mathbb {R}}$ is called "descending" if $E_{i}\supseteq E_{i+1}$ for $1\leq i$.

Definition: continuity of measure

Let $E_{1},E_{2},\dots \in {\mathcal {M}}$ be a descending sequence of measurable sets, and assume $E_{1}$ has finite measure. The property that

$\lim _{n\to \infty }\lambda (E_{n})=\lambda \left(\bigcap _{i=1}^{\infty }E_{i}\right)$

is called downward continuity of measure.

Proof

We now set for ourselves the proof of the theorem.

Theorem: Continuity of Measure $\lambda$ satisfies both upward and downward continuity of measure.

Exercise 7. Prove the Upward Continuity of Measure To prove the upward continuity of measure, define the sequence of disjoint measurable sets, $F_{1}=E_{1},\quad F_{2}=E_{2}\smallsetminus E_{1},\quad F_{n}=E_{n}\smallsetminus E_{n-1}\quad {\text{ for }}2\leq n$ Show that $\bigcup _{i=1}^{\infty }E_{i}=\bigsqcup _{i=1}^{\infty }F_{i}$. Next apply countable additivity. Finally, justify and then use the fact that $\sum _{i=1}^{n}\lambda (F_{i})=\lambda (E_{n})$ for each $1\leq n$.

Exercise 8. Prove the Downward Continuity of Measure To prove the downward continuity of measure, let $E_{1},E_{2},\dots$ be as in the statement of the definition. Define the ascending sequence of sets $F_{1}=E_{1}\smallsetminus E_{2},\quad F_{n}=E_{1}\smallsetminus E_{n+1}\quad {\text{ for }}1\leq n$. 1. Prove that this sequence is ascending, and then apply the upward continuity of measure. 2. Infer downward continuity of measure.

Exercise 9. Counterexample Give an example descending countable sequence of measurable sets, $E_{1},E_{2},\dots$, such that $\lim _{n\to \infty }\lambda (E_{n})\neq \lambda \left(\bigcap _{n=1}^{\infty }E_{n}\right)$.