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