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. Template:Robelbox $\lambda$ is countably additive. Template:Robelbox/close

Pairwise Additive

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

We would like to show that $\lambda (E\bigsqcup F)=\lambda (E)+\lambda (F)$. Template:Robelbox Prove the desired result by applying the measurability of F to $E\bigsqcup F$. Don't forget to use the fact that E and F are disjoint. Template:Robelbox/close Template:Robelbox Notice that the proof did not actually require E to also be measurable. Therefore state a generalization of the above result.

Template:Robelbox/close Template:Robelbox Prove by induction that $\lambda$ is therefore finitely additive. As part of the exercise, state what "finitely additive" should mean.

Template:Robelbox/close Template:Robelbox Let $E_1,E_2,\dots \in ℳ$ be any countable collection of disjoint measurable sets. We would like to show

$\lambda \left(\underset{i=1}{\overset{\mathrm{\infty }}{⨆}}{E}_i\right)=\sum _{i=1}^{\mathrm{\infty }}\lambda (E_i)$

To do so, state the result which you just proved for finite additivity. Then apply monotonicity and then take the limit as $n\to \mathrm{\infty }$.

Finally, use the above to prove countable additivity. Template:Robelbox/close Template:Robelbox Find $\lambda ((0,1)\bigsqcup (2,3))$ and then find $\lambda (\underset{i=1}{⨆}(i,i+1))$.

Also find $\lambda (\bigcup _{k=1}^{\mathrm{\infty }}(k,k+1/2^k))$.

Excision

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

Template:Definition Template:Robelbox Prove that $\lambda$ satisfies excision.

Continuity of Measure

Recall that, roughly stated, if a function f is continuous then $\underset{x\to x_0}{\lim }f(x)=f\left(\underset{x\to x_0}{\lim }x\right)=f(x)$. 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 $1\le i$) then

$\underset{n\to \mathrm{\infty }}{\lim }\lambda \left(\bigcup _{i=1}^n{E}_i\right)=\lambda \left(\underset{n\to \mathrm{\infty }}{\lim }\bigcup _{i=1}^n{E}_i\right)$

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

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

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

A sequence of sets $E_1,E_2,\dots \subseteq ℝ$ is called "descending" if $E_i\supseteq E_{i+1}$ for $1\le i$.

Proof

We now set for ourselves the proof of the theorem. Template:Robelbox $\lambda$ satisfies both upward and downward continuity of measure. Template:Robelbox/close Template:Robelbox

To prove the upward continuity of measure, define the sequence of disjoint measurable sets,

$F_1=E_1,F_2=E_2\setminus E_1,F_n=E_n\setminus E_{n-1}\text{ for }2\le n$

Show that $\bigcup _{i=1}^{\mathrm{\infty }}{E}_i=\underset{i=1}{\overset{\mathrm{\infty }}{⨆}}{F}_i$.

Next apply countable additivity.

Finally, justify and then use the fact that $\sum _{i=1}^n\lambda (F_i)=\lambda (E_n)$ for each $1\le n$.

Template:Robelbox/close Template:Robelbox 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 $F_1=E_1\setminus E_2,F_n=E_1\setminus E_{n+1}\text{ for }1\le n$.

1. Prove that this sequence is ascending, and then apply the upward continuity of measure.

2. Infer downward continuity of measure.

Template:Robelbox/close Template:Robelbox Give an example descending countable sequence of measurable sets, $E_1,E_2,\dots$, such that $\underset{n\to \mathrm{\infty }}{\lim }\lambda (E_n)\ne \lambda \left(\bigcap _{n=1}^{\mathrm{\infty }}{E}_n\right)$.

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