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.

$$\lambda$$ 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 $$\lambda(E\sqcup F)=\lambda(E)+\lambda(F)$$.

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.

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

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

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.

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.

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 $$1\le i$$) 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 $$\lim_{n\to\infty}\bigcup_{i=1}^nE_n$$ 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 $$E_n$$.)

A sequence of sets $$E_1,E_2,\dots\subseteq\Bbb R$$ 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.

$$\lambda$$ satisfies both upward and downward continuity of measure.

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


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

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 $$1\le n$$.

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 $$F_1=E_1\smallsetminus E_2, \quad F_n = E_1\smallsetminus E_{n+1} \quad \text{ for } 1\le n$$.

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

2. Infer downward continuity of measure.

Give an example descending countable sequence of measurable sets, $$E_1,E_2,\dots$$, such that $$\lim_{n\to\infty}\lambda(E_n) \ne \lambda\left(\bigcap_{n=1}^\infty E_n\right)$$.