Measure Theory

Definition

For the following we first denote that
• $E_{k}$ increases to $E$ means $E_{k} \subset E_{k + 1}$ and $E = ⋃_{k = 1}^{\infty} E_{k}$ which we write as
$E_{k} e a r r o w E$
• $E_{k}$ decreases to $E$ means $E_{k + 1} \subset E_{k}$ and $E = ⋂_{k = 1}^{\infty} E_{k}$ which we write as
$E_{k} ↘ E$

Corollary

(continuity of measure) Suppose $E_{1}, E_{2} . . .$ are measurable subsets of $R^{d}$ .
(i) If $E_{k} e a r r o w E$ or
(ii) If $E_{k} ↘ E$ and $m (E_{k}) < \infty$ for some $k$
Then:

m (E) = lim_{n \to \infty} m (E_{n})

Proof. For (i) Construct disjoint measurable subsets $G_{k} = E_{k} - E_{k - 1}$ . Then we have by countable additivity

m (E) = lim_{n \to \infty} m (⋃_{k = 1}^{N} G_{k}) = lim_{n \to \infty} \sum_{k = 1}^{N} m (G_{k})

For (ii) Construct disjoint sets $G_{k} = E_{k} - E_{k - 1}$ for each k. Because every subsequent $E_{k}$ is a subset of $E_{n}$

⋂_{k = n}^{\infty} E_{k} = ⋂ E_{k} = E

which is the set common between all subsequent subsets and $E_{n}$ and by countable additivity

m (⋃_{k = n}^{\infty} G_{k}) = \sum_{k = n}^{\infty} m (G_{k})

which is the set difference of all subsequent subsets with $E_{n}$ Putting everything together we have

m (E_{n}) = m (E) + m (\sum_{k = n}^{\infty} G_{k})

(S) Since there exists some n where $m (E_{n}) \leq \infty$ as assumed in the proposition this implies:

by monotonicy we have $m (E) < m (E_{n}) \leq \infty$
$m (\sum_{k = n}^{\infty} G_{k})$ converges for that n to some finite value. Hence it is clear from

\sum_{k = n}^{\infty} m (G_{k}) = \sum_{k = 1}^{\infty} m (G_{k}) - \sum_{k = 1}^{n} m (G_{k})

that $\sum_{k = n}^{\infty} G_{k}$ converges since all terms on both RHS and LHS here must be finite. Thus there exists an N in which for all $n > N$ ,

\sum_{k = n}^{\infty} m (G_{k}) = \sum_{k = 1}^{\infty} m (G_{k}) - \sum_{k = 1}^{n} m (G_{k}) \leq ε

Taking limits on all sides of (1):

lim_{n \to \infty} m (E_{n}) = lim_{n \to \infty} m (E) + lim_{n \to \infty} \sum_{k = n}^{\infty} m (G_{k})

= m (E) + 0