Problem 3

Exercise 5.3 (A summatory condition for decaying measure) Prove that

$\displaystyle \sum_{n=1}^\infty \mu(E_n) < \infty \implies \mu\left(\limsup_{n\to\infty} E_n\right)=0$

(5.8)

Proof. Recall the definition of $\limsup$ for sets:

$\displaystyle \limsup_{n\to\infty} E_n = \bigcap_{n=1}^\infty\bigcup_{m=n}^\infty E_m.$

(5.9)

It follows that

$\displaystyle \mu\left(\limsup_{n\to\infty} E_n\right) \leq \mu\left(\bigcup_{m=n}^\infty E_m\right) \quad \forall n\geq 1$

By countable subadditivity, we know

$\displaystyle \mu\left(\bigcup_{m=n}^\infty E_m\right) \leq \sum_{m=n}^\infty \mu(E_m) \quad \forall n\geq 1$

(5.11)

But $\sum_{n=1}^\infty \mu(E_n) < \infty$ implies

$\displaystyle \lim_{n\to\infty} \sum_{m=n}^\infty \mu(E_m)=0$

(5.12)

Applying this limit to equation $% latex2html id marker 8946 $ \ref{limsupkill}$$ then shows $\mu(\limsup E_n) = 0$ :

$\displaystyle \mu\left(\limsup_{n\to\infty} E_n\right) \leq \lim_{n\to\infty} \sum_{m=n}^\infty \mu(E_m)=0$

(5.13)

$\qedsymbol$