Spring 2021 - Problem 2

Solution.

1. Let $A = \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty A_k$, which is Borel as a countable union and countable intersection of Borel sets. By deleting sets if necessary, we may assume without loss of generality that $\nu\p{A_k^\comp} \leq \frac{1}{2^k}$. Then

   $$\begin{aligned} \mu\p{A} &= \lim_{N\to\infty} \mu\p{\bigcup_{n=1}^N \bigcap_{k=n}^\infty A_k} \\ &\leq \lim_{N\to\infty} \sum_{n=1}^N \mu\p{\bigcap_{k=n}^\infty A_k} \\ &= 0, \end{aligned}$$

   since whenever $m \geq n$, we have $\mu\p{\bigcap_{k=n}^\infty A_k} \leq \mu\p{A_m} \xrightarrow{m\to\infty} 0$. Similarly,

   $$\begin{aligned} \nu\p{A^\comp} &= \lim_{N\to\infty} \nu\p{\bigcap_{n=1}^N \bigcup_{k=n}^\infty A_k^\comp} \\ &= \lim_{N\to\infty} \nu\p{\bigcup_{k=N}^\infty A_k^\comp} \\ &\leq \lim_{N\to\infty} \sum_{k=N}^\infty \frac{1}{2^k} \\ &= 0, \end{aligned}$$

   since the infinite sum converges. Thus, $\mu\p{A} = \nu\p{A^\comp} = 0$, so $\mu \perp \nu$.

2. Since $f_n$ is Borel, the set $A_n = \set{f_n > 1} = \set{\frac{1}{f_n} < 1}$ is Borel. Then by Chebyshev's inequality,

   $$\begin{aligned} \mu\p{A_n} &\leq \int f_n \,\diff\mu \xrightarrow{n\to\infty} 0 \\ \nu\p{A_n^\comp} &\leq \int \frac{1}{f_n} \,\diff\nu \xrightarrow{n\to\infty} 0. \end{aligned}$$

   Thus, by (1), we see that $\mu \perp \nu$, as required.