Counterexample for downward monotone convergence theorem on measurable set

Question

This is exercise on tao's note. What is counterexample? I can't find it. What I trying to prove Downward monotone convergence is that Think about complement of $E_{n}$ and apply upward monotone convergence theorem. However, I think I don't use condition for finite measure. Anyone who can help me?

Exercise 11 (Monotone convergence theorem for measurable sets)

(Upward monotone convergence) Let ${E_1 \subset E_2 \subset \ldots \subset {\bf R}^n}$ be a countable non-decreasing sequence of Lebesgue measurable sets. Show that ${m( \bigcup_{n=1}^\infty E_n ) = \lim_{n \rightarrow \infty} m(E_n)}$. (Hint: Express ${\bigcup_{n=1}^\infty E_n}$ as the countable union of the lacunae ${E_n \backslash \bigcup_{n'=1}^{n-1} E_{n'}}$.)

(Downward monotone convergence) Let ${{\bf R}^d \supset E_1 \supset E_2 \supset \ldots}$ be a countable non-increasing sequence of Lebesgue measurable sets. If at least one of the ${m(E_n)}$ is finite, show that ${m( \bigcap_{n=1}^\infty E_n ) = \lim_{n \rightarrow \infty} m(E_n)}$.

Give a counterexample to show that the hypothesis that at least one of the ${m(E_n)}$ is finite in the downward monotone convergence theorem cannot be dropped.

Answer 1

A counter-example for how the Downward Monotone Convergence theorem can fail is given by; Let $n = 1$ so we are working in $\mathbb{R}^1$ and let $E_n = [n,\infty)$. Then $m(E_n) = \infty$ for all $n$. And $E_n$ satisfies the criterion $E_1 \supset E_2 \supset E_3 \supset \cdots$. However, $m(\cap_{n=1}^{\infty} E_n) = m(\emptyset) = 0 \neq \infty = \lim \, m(E_n)$.

So let's go over the proof of the theorem to see exactly where the finiteness of an $E_n$ is used. First, using the same idea from the first problem we can write each $E_n$ as a disjoint union:

$$ E_n = \cup_{k=n}^{\infty} F_n \, \bigcup \, S $$

Where each $F_n$ is given by $F_n = E_n - E_{n+1}$, and $S$ is given by $S = \cap^{\infty} E_n$. It should be clear that these sets are mutually disjoint. Now the additivity of measures gives

$$ m(E_n) = m(S) + \Sigma_{k=n}^{\infty} m(F_k) $$

The statement that $m(E_n)$ is finite for some $n$ is exactly the statement that $\Sigma_{k=n}^{\infty} m(F_k)$ converges for some $n$ and that $m(S)$ is finite.

That $m(S)$ is finite is immediate from the monotonicity of measures and that $S \subset E_n$ for all $n$ along with that $E_n$ has finite measure for some $n$. And since the series $\Sigma_{k=n}^{\infty} m(F_k)$ converges for some $n$ we have:

$$ \lim_n \, \Sigma_{k=n}^{\infty} m(F_k) = 0. $$

We get:

$$ \lim_n \, m(E_n) = m(S) + \lim_n \, \Sigma_{k=n}^{\infty} m(F_k) = m(S). $$