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.

  • 1
    $\begingroup$ Consider $\mathbb{R}^1$ and let $E_n = [n,\infty)$. $\endgroup$ – breeden Jan 20 '14 at 20:09
  • $\begingroup$ @breeden post it as an answer, though it is really appalling the OP made no attempt. almost any sensible sort of set would work... $\endgroup$ – Lost1 Jan 20 '14 at 20:11
  • $\begingroup$ Wow, it's pretty simple. Thanks, Breeden. $\endgroup$ – user122655 Jan 20 '14 at 20:11
  • $\begingroup$ @Lost1 OK. I decided to provide a proof for the Downward Monotone Convergence theorem that made finiteness requirement very obvious. $\endgroup$ – breeden Jan 20 '14 at 20:36
  • 1
    $\begingroup$ @Lost1 Sorry, I just saw your comment. Thank you for letting me know. $\endgroup$ – user122655 Jan 21 '14 at 9:17

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:

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

| cite | improve this answer | |
  • 1
    $\begingroup$ Wow, Thank you for kind explanation! $\endgroup$ – user122655 Jan 21 '14 at 9:09
  • $\begingroup$ This idea of splitting E into a countable union AND a limit set is something I'll have to ponder over for a while :-) $\endgroup$ – Thomas Ahle Aug 20 '14 at 16:36
  • $\begingroup$ @ThomasAhle When working with measures, it is often quite convenient to be working with disjoint sets (this is so you get additivity). So your first attempt should be to write each $E_n$ into disjoint "parts", and your first naive guess should be: I'll just make each "part" $E_n - E_{n+1}$! Then when you take the union, you realize that if $x$ is in every $E_k$ then it is subtracted every time. So you say, crap, and then add those $x$ that are in every set: ie, $S = \cap^{\infty} E_n$. Does that help your pondering? $\endgroup$ – breeden Aug 20 '14 at 17:11
  • $\begingroup$ The first display equation should be $F_k$ instead of $F_n$. $\endgroup$ – Troy Woo Aug 15 '15 at 9:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.