5
$\begingroup$

This doesn't seem to have been asked on this site before. I've been self-studying measure theory and came across this problem.

Given a sequence of measurable functions $\{f_n\}_{n \in \mathbb{N}}$ such that for all $\epsilon \gt 0$

$$\sum_{n=1}^\infty \mu (\{x: |f_n (x)| \gt \epsilon \}) \lt \infty $$

Prove that $f_n \to 0$ a.e.

I've given the problem a try and clearly the sum being finite implies the summand converges to 0 which satisfies the definition of convergence in measure. Also, I know that there exists a subsequence $f_{n_j}$ converging a.e to 0. In fact, were this a finite measure space, I could prove the problem statement. However, in this more general setting, I'm unable to prove the statement.

Any help would be appreciated. This is my first time posting here so I hope I've met all conventions.

$\endgroup$
  • $\begingroup$ Where did you find this problem? $\endgroup$ – Wolfy Nov 24 '16 at 20:22
  • $\begingroup$ @Wolfy I found this problem in a set of old problem sheets a friend of mine provided me with for self study. $\endgroup$ – anktsdmcknsy Nov 25 '16 at 1:21
  • $\begingroup$ @Wolfy Turns out this problem is also in Mathematical Analysis (1996) by Andrew Browder, pp 249 #12. $\endgroup$ – anktsdmcknsy Nov 25 '16 at 21:41
  • $\begingroup$ I see, I will try to give you a proof of this I believe I have done it before. Are you still stuck? $\endgroup$ – Wolfy Nov 26 '16 at 0:59
  • $\begingroup$ @Wolfy I'm not quite convinced by the current answer. Thanks . $\endgroup$ – anktsdmcknsy Nov 26 '16 at 1:06
2
$\begingroup$

WLOG all $f_n$ are nonnegative. I am assuming $\mu$ is a positive measure.

If the result fails, then $\limsup f_n(x) >0$ for all $x$ in a set of positive measure $A.$ That in turn implies that for some $\epsilon>0,$ $\limsup f_n(x) >\epsilon$ for all $x$ in a set of positive measure $B, B\subset A.$ Thus for every $x\in B,$ $f_n(x) > \epsilon$ for infinitely many $n.$ That tells us $\sum_n \chi_{\{f_n > \epsilon\}}(x) = \infty$ for all $x\in B.$ Hence

$$\infty = \int_B \left (\sum_n \chi_{\{f_n > \epsilon\}}\right )\,d\mu = \sum_n \int_B\chi_{\{f_n > \epsilon\}}\,d\mu = \sum_n \mu(B\cap \{f_n > \epsilon\}).$$

But $\mu(B\cap \{f_n > \epsilon\}) \le \mu(\{f_n > \epsilon\})$ for each $n.$ Thus $\sum_n \mu(\{f_n > \epsilon\})= \infty,$ contradicting the given hypothesis.

$\endgroup$
  • $\begingroup$ zhw: how do you get the second equality. I mean how do you interchange the integral and the summation? $\endgroup$ – serenus Jan 8 at 17:17
  • $\begingroup$ @serenus If all $g_n$ are nonnegative, then $\int (\sum g_n) = \sum \int g_n.$ Proof: MCT. $\endgroup$ – zhw. Jan 8 at 17:19
  • $\begingroup$ Yes, I agree, I would like to use MCT, but the function $\sum g_n$ takes $\infty$ at any $x\in B$. In this case can we use MCT? Does the MCT include this case? $\endgroup$ – serenus Jan 8 at 17:22
  • $\begingroup$ @serenus Certainly. In the MCT the functions are measurable and take values in $[0,\infty].$ $\endgroup$ – zhw. Jan 8 at 17:25
  • $\begingroup$ zhw: It seems that your answer can be easily modified to proved that the sequence of indicators $(\chi_{E_n})_{n\in\mathbb{N}}$ converges to zero a.e. whenever $\sum_{n=1}^\infty \mu(E_n)<\infty$, where $(E_n)_{n\in\mathbb{N}}$ is a sequence of measurable subsets of a measurable set $E$. $\endgroup$ – serenus Jan 8 at 17:39
1
$\begingroup$

For $m\in\mathbb{N}$, let $$F_m=\sum_{n=1}^{\infty}1_{\{|f_n|>1/m\}}.$$ Then $$\int F_m\,d\mu=\sum_{n=1}^{\infty}\mu(\{|f_n|>1/m\})<\infty,$$which implies $\mu(\{F_m=\infty\})=0$. For each $m$, there exists a null set $N_m$ such that $F_m(x)<\infty$ for all $x\notin N_m$. Let $N=\cup_m N_m$, which is null. We have that, for all $x\notin N$ and for all $m$, $F_m(x)<\infty$. Then, for all $x\notin N$ and for all $m$, there exists $n_0(m,x)\in\mathbb{N}$ such that, for all $n\geq n_0(m,x)$, $|f_n(x)|\leq 1/m$. Apply limits: for all $x\notin N$ and for every $m$, $\limsup_n |f_n(x)|\leq 1/m$. Now let $m\rightarrow\infty$: for all $x\notin N$, $\limsup_n|f_n(x)|=0$.

$\endgroup$
  • $\begingroup$ Do you mind having a look at my solution and giving your thoughts on it? $\endgroup$ – anktsdmcknsy Nov 28 '16 at 21:24
  • $\begingroup$ @anktsdmcknsy Your solution is correct. Notice that the solution I wrote is just another proof of Borel-Cantelli lemma. $\endgroup$ – user39756 Nov 29 '16 at 15:00
1
$\begingroup$

Looking at @user39756 's answer, I believe I have the answer. My hunch is that this is basically Borel-Cantelli I.

Borel Cantelli I states that:

Given measurable sets $\{E_n\}_{n \in \mathbb{N}}$ with $\sum_{n=1}^\infty \mu(E_n) \lt \infty$, we have $\mu(\limsup_{n \to \infty} E_n) = 0$.

Recall $\limsup_{n \to \infty} E_n = \bigcap_{i=1}^\infty \bigcup_{j=i}^\infty E_j$

Defining our appropriate measurable sets for this problem.

For $m \in \mathbb{N}$, let $A_{m,n} = \{x : |f_n(x)| \gt \frac{1}{m}\}$

Thus define $$A := \bigcup_{m \in \mathbb{N}} \bigcap_{i=1}^\infty \bigcup_{j=i}^\infty A_{m,j} = \bigcup_{m \in \mathbb{N}}\limsup_{n \to \infty} A_{m,n} = \{x : |\lim_{n\to \infty} f_n(x)| \gt 0 \} = \{x : \lim_{n\to \infty} f_n(x) \neq 0 \}$$

We want $\mu (A) = 0$, that is, $f_n \to 0$ a.e.

Now, $0 \le \mu (A) = \mu (\bigcup_{m \in \mathbb{N}}\limsup_{n \to \infty} A_{m,n}) \le \sum_{m \in \mathbb{N}} \mu(\limsup_{n \to \infty} A_{m,n}) = \sum_{m \in \mathbb{N}} 0 = 0$.

The above follow from subadditivity, and the fact that for each $m \in \mathbb{N}$, $\mu(\limsup_{n \to \infty} A_{m,n}) = 0$ by the finiteness of the sum in the problem statement for arbitrary $\epsilon \gt 0$.

Thus, $\mu(A) = 0 $.

Would appreciate any feedback.

$\endgroup$
  • $\begingroup$ I was just going to prove this using the same approach, this looks completely correct to me, nice job $\endgroup$ – Wolfy Nov 26 '16 at 23:20

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.