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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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 $x\notin N$, $\limsup_n|f_n(x)|=0$.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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