My concern arises when I was going over the proof that if $f_n\rightarrow_\mu f$, then for some subsequence $n_k$ we have $f_{n_k}\rightarrow_{a.e.}f$, provided that $\{f_n\}$ are measurable and finite a.e.

To me, I don't quite get the intuitive difference between convergence in measure and convergence almost everywhere very well. My characterization of convergence almost everywhere will be something like

$$\mu(\{\lim f_n\neq f\})=0 \Leftrightarrow \mu(\lim |f_n-f|\neq0)=0$$

and convergence in measure would be

$$\lim \mu(|f_n-f|\geq\epsilon)=0$$

In short, I distinguish the two concepts by thinking one has the limit operator inside the measure one is outside, so intuitively converge almost everywhere must be stronger. (I know it is only the case when the measure is finite, which I don't quite get yet.)

However, when attempted to prove the statement, I confused myself when writing down the condition:

$$\forall \epsilon>0, \mu(|f_n-f|\geq\epsilon)\rightarrow0,n\rightarrow\infty$$

It is clear that to attempt this question, I should construct a subsequence that is convergent almost everywhere. What I first think of is to write something like $$\mu(|f_n-f|\geq1/2^k)<1/2^k$$

but then I got stuck. I have no intuitive understanding of what kind of subsequence indexed by $k$ should I obtain from this condition. If I pick up $\{f_n\}$ such that the condition hold, find the $f_n$ with smallest $n$ and indext it by $k$, let $k\rightarrow \infty$, the choice of sequence seems more likely to follow the pattern of the original sequence. I looked at the textbook and here is the full step of the proof:

for every positive integer $k$, $\exists n_k,$ when $n\geq n_k$ $$\mu(|f_n-f|\geq1/2^k)<1/2^k$$ Suppose $n_1<n_2....<n_k<...$, $f'_k\equiv f_{n_k},k=1,2,....$ is a subsequence of $\{f_n\}$ and it is convergent a.e. to $f$. In fact, $\forall \epsilon>0$ \begin{align} \mu(\bigcap_k\bigcup_v|f'_{k+v}-f|\geq\epsilon) & \leq \mu(\bigcup_v|f'_{k_0+v}-f|\geq\epsilon)\text{ for every $k_0$} \\ & \leq \sum_v\mu(|f'_{k_0+v}-f|\geq\epsilon) \\ & \leq \sum_v \mu(|f'_{k_0+v}-f|\geq\frac{1}{2^{k_0+v}})\ (\frac{1}{2^{k_0}}<\epsilon) \\ & \leq \sum \frac{1}{2^{k_0+v}} \\ & = \frac{1}{2^{k_0}} \end{align} Since the inequality holds for any $k_0$ sufficiently large, $\mu(\bigcap_k\bigcup_v|f'_{k+v}-f|\geq\epsilon)=0.$

I quite enjoy the proof as it is clearly written and easy to follow, however, it doesn't very intuitive to me what has been done to the condition that allows the subsequence to have this nice property. What kind of constraint that I have made so that the subsequence convergent a.e. to $f$?Is there any intuitive way to understand the chosen subsequence?

  • $\begingroup$ Remark: The subsequence is far from unique. Even just changing $1/2^k$ to some other $a_k>0$ such that $\sum a_k < \infty$ yields the same result. Not to mention the fact that once you have a convergent subsequence, you can always refine it over and over again. $\endgroup$ – zugzug Nov 11 '20 at 1:58
  • $\begingroup$ @zugzug Hi, I know that the sequence is not unique, what I am asking is the intuition of the difference between two definitions because I am having trouble seeing why the constructed subsequence should convergent almost everywhere,, intuitively. $\endgroup$ – JoZ Nov 11 '20 at 10:27

Your Answer

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

Browse other questions tagged or ask your own question.