I am studying the following theorem: Consider $E\in\mathscr{F}$(sigma-algebra), and $E\in\Omega$ defined on a measure space $(\Omega,\mathscr{F},\mu)$. Suppose $\mu(E)<\infty$, and $\{f_n\}$ is a sequence of measurable functions on $E\to\mathbb{R}$ which are finite almost everywhere and converge almost everywhere to a function $f:E\to\mathbb{R}$ which is also finite almost everywhere. Then $f_n\to f$ almost uniformly in $E$. The proof starts with the following: $\epsilon$>0

What does this mean? How do I read this intersection and union signs in this case? Thanks!$\mu\{x:\bigcap_\limits{k\geqslant 1}^{}\bigcup_\limits{n}^{}\bigcap_\limits{l\geqslant n}^{}{|f_l-f|}\leqslant \frac{1}{k}\}$

  • 4
    $\begingroup$ Intersections correspond to $\forall$ and unions correspond to $\exists$. So you can simply read as $$ \textstyle \Big\{ x : \bigcap_k \bigcup_n \bigcap_{l \geq n} |f_l(x) - f(x)| \leq \frac{1}{k} \Big\} = \{ x : \forall k, \exists n, \forall l \geq n, |f_l(x) - f(x)| \leq \frac{1}{k} \}. $$ Since $\Bbb{R}$ is first countable, this is equivalent to the following more friendly expression $$ \textstyle \Big\{ x : \bigcap_k \bigcup_n \bigcap_{l \geq n} |f_l(x) - f(x)| \leq \frac{1}{k} \Big\} = \{x : f_n (x) \text{ converges to } f(x) \}. $$ $\endgroup$ – Sangchul Lee Apr 6 '17 at 14:45
  • $\begingroup$ That "x:" is not standard notation, nor using "cup" for "$\exists$" or "cap" for "$\forall$." The standard notation would be $\mu\{\bigcap_\limits{k\geqslant 1}^{}\bigcup_\limits{n}^{}\bigcap_\limits{l\geqslant n}^{}{(|f_l-f|}\leqslant \frac{1}{k})\}$. As @SangchulLee explains, this is the $\mu$ measure of the set where $f_n\to f$ (pointwise). $\endgroup$ – VictorZurkowski Apr 6 '17 at 15:12
  • $\begingroup$ Also, in your statement you wrote "E\in\Omega", it should be $E\subseteq\Omega$. $\endgroup$ – VictorZurkowski Apr 6 '17 at 15:16
  • $\begingroup$ @VictorZurkowski Do not forget that $\mu$ compromises the measure of $(\Omega,\mathscr{F})$ and the dominion of the function is $E$ and $x\in E$ that is why you write the $x$ as Sangchul Lee did. $\endgroup$ – Pedro Gomes Apr 6 '17 at 18:16
  • $\begingroup$ Right. Then the notation I have seen is $\mu\{\bigcap_\limits{k\geqslant 1}^{}\bigcup_\limits{n}^{}\bigcap_\limits{l\geqslant n}^{}{E(|f_l-f|}\leqslant \frac{1}{k})\}$ $\endgroup$ – VictorZurkowski Apr 6 '17 at 18:39

Define $E(k,l) = \{x \in \Omega: |f_l(x) -f(x)| \le \frac{1}{k}\}$.

This is a well-defined measurable set in $(\Omega,\mathscr{F})$ whenever all $f_n$ and also $f$ are measurable.

Then reformulate $f_n(x) \rightarrow f(x)$ for a certain $x \in \Omega$. The statement of pointwise convergence at $x$ means by definition:

$$\forall \varepsilon > 0: \exists n \in \mathbb{N} \forall l \ge n: |f_l - f(x)| < \varepsilon$$

WE cna uee $\le$ instead of $<$ as well,and as the values of $\frac{1}{k}$ get as small as we like we can replace the $\forall \varepsilon>0$ by $\forall k$ and $\varepsilon$ by $\frac{1}{k}$, to make everything countable (Which is what you want in measure theory). So $f_n(x) \rightarrow f(x)$ means for some $x$:

$$\forall k \in \mathbb{N} :\exists n \in \mathbb{N}: \forall l \ge n: |f_l(x) -f(x)| \le \frac{1}{k}$$

We can write that last statement using the definition of $E(k,l)$ as:

$$\forall k \in \mathbb{N} :\exists n \in \mathbb{N}: \forall l \ge n: x \in E(k,l)$$

And as we have sets now, we can also reformulate as intersections and unions, so $f_n(x) \rightarrow f(x)$ iff (indices implicitly over $\mathbb{N}$):

$$x \in \bigcap_{k} \bigcup_{n} \bigcap_{l \ge n} E(k,l)$$

which is a measurable set $P = P((f_n, f))$ say (it depends only on the sequence and its pointwise limit a.e.),as all $E(k,l)$ are, which is the whole point of writing it like this. So saying that $f_n$ converges to $f$ almost everywhere on $E$, means that the complement of $P$ in $\Omega$ has measure $0$, which seems to be used in the remainder of the proof, I'm guessing. It implies that $\mu(P \cap E) = \mu(E)$.

Your notation of the set sort of confuses the quantor definition with the operations write-up in terms of unions and intersections.

  • $\begingroup$ Thank you a lot for your replies, you have been very helpful. There is something I do not quite understand in your last equation $\forall \varepsilon > 0: \mu(E \cap P) \ge (1-\varepsilon)\mu(E)$, On this equation you are implying that $\mu(P)\geqslant\epsilon\mu (E)$. How can this be if the $\epsilon$ is defined on function codomain? $\endgroup$ – Pedro Gomes Apr 7 '17 at 13:13
  • $\begingroup$ What is the interpretation of $\forall \varepsilon > 0: \mu(E \cap P) \ge (1-\varepsilon)\mu(E) $? Where did it come from? $\endgroup$ – Pedro Gomes Apr 7 '17 at 13:58
  • 1
    $\begingroup$ @PedroGomes I think we know that $\mu(E\cap P) = \mu(E)$. Almost all points of $E$ are points of pointwise convergence, i.e. are in $P$. $\endgroup$ – Henno Brandsma Apr 7 '17 at 14:08

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.