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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!$\mu\{x:\bigcap_\limits{k\geqslant 1}^{}\bigcup_\limits{n}^{}\bigcap_\limits{l\geqslant n}^{}{|f_l-f|}\leqslant \frac{1}{k}\}$

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    $\begingroup$ 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, 2017 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, 2017 at 15:12
  • $\begingroup$ Also, in your statement you wrote "E\in\Omega", it should be $E\subseteq\Omega$. $\endgroup$
    – VictorZurkowski
    Apr 6, 2017 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, 2017 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, 2017 at 18:39

1 Answer 1

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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.

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  • $\begingroup$ 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, 2017 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, 2017 at 13:58
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    $\begingroup$ @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, 2017 at 14:08

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