What does this mean $\mu\{x:\bigcap_\limits{k\geqslant 1}^{}\bigcup_\limits{n}^{}\bigcap_\limits{l\geqslant n}^{}{|f_l-f|}\leqslant \frac{1}{k}\}$ 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!$\mu\{x:\bigcap_\limits{k\geqslant 1}^{}\bigcup_\limits{n}^{}\bigcap_\limits{l\geqslant n}^{}{|f_l-f|}\leqslant \frac{1}{k}\}$ 
 A: 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.
