The set of $x$ where a sequence convergences in terms of set operations I'm befuddled by this.
Suppose $f:\mathbb{R}\to\mathbb{R}$, $f_n:\mathbb{R}\to\mathbb{R}$, $n=1,2,\dots$, and consider the set $$\bigcap_{k\geq 1}\bigcup_{p\geq 1}\bigcap_{m\geq p}\{x\in\mathbb{R} \ | \ |f(x)-f_m(x)| <1/k\}.$$
Is this the set of $x$ where $f_n(x)$ converges to $f(x)$, or the set of $x$ where it converges uniformly? What would need to be modified to change from one type of convergence to the other?
This comes from an effort to understand the last step in the proof of Egorov's theorem.
Thank you.
 A: First, you write "the set of $x$ where [$f_n$] converges uniformly [to $f$]".  This isn't actually meaningful.  Uniform convergence isn't something that happens or doesn't happen at a point, so it doesn't make sense to consider the set of points where it happens.  Rather, uniform convergence is something that happens or doesn't happen on a set.
So, the right question is whether the set you gave is a set on which $f_n$ converges uniformly to $f$.  The answer to that is, not necessarily, because the witness for $p$ can depend on $x$.  What I mean is maybe clearer if you rewrite your expression using logical quantifiers instead:
$$ \{ x : (\forall k\ge 1: \exists p_k : \forall m\ge p_k : |f(x)-f_m(x)| < \tfrac1k) \} $$
So for $x$ to be a member of this set, there have to exist some values $p_k$ doing a certain job.  But which $p_k$ do that job can be different for different $x$, which is exactly the kind of non-uniformity proscribed by uniform convergence.  (For a concrete example, take any example showing that uniform convergence is really stronger than pointwise convergence.  Say, $f_n(x) = x^n$, on $[0,1)$.)
In the Wikipedia proof of Egorov's theorem that you linked in comments, on the other hand, we see
\begin{align*}
A\setminus B
&= \bigcap_{k\ge 1} \bigcap_{m\ge n_k} \{x : |f(x)-f_m(x)| < \tfrac1k\} \\
&= \{x : \forall k\ge 1 : \forall m\ge n_k : |f(x)-f_m(x)| < \tfrac1k\}
\end{align*}
Here there is no "$\exists p_k$"; the job done by the $p_k$ in your example is here being done by $n_k$, which has been constructed earlier in the proof and so doesn't depend on $x$.  This set is, in fact, the set of $x$ for which that choice of values $n_k$ does the job.
A: A point $x$ belong to the above set if and only if for all $k>0$, there exists $p$ that depends on $x$ and $k$ such that we have the strict inequality inside the set, for all $m$ bigger than or equal to $p$. Since $p$ depends on $x$ the convergence seems not to be uniform.
