Cauchy in Measure Implies Subsequential Almost Uniform Convergence Confusion I am confused when reading the proof for the statement from my lecture note: "Cauchy in Measure implying subsequential almost uniform convergence".
Definitions:
Let $(X, \mathcal{F}, \mu)$ be a measure space. We say $\{ f_n \}_{n = 1} ^\infty$ is Cauchy in measure is for all $\epsilon > 0$, we have
$$
\lim_{n, m \to \infty} \mu(\{ x \in X: |f_n(x) - f_m(x)| \geq \epsilon \}) = 0.
$$
Moreover, we say $\{ f_n \}_{n = 1}^\infty$ almost uniformly converge to $f$ if for all $\epsilon > 0$, there exists $E \in \mathcal{F}$ such that $\mu(E) < \epsilon$ and
$$
\lim_{n \to \infty} \sup_{X - E} |f_n(x) - f(x)| = 0.
$$
Here is the proof for the statement:
Recursively we can find $n_1 < n_2 < ...$ such that for all $k \in \mathbf{N}$ and $m \geq n_k$, $\mu(\{ x \in X: |f_m(x) - f_{n_k}(x)| \geq 2^{-k} \}) < 2^{-k}$. Let $g_k = f_{n_k}$ and $F_k = \{ x \in X: |g_{k + 1}(x) - g_k(x)| \geq 2^{-k} \}$, so $\mu(F_k) < 2^{-k}$. Put $\hat{F}_m = \bigcup_{k = m} ^\infty F_k$, then $\mu(\hat{F}_m) < \sum_{k = m} ^\infty 2^{-k} = \frac{2}{2^m}$. We note if $x \in \hat{F}_m^c$, then $x \not \in F_k$ for all $k \geq m$, so for all $i \geq j \geq m$, $|g_i(x) - g_j(x)| \leq \sum_{k = j} ^{i - 1} |g_{k + 1}(x) - g_k(x)| < \sum_{k = j} ^\infty 2^{-k} = \frac{2}{2^j}$.It follows that $\{ g_j(x) \}_{i = 1} ^\infty$ is uniformly Cauchy on $\hat{F}_m^c$. Therefore, it is pointwise Cauchy on $E = \bigcup_{m = 1} ^\infty \hat{F}_m^c$. This prompts us to define
$$
f(x) = \begin{cases}
\lim_{j \to \infty} g_j(x) &, \mbox{ if } x \in E \\
0 &, \mbox{ if } x \in E^c.
\end{cases}
$$
It is easy to see $f$ is measurable and $\mu(E^c) = 0$. Since
$$
|f(x) - g_j(x)| = \lim_{i \to \infty} |g_i - g_j(x)| \leq \frac{2}{2^j}
$$
for all $x \in \hat{F}_k^c$, we have $g_j \to f$ almost uniformly and hence a.e.
My Question:
I think I am convinced that $g_j \to f$ a.e. as this is pretty much just by definition of $f$. However, how exactly did we even come to the conclusion that $g_j \to f$ almost uniformly in the end? I failed to see how the last inequality give us this implication. In particular, it seems like this inequality is dependent on the choice of $x$ and taking the supremum might ruin this. Could anyone please explain this from the definition of almost uniform convergence? I am starting to question if this is even a correct statement as I haven't seen such a statement in the textbooks such as Folland (The most comparable statement being Folland Proposition 2.30: Real Analysis, Folland Proposition 2.30 Modes of Convergence). However, Folland simply inferred pointwise a.e. convergence instead of almost uniformly convergence.
 A: To begin matching the definition of almost-uniform convergence, let $\epsilon > 0$, and choose $m$ so large that $2/2^m < \epsilon$. It was noted in the proof that $\mu(\hat F_m)<2/2^m$. Set "$E$" in the definition of almost-uniform convergence equal to $\hat F_m$. By our choice, $\hat F_m$ already has small measure, so the ultimate claim is that $g_j$ converges uniformly to $f$ on $\hat F_m^c$. (Note it was already proven that $g_j\to f$ on $\hat F_m^c$; now we are going to see how to upgrade this to uniform convergence.)
To see this, the definition $\hat F_m^c = \bigcap_{k=m}^\infty F_k^c$ tells us for each $k\ge m$, $|g_{k+1}-g_k|<2^{-k}$ holds on $\hat F_m^c$. Hence, by the Weierstrass $M$-test, $\sum_{k=m}^\infty (g_{k+1}-g_k)$ is a uniformly and absolutely convergent series on $\hat F_m^c$. I leave it to you to check $g_m+\sum_{k=m}^\infty(g_{k+1}-g_k)$ actually converges pointwise (and hence uniformly by the last sentence) to $f$, and consequently that $g_j\to f$ uniformly on $\hat F_m^c$. (Hint: the series is telescoping.)
