real analysis - Continuity and almost everywhere convergence

(Folland 2.37) Suppose that $f_{n}$ and $f$ are measurable complex-valued functions,
and $\phi: \mathbb{C} \to \mathbb{C}$. If $\phi$ is continuous and
$f_{n} \to f$ a.e., then $\phi \circ f_{n} \to \phi \circ f$ almost
everywhere.

My Proof:

Since $f_{n} \to f$ a.e. it follows that $$N= \{x: \lim_{n \to \infty}
(f_{n} - f) \neq 0 \}$$ is a null set.

Since $\phi$ is continuous, it follows that $\forall \varepsilon >0$,
$\exists \delta >0$ s.t. $\forall x$, $\forall f_{k}(x) \in
> (f_{n}(x))_{n\geq1}$ $$||f_{k}(x) - f(x)|| < \delta \implies || (\phi
> \circ f_{k})(x) - (\phi \circ f)(x)|| < \varepsilon$$ Now suppose

$\exists w$ s.t. $\exists \varepsilon >0$ s.t. $\forall K \in
> \mathbb{N}$, $\exists k \geq K$ with $|| (\phi \circ f_{k})(x) - (\phi
> \circ f)(x)|| \geq \varepsilon$.

By continuty of $\phi$ it follows that $\exists \delta > 0$ such that
$\forall K \in \mathbb{N}$ $\exists k \geq K$ with $||f_{k}(w) -
> f(w)|| \geq \delta$.

Thus $w \in N$, and so $\phi \circ f_{n} \to \phi \circ f$ everywhere
except on a set of measure zero.

My question is this: I am trying to be very specific with my definition of continuity and limits, and I am not sure if I have been successful. In particular, I am not sure if the section:

Now suppose $\exists w$ s.t. $\exists \varepsilon >0$ s.t. $\forall K
> \in \mathbb{N}$, $\exists k \geq K$ with $|| (\phi \circ f_{k})(x) -
> (\phi \circ f)(x)|| \geq \varepsilon$.

By continuty of $\phi$ it follows that $\exists \delta > 0$ such that

$\forall K \in \mathbb{N}$ $\exists k \geq K$ with $||f_{k}(w) -
> f(w)|| \geq \delta$.

is valid. Any advice on this proof would be useful.

Answer

I'm not really following the notation in your proof. Here is how I would write it:

Suppose $\lim_{n\to\infty}f_n(x) = f(x)$ and let $\varepsilon>0$. By continuity of $\varphi$, we can choose $\delta>0$ such that $|\varphi(f(x))-\varphi (y)|<\varepsilon$ when $|f(x)-y|<\delta$. Then we can choose $N$ such that $n\geqslant N$ implies $|f(x)-f_n(x)|<\delta$. It follows that $$|\varphi(f(x))-\varphi(f_n(x))| < \varepsilon $$
for $n\geqslant N$, as desired.