If $f_{k}\overset{m}{\to}f$ on $E\subset \mathbb{R}^{n}$, there is a subsequence $f_{k_{j}}$ such that $f_{k_{j}}\to f$ a.e in $E$

I'm currently reading Zygmund's "Measure and Integral". Theorem (4.22) states that: If $f_{k}\overset{m}{\to}f$ on $E\subset \mathbb{R}^{n}$, there is a subsequence $f_{k_{j}}$ such that $f_{k_{j}}\to f$ a.e in $E$. (The measure used here is the Lebesgue measure, denoted $m$)

The proof of the book is as follows:

Since $f_{k} \overset{m}{\to} f$, given $j=1,2,....$, there exists $k_{j}$ such that $$m(\{|f-f_{k}|>\frac{1}{j}\})<\frac{1}{2^{j}}$$ for $k\leq k_{j}$. We may assume that $k_{j}\nearrow$. Let $E_{j}=\{|f-f_{k_{j}}|>\frac{1}{j}\}$, and $H_{m}=\bigcup_{j=m}^{\infty}E_{j}$. Then $m(E_{j})<2^{-j}$, $m(H_{m})\leq \sum_{j=m}^{\infty} 2^{-j} = 2^{-m+1}$, and $|f-f_{k_{j}}|\leq \frac{1}{j}$ in $E-E_{j}$. Thus, if $j\geq m$, $|f-f_{k_{j}}|\leq \frac{1}{j}$ in $E-H_{m}$, so that $f_{k_{j}}\to f$ in $E-H_{m}$. Since $m(H_{m})\to 0$, it follows that $f_{k_{j}}\to f$ a.e in $E$. This completes the proof.

I have 2 problems about the proof:

1. To prove $f_{k_{j}}\to f$ a.e, we should find a set $Z$ such that $f_{k_{j}}\to f$ on $E-Z$. What is the set $Z$ in this proof? It seems like it is the $Z$ is the limit of $H_{m}$ but I don't know how to state this more clearly.
2. In the proof, it mentions "Thus, if $j\geq m$, $|f-f_{k_{j}}|\leq \frac{1}{j}$ in $E-H_{m}$, so that $f_{k_{j}}\to f$ in $E-H_{m}$". Does this imply $f_{k_{j}}$ converges uniformly to $f$ in $E-H_{m}$? Since given $\varepsilon>0$, we can find $j\geq m$ and $\frac{1}{j}<\varepsilon$, therefore, for all $j'\geq j$, we have $|f-f_{k_{j'}}|<\frac{1}{j'}<\frac{1}{j}<\varepsilon$.

Any help is appreciated.

1. $Z=\bigcap_n H_m$. [If $x \in E-Z$ then $x \in E-H_m$ for some $m$, so $f_{k_j} (x) \to f(x)$].

2. Yes, $f_{k_j} \to f$ uniformly on $E-H_m$. (However, uniform convergence on $Z$ need not hold).