For $k, N \in \mathbb{N}$, define $E_N^{(k)} := \bigcap_{p \ge N} \{x \in A : |f_p(x) - f(x)| \le 1/k\}$. Then: (i) $E_N^{(k)} \subseteq E_{N+1}^{(k)}$ for each $k$ (fewer conditions in the intersection). (ii) For each fixed $k$, $\bigcup_{N \ge 1} E_N^{(k)} = A$ by pointwise convergence (every $x \in A$ is eventually within $1/k$ of $f(x)$). Fix $\varepsilon > 0$. For each $k$, since $(E_N^{(k)})_N$ increases toward $A$ and $\mu(A) < \infty$, choose $N_k$ with $\mu(A \setminus E_{N_k}^{(k)}) \le \varepsilon / 2^k$. Set $\Delta := \bigcup_{k \ge 1} (A \setminus E_{N_k}^{(k)})$. Then $\mu(\Delta) \le \sum_{k=1}^\infty \varepsilon / 2^k = \varepsilon$. Let $A_\varepsilon := A \setminus \Delta$. Then $\mu(A \setminus A_\varepsilon) = \mu(\Delta) \le \varepsilon$, and $A_\varepsilon \subseteq E_{N_k}^{(k)}$ for every $k \ge 1$. So for any $x \in A_\varepsilon$ and any $k \ge 1$, $|f_p(x) - f(x)| \le 1/k$ for all $p \ge N_k$. Since $N_k$ is independent of $x \in A_\varepsilon$: \begin{align*} \sup_{x \in A_\varepsilon} |f_p(x) - f(x)| \le 1/k \quad \text{for all } p \ge N_k \end{align*} which gives [uniform convergence](/page/Uniform%20Convergence) on $A_\varepsilon$.