My proof to the above problem is:
Define $f(x) := \lim_{n}f_{n}(x)$ for all $x \in A$. Since $f_{n} \rightarrow f$ on $A$, for any $\epsilon > 0$, $x \in A$, there exists $n_{x} \in \mathbb{N}$ such that $|f_{n_{x}}(x) - f(x)| < \epsilon$. Thus, fixing $\epsilon$, each $a\in A$ belongs to one of the countably many sets $$A(n_{a}) := \{x : |f_{n_{a}}(x) - f(x)| < \epsilon\} = f_{n_{a}}^{-1}\Big( B\big(f(x), \epsilon\big)\Big).$$ Conversely, any $x \in A(n_{a})$ must be in $A$ by definition of convergence. So we can write \begin{align*} A = \bigcup_{n_{a}\in\mathbb{N}} A(n_{a}) = \bigcup_{n_{a} \in \mathbb{N}} f_{n_{a}}^{-1}\Big( B\big(f(x), \epsilon\big)\Big), \end{align*} which is a countable union of measurable sets, as the $f_{n}$ are measurable. Hence $A$ is measurable.
I feel a little weird about this "proof". Specifically the preimages $f_{n_{a}}^{-1}(B(f(x), \epsilon))$, since I've restricted each $f_{n}$ to $A$. But I can't quite tell if what I am saying is valid - that these are measurable in the ambient space that $A$ lies in or not necessarily.
I have an issue with the statement $$ \{x : \lvert f_{n_a}(x) - f(x)\rvert < \epsilon\} = f_{n_a}^{-1}\left(B(f(x), \epsilon)\right) $$ We've ended up with a free $x$ on the RHS that's not defined anywhere. A quite pretty proof exists using the fact that $f_n(x)$ converges if and only if $f_n(x)$ forms a Cauchy sequence. This gives us that $$ x \in A \iff (\forall \epsilon > 0) (\exists n_0)(\forall n \geq n_0)\ \lvert f_n(x) - f_{n_0}(x)\rvert < \epsilon$$ This characterisation of $A$ means we can ignore what $f_n$ is converging to (i.e. we don't need to define the limit $f$). From this we directly get that $$ A = \cap_{k = 1}^{\infty} \cup_{n_0=1}^{\infty} \cap_{n=n_0}^{\infty} (f_n - f_{n_0})^{-1} (-\tfrac{1}{k}, \tfrac{1}{k})$$ where we've replaced $\epsilon$ with $\frac{1}{k}$ to keep things countable. As a difference of measurable functions $f_n - f_{n_0}$ is measurable. Therefore $(f_n-f_{n_0})^{-1}(-\tfrac{1}{k}, \tfrac{1}{k})$ is measurable. Therefore $A$ is measurable as a union of measurable sets.