Question: Let $\Omega$ be a non-empty open set in $\mathbb{C}$ and let $f$ be a continuous function on $\Omega$. Suppose $\{f_n\}$ is a sequence of holomorphic functions on $\Omega$ such that $$\lim_{n\to \infty} \int_D |f_n(x+iy) - f(x+iy)|\, dx \, dy =0$$ for all closed disk $D \subset \Omega$. Show that $f$ is holomorphic in $\Omega$ and both $f_n \to f$ and $f_n' \to f'$ uniformly on compact subsets of $\Omega$.
My attempt: To prove $f$ is holomorphic in $\Omega$, I aim to show that $f$ satisfy the Cauchy integral formula, that is $$f(z_0) = \int_{|z-z_0| = R} \frac{f(z)}{z - z_0}\, dz$$ for all $z \in \Omega$ where $\{z\colon |z- z_0| \leq R\} \subset \Omega$. Given that $f_n$ is holomorphic in $\Omega$ by the use of polar coordinates $$f_n(z_0) =\int_D\frac{f_n(z)}{z - z_0}\, dz = \frac{1}{\pi R^2} \int_0^{2\pi} \int_0^R f_n(z_0 + R e^{i\theta})r \, d r \, d \theta$$ where the compact closure of $D = D(z_0, R)$ is contained in $\Omega$ whereas $$\int_D f_n(x+iy) \, d x\, dy = \int_0^{2\pi} \int_0^R f_n(z_0 + r e^{i\theta}) r \, d r\, d \theta$$ so $\int_D f_n(x+iy)\, dx \, dy $ and $f_n(z_0)$ are not directly comparable. are there any ways to resolve this issue? Or is there an alternative method to solve this problem?
Here is my answer following the prompts in the comments.
Let $z_0 \in \Omega, \epsilon > 0$ be given and choose $R > 0$ such that $\overline{D_{R + \epsilon}(z_0)} \subset \Omega$. Then, there exists $N \in \mathbb{N}$ so that
$$n \geq N \implies \int_{D_{R + \epsilon}(z_0)}|f_n(x+iy) - f(x+iy)|\, dx \, dy < \frac{|D_{\epsilon}(z_0)|\epsilon}{2}.$$
and in particular $$n,m \geq N \implies \int_{D_{R + \epsilon}(z_0)} |f_n(x+iy) - f_m(x+iy)|\, dx \, dy < |D_{\epsilon}(z_0)|\epsilon.$$ Also given $f_n, f_m$ are holomorphic in $\Omega$ then by mean-value property $$f_n(z_0) = \frac{1}{2\pi} f_n(z_0 + r e^{i \theta})\, d \theta, \quad f_m(z_0) =f_m(z_0 + r e^{i \theta})\, d \theta$$ for all $0 < r \leq R$. Then by the use of polar coordinates whenever $n, m \geq N$ \begin{align} |D_R(z_0)| |f_n(z_0) - f_m(z_0)| &= \pi R^2 |f_n(z_0) - f_m(z_0)|\\ &= \left| \int_{D_R(z_0)}[f_n(x+iy) - f_m(x+iy)] \, dx \, dy \right| \\ &\leq \int_{D_{R + \epsilon}(z_0)}\left| f_n(x+iy) - f_m(x+iy) \right|\, dx \, dy \\ &\leq |D_\epsilon(z_0)| \epsilon \end{align} Thus, $n,m \geq N \implies |f_n(z_0) - f_m(z_0)| \leq \frac{|D_\epsilon(z_0)|}{|D_R(z_0)|}\epsilon < \epsilon.$ Given any $z_1 \in \overline{D_R(z_0)}$ then $\overline{D_\epsilon(z_1)} \subset \overline{D_{R + \epsilon}(z_0)}$ so \begin{align} |D_\epsilon(z_1)| |f_n(z_1) - f_m(z_1)| &= \pi \epsilon^2 |f_n(z_1) - f_m(z_1)|\\ &= \left| \int_{D_\epsilon(z_1)}[f_n(x+iy) - f_m(x+iy)] \, dx \, dy \right| \\ &\leq \int_{D_{R + \epsilon}(z_0)}\left| f_n(x+iy) - f_m(x+iy) \right|\, dx \, dy \\ &\leq |D_\epsilon(z_0)| \epsilon \end{align} Thus, $n,m \geq N \implies |f_n(z_1) - f_m(z_1)| \leq \epsilon.$ Therefore, $\{f_n\}$ is uniformly Cauchy in $\overline{D_R(z_0)}$ so $\{f_n\}$ is uniformly convergent in $\overline{D_R(z_0)}$. As such closed disks are arbitrary we have $f_n$ converges uniformly on every compact subsets of $\Omega$ implying that $f_n$ converges uniformly to $F$, a holomorphic function in $\Omega$. Now, we are left to show that $F(z) = f(z), \forall z \in \Omega.$ Given any closed disk $D \subset \Omega$, we have $$0 = \lim_{n \to \infty} \int_D |f_n(x+iy) - f(x+iy)|\, dx \, dy = \int_D |F(x+iy) - f(x+iy)|\, dx \, dy$$ thus $F = f$ a.e. in $D$. If for a contradiction that the set $\{z \in D \colon f(z) \neq F(z)\} = (f - F)^{-1}(D - \{0\})$ is nonempty then it being open must have positive measure, contrary to our conclusion. Thus, $f(z) = F(z), \forall z \in D$. Again, $D$ is an arbitrary closed disk in $\Omega$ then $f(z) = F(z), \forall z \in \Omega$, this concludes the proof. As for the convergence of the derivatives, it follows a standard argument available in most complex analysis text.
Please do let me know if there any gaps in my proof. Thank you!