Approximate a complex measurable function pointwisely almost everywhere by polynomials This is Exercise 13.12 in Rudin's Real and Complex Analysis:

Let $f$ be a complex-valued measurable function defined in $\mathbb{C}$. Then there is a sequence of polynomials $P_n$ such that $\lim_{n\to\infty}P_n(z) = f(z)$ almost everywhere.

I and my friend find this exercise interesting. We can prove this for the case $f$ is the characteristic function of a open (or closed) disc. But we don't know how to go further from here. Could anyone give some hints for us?
Edit: I give here some of my thoughts:


*

*At first, we should solve for the case $f$ is the characteristic function of a open dics (or rectangle). Done.

*Second, solve for the case $f$ is the characteristic function of a measurable set. (I get stuck at this step.)

*Third, solve for the case $f$ is a combination of two characteristic functions, hence I think the argument will work for a simple function. (I still have difficulties here, a polynomial approximates one characteristic function may be very different to another characteristic function.)

*...
 A: First, by standard measure theory constructions there exists a sequence of simple functions 
$$\phi_n = \sum_{k=1}^{m_n} a_{k,n} \chi_{R_{k,n}}$$
converging to $f$ a.e., where $R_{k,n}$ are open rectangles, $R_{k,n} \cap R_{j,n} = \emptyset$ for $k \ne j$, and such that $\bigcup_{k=1}^{m_n} \overline{R_{k,n}} = [-n,n]^2$. Let
$$\Omega_n := \bigcup_{k=1}^{m_n} R_{k,n}.$$
Then $\Omega_n$ is open with connected complement, $\phi_n$ is analytic on $\Omega_n$, since it is constant on each connected component, and the measure of $[-n,n]^2 \setminus \Omega_n$ is zero. By Runge's approximation theorem there exists a sequence of complex polynomials $P_{j,n}$ with $P_{j,n} \to \phi_n$ locally uniformly on $\Omega_n$ as $j\to\infty$. So there exists a polynomial $P_n = P_{j_n,n}$ such that the measure of the set $$E_n := \left\{ z \in [-n,n]^2 : |P_n(z) - \phi_n(z)| \ge \frac1n\right\}$$ is $\le 2^{-n}$. By a Borel-Cantelli argument this implies that for almost all $z\in \mathbb{C}$ there exists $N=N(z)$ such that $|P_n(z)-\phi_n(z)| < 1/n$ for $n \ge N$. Since $\phi_n \to f$ a.e., this shows $P_n \to f$ a.e.
