Why does the "$i$" in $f=u+vi$ not affect the measurability of $f$? I stumbled across the following result in Rudin's Real and Complex Analysis:

He says in part (c) that "the complex case then follows from (a) and (b)."  To me, this is saying that because $f$ and $g$ are sums of real-valued functions, i.e. $$f=u_1+iv_1, \quad g=u_2+iv_2$$ for real valued $u_1,u_2,v_1$ and $v_2$, the sum of $f$ and $g$ must also be measurable, i.e. $$f+g=
\underbrace{(u_1+u_2)}_{\text{measurable}}+i\underbrace{(v_1+v_2)}_{\text{measurable}}$$ 
What I can't seem to get past is how the $i$ was seemingly ignored!  Why are we allowed to ignore it? 

Per part (b), $u_1, u_2, v_1$, and $v_2$ are real-measurable functions on $X$, and thus $u_1 + u_2$ and $v_1 + v_2$ are real-measurable functions on $X$. Per part (a), this implies that $f+g = (u_1 + u_2) + i(v_1 + v_2)$ is a complex-measurable function on $X$. In particular, the $i$ is baked in to the statement of part (a).
