(The integral is Lebesgue with $\sigma$-finite measures.)
I've shown that the result holds if $f$ is simple and if $f$ is bounded and supported on a set of finite measure. In showing the result when $f$ is nonnegative and integrable, we have
$$\int f\,d(\mu+\nu)\equiv\sup_g\int g\,d(\mu+\nu),$$ taking the $\sup$ over all $0\le g\le f$ so that $g$ is bounded and supported on a set of finite measure.
If we can show $\sup_g(\int g\,d\mu+\int g\,d\nu)=\sup_g\int g\,d\mu+\sup_g\int g\,d\nu$, we're done. But I'm not sure how to show $\ge$.
Any hints (or alternative methods to show the result) would be greatly appreciated. Thanks in advance!
Your approach is the correct one, using the path
indicator function $\rightarrow$ simple function $\rightarrow$ nonnegative measurable function.
It is a simple but important fact that a nonnegative measurable function can be written as limit of an increasing sequence of simple functions. So, assume $f$ is nonnegative and measurable and let $\{f_n\}$ be a sequence of simple functions with $f_n \nearrow f$. This can be done everywhere so there is no fuss about $\mu$ or $\nu$ null sets.
You've proved already that $$\int f_n \, d\mu + \int f_n \, d\nu = \int f_n d(\mu + \nu).$$ You get the result for $f$ by applying the Monotone Convergence Theorem.