Interchanging an integrand and a measure Suppose that $f$ and $g$ are continuous functions on a compact set $K$ and $\mu$ is a (complex) measure. How does the equality
$$\int_K f\ \mathrm dg \cdot \mu = \int_K fg\ \mathrm d \mu$$
can be justified, i.e., what does "$\cdot$" mean here? I stumbled upon this issue when studying the proof of the auxiliary result d) here. What I already found is that it is possible to multiply a function with a measure, as explained in this question. Am I right that the above equality is a consequence of this?
 A: As already mentioned in the comments $g \cdot \mu$ is the complex measure $A \mapsto \int g \chi_A d\mu$ where $A$ is a $\mu$-measurable set and $\chi_A$ is the characteristic function of $A$. Suppose $f$ is simple, that is $f = \sum f_k \chi_{E_k}$ for a finite disjoint collection of measurable sets $\{E_k\}$ and coefficients $f_k \in \mathbb{C}$. Now note, that 
\begin{eqnarray}
\int f d(g \cdot \mu) = \sum f_k \int g \chi_{E_k} d\mu = \int \sum f_k \chi_{E_k} g d\mu = \int fg d\mu \tag{$\star$}
\end{eqnarray}
Since our result thus holds for simple functions, by the usual approximation argument, the general result is obtained.

$\textbf{Update}:$
  If $\mu$ is a complex measure, we can consider the real and imaginary part of $\mu$, say $\mu_1 = \mathfrak{Re}(\mu)$ and $\mu_2 = \mathfrak{Im}(\mu)$, then you can check that the set functions $\mu_i$ for $i=1,2$ define signed measures. Now consider their relative Hahn-Jordan decomposition, so $\mu_1 = \mu_1^+ -\mu_1^-$ and $\mu_2 = \mu_2^+-\mu_2^-$ and note, that $\mu_i^+$ and $\mu_i^-$ are positive measures for $i=1,2$. In general integration with respect to complex measures is defined by (if the RHS exists)
  \begin{eqnarray}
\int f d\mu = \bigg(\int f d\mu_1^+ - \int f d\mu_1^-\bigg) + i \bigg(\int f d\mu_2^+ - \int f d\mu_2^- \bigg) \tag{$\star\star$}
\end{eqnarray}
  where the integral of a complex valued function $f$ with respect to a positive measure $\nu$ is defined by (if the RHS exists)
  \begin{eqnarray} 
\int f d\nu = \int \mathfrak{Re}(f) d\nu + i \int \mathfrak{Im}(f) d\nu \\ = \int \mathfrak{Re}(f)^+ d\nu - \int \mathfrak{Re}(f)^- d\nu + i \big(\int \mathfrak{Im}(f)^+ d\nu - \int \mathfrak{Im}(f)^- d\nu \big) \tag{$\star\star\star$}
\end{eqnarray}
  where we have used the notation $h^+ = \max(0,h)$ and $h^- = -\min(0, h)$ for a function $h$.
  Now plug in $(\star\star\star)$ into $(\star\star)$ and all you end up with is a sum of integrals of non-negative functions with respect to positive measures. So to conclude, let $f$ be arbitrary again and let $f_n^\pm \nearrow \mathfrak{Re}(f)^\pm$ and $g_n^\pm \nearrow \mathfrak{Im}(f)^\pm$ be sequences of non-negative simple functions. Now set $F_n = (f_n^+ - f_n^-) + i(g_n^+-g_n^-)$, then clearly $F_n \to f$ and since $F_n$ is simple by $(\star)$ we have $$\int F_n d(g \cdot \mu) = \int F_n g d\mu$$ But the LHS and the RHS are just sums of integrals over non-negative simple functions with respect to positive measures, so since each individual integral converges, the sum does so too and so we conclude $$\int f d(g \cdot \mu) = \int fg d\mu$$

