A variant of Dominated Convergence Theorem Let $(a_1,a_2,\ldots)$ be a sequence of nonnegative numbers summing up to $1$.
Given a measurable space $(X,\mathscr{F})$, assume also the $f:X\to \mathbb{R}$ is a measurable function, and let $(\mu_1,\mu_2,\ldots)$ be a sequence of probability measures $\mathscr{F} \to \mathbb{R}$.
Then, is it true that
$$
\int_X f \mathrm{d}\left(\sum_{i\ge 1} a_i\mu_i\right)=\sum_{i\ge 1}a_i \left(\int_X f \mathrm{d}\mu_i\right)\,\,\,?
$$
 A: Since $(a_1,a_2,\ldots)$ be a sequence of nonnegative numbers summing up to $1$ and $(\mu_1,\mu_2,\ldots)$ be a sequence of probability measures $\mathscr{F} \to \mathbb{R}$, it is easy to see that $\sum_{i\ge 1} a_i\mu_i$ is in fact a (probability) measure.
Note that, for each $j$, $\mu_j \ll \sum_{i\ge 1} a_i\mu_i$. Since both measures are finite, we can apply Radon-Nikodym Theorem. So there is $h_j$ a measurable finite non-negative function such that 
$$ \int_X f \mathrm{d}\mu_j = \int_X f h_j \mathrm{d}\left(\sum_{i\ge 1} a_i\mu_i\right) \tag{1}$$
Since $\sum_{i\ge 1} a_i\mu_i \ll \sum_{i\ge 1} a_i\mu_i$, we can use Radon-Nikodym Theorem again (uniqueness a.e. of Radon-Nikodym derivative) to conclude that 
$$\sum_{i\ge 1} a_j h_j = 1 \textrm{ a.e. }$$
Then, from $(1)$, either using Dominated Convergnece Theorem (if $f$ is integrable) or Monotone Convergence Theorem (if $f$ is non-negative), we have
$$ \sum_{j\ge 1} a_j\int_X f \mathrm{d}\mu_j = \sum_{j\ge 1} a_j\int_X f h_j \mathrm{d}\left(\sum_{i\ge 1} a_i\mu_i\right) = \int_X f \left(\sum_{j\ge 1} a_j h_j\right) \mathrm{d}\left(\sum_{i\ge 1} a_i\mu_i\right) = \\=\int_X f  \mathrm{d}\left(\sum_{i\ge 1} a_i\mu_i\right)$$
A: There are many ways to prove that. Ramiro's solution is beautiful, but, in fact, Radon-Nikodym theorem is not needed. 
This alternative solution resembles (one possible) proof of the Dominated Convergence theorem and relies on absolute continuity of Lebesgue integral.
Denote $\sum_{i \ge 1} a_i \mu_i$ by $\mu$ and suppose $f$ is $\mu$-integrable. Also, in $\int_X$ letter $X$ is omitted.
Let us estimate
$$ \left| \int f\,\mathrm{d}\mu - \sum_{i=1}^N a_i \int f\,\mathrm{d}\mu_i \right| = \left| \int f\,\mathrm{d} \!\left( \sum_{i \ge N+1} a_i \mu_i \right) \right| \le \int \lvert f \rvert \,\mathrm{d} \!\left( \sum_{i \ge N+1} a_i \mu_i \right) \tag{1}$$
Take $\varepsilon > 0$. By absolute continuity of the integral, there exists $\delta > 0$, s.t. $\int_A \lvert f \rvert \,\mathrm{d}\mu < \varepsilon$, whenever $\mu(A) < \delta$.
Sets $A_M =  \{ x\in X : \lvert f(x) \rvert > M \}$ decrease to $\varnothing$ as $M\rightarrow \infty$. So we can find $M$, s.t. $\mu(A_M) < \delta$. Therefore,
\begin{align*}
\int \lvert f \rvert \,\mathrm{d} \!\left( \sum_{i \ge N+1} a_i \mu_i \right) &= \int_{A_M} \lvert f \rvert \,\mathrm{d} \!\left( \sum_{i \ge N+1} a_i \mu_i \right) + \int_{X \setminus A_M} \lvert f \rvert \,\mathrm{d} \!\left( \sum_{i \ge N+1} a_i \mu_i \right) \le \\ &\le \int_{A_M} \lvert f \rvert \,\mathrm{d} \mu + M \!\left( \sum_{i \ge N+1} a_i \mu_i(X \setminus A_M) \right) < \varepsilon + M \sum_{i \ge N+1} a_i
\tag{2}
\end{align*}
Combining $(1)$ with $(2)$, we conclude that for sufficiently large $N$
$$
\left| \sum_{i=1}^N a_i \int f\,\mathrm{d}\mu_i - \int f\,\mathrm{d}\mu \right| < 2\varepsilon,
$$
hence the series $\sum_{i \ge 1} a_i \int f\,\mathrm{d}\mu_i$ converges to $\int f\,\mathrm{d}\mu$.

If $f$ is positive but its integral is infinite you may approximate $f$ by $f^M (x) = \max\{f(x), M\}$ and exchange summation with the limit as $M \rightarrow \infty$, using uniformness of the latter, to show that the series
$\sum_{i \ge 1} a_i \int f\,\mathrm{d}\mu_i$ diverges to infinity. Though I would call this «a variant of Monotone Convergence theorem».
Alternatively, you may begin with simple functions, as was suggested in other answers.

Note, that your claim is formally not true for all measurable functions. If the series $\sum_{i \ge 1} a_i \int f\,\mathrm{d}\mu_i$ converges, even absolutely, $\mu$-integrability does not follow: you may have $\int f \,\mathrm{d} \mu_i = 0$, but $a_i \int \lvert f \rvert \,\mathrm{d} \mu_i$ summing up to infinity. For instance, this is possible when $\mu_i$ are supported on disjoint sets, you may think of a counterexample on $\mathbb R$.
A: Here's another approach. One minor but important detail to verify is that $\mu$ is indeed a measure. As will be shown below, once this is verified, the result immediately follows. 
We begin by simplifying notation. Suppose that $\nu_1,\nu_2,\dots$ are measures on $(X,{\cal F})$. Define 
$$ \nu: {\cal F}\to [0,\infty)$$ 
by letting 
$$\nu(A) = \sum_{i} \nu_i(A).$$ 
We will show that $\nu$ is a measure on $(X,{\cal F})$. 
Indeed, $\nu(\emptyset)=0$. 
Now suppose that $A_1,A_2,\dots$ are disjoint elements of ${\cal F}$. 
Then 
$$ \nu(\cup A_j) = \sum_i \nu_i(\cup A_j) =\sum_i \lim_{n\to\infty} \nu_i (\cup_{j\le n}  A_j)=(*)$$ 
Therefore by the monotone convergence theorem, 
$$ (*) = \lim_{n\to\infty} \sum_i \nu_i (\cup_{j \le n} A_j)= \lim_{n\to\infty} \sum_{j \le n} \sum_i \nu_i(A_j)= \lim_{n\to\infty} \sum_{j\le n} \nu(A_j)=\sum_{j} \nu(A_j).$$
Now we can integrate with respect to $\nu$: 
$$ \int_A d \nu = \nu (A) = \sum_{i} \nu_i(A) = \sum_i \int_A d \nu_i .$$ 
This clearly extends to any simple function: 
$$ \int \phi d\nu  = \sum_i \int \phi d\nu_i,$$ 
Now for a nonnegative measurable function $f$, we have a sequence of nonnegative simple functions $\phi_n$ such that $\phi_n \nearrow f$. It the follows from monotone convergence that 
$$ \int f d \nu = \lim_{n\to\infty} \int \phi_n d \nu = \lim_{n\to\infty} \sum_i \int \phi_n d \nu_i=(**)$$ 
Now apply monotone convergence to the righthand side to obtain 
$$(**) = \sum_i \lim_{n\to\infty} \int \phi_n d \nu_i =\sum_i \int f d\nu_i.$$
A: I like Ramiro's elegant answer, but this one is more basic. First, define $\mu:=\sum_ia_i\mu_i$. By definition, we have that $\int f\mathrm d\mu=\sum_ia_i\int f\mathrm d\mu_i$ when $f=\mathbf1_A$, and so this extends by linearity to simple functions. If $(f_n)$ is an increasing sequence of simple functions with $0\le f_n\uparrow f$ then $0\le a_i\int f_n\mathrm d\mu_i\uparrow a_i\int f\mathrm d\mu_i$ for each $i$ by the monotone convergence theorem, so by two more applications of the monotone convergence theorem
$$
\int f\ \mathrm d\mu=\lim_{n\to\infty}\int f_n\ \mathrm d\mu=\lim_{n\to\infty}\sum_{i\ge1}a_i\int f_n\ \mathrm d\mu_i=\sum_{i\ge1}a_i\int f\ \mathrm d\mu_i.
$$
For general $f\in L^1(\mu)$, split into positive and negative parts and we are done.
