Let $X$ be a set and $\mathcal{A} \subseteq P(X)$ a $\sigma$-algebra. Assume $\nu : \mathcal{A} \to [0,\infty]$ is a finitely additive measure. If $f : X \to [0,\infty]$ is a measurable function, we can define $$ \int\_{X}f\,d\nu$$ in the standard way. If $f,g :X \to [0,\infty]$ are simple measurable functions then it is easy to prove that $$\int f + g\,d\nu = \int f\,d\nu + \int g\,d\nu. $$ However, if $f$ and $g$ are just measurable functions, then it is only obvious that $$ \int f\,d\nu + \int g\,d\nu \leq \int f + g\, d\nu. $$



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> Question : Does integration with respect to a finitely additive measure respect addition?
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Note, that if $\nu$ is countably additive, then the standard way to prove that integration respects addition is to appeal to the monotone convergence theorem.



Does integrating with respect to a finitely additive measure respect addition?
As mentioned in the question, the inequality
$$
\begin{align}\int(f+g)\, d\nu\ge\int f\,d\nu+\int g\,d\nu&&{\rm(1)}\end{align}
$$
follows easily from the definition of the integral $\int f\,d\nu$ (as the supremum of the integrals of nonnegative simple functions bounded by $f$). So, I'll just show the reverse inequality which will establish additivity of the integral.


Choose any nonnegative simple function $h\le f+g$. We need to show that $\int f\,d\nu+\int g\,d\nu\ge\int h\,d\nu$. However, we can write $h=\sum\_{k=1}^nc\_k1\_{A\_k}$ for $c\_k\in(0,\infty)$ and pairwise disjoint $A\_k\in\mathcal{A}$. As long as it can be shown that $\int\_{A\_k}f\,d\nu+\int\_{A\_k}g\,d\nu\ge c\_k\nu(A\_k)$, then the required inequality will follow by summing over $k$ and applying (1). So, replacing $f,g$ by $1\_{A\_k}f,1\_{A\_k}g$ respectively (for a fixed $k$), we reduce to the case where $n=1$. Dividing through by $c\_k$ reduces to $c\_k=1$.


So, we have reduced to the situation with $f+g\ge1\_A$ and just need to show that $\int f\,d\nu+\int g\,d\nu\ge\nu(A)$. Without loss of generality (capping $f,g$ by 1 if necessary), we further reduce to the case with $0\le f,g\le1$. Then, for each positive integer $N$, consider the simple functions 
$$
\begin{align}
f\_N&=\sum\_{j=0}^{\lfloor N\rfloor}1\_{f^{-1}((j/N,(j+1)/N])}\frac jN\le f,\\\\
g\_N&=\sum\_{j=0}^{\lfloor N\rfloor}1\_{g^{-1}((j/N,(j+1)/N])}\frac jN\le g.
\end{align}
$$
We have $f\_N+g\_N\ge(1-\frac2N)1\_A$. So, using additivity for simple functions
$$
\int f\,d\nu+\int g\,d\nu\ge\int f\_N\,d\nu+\int g\_N\,d\nu\ge\left(1-\frac2N\right)\nu(A).
$$
Letting $N$ increase to infinity gives the required inequality.


