I found an explanation of how to integrate with respect to a finitely-additive measure (taken from a book) :
If $M$ is an algebra (not necessarely a $\sigma$-algebra!) on $X$ and $\mu : M → [0, ∞]$ is a finitely additive (not necesseraly countabley additive!) measure on $X$, we can still define a notion of integral as follows: we say that a measurable function $u : X → [−∞, ∞]$ is integrable if there exists a sequence ${s_n }$ of measurable simple functions, each of them bounded and vanishing outside a set of finite measure (depending on $n$), such that $$\lim\limits_{n\to\infty} \mu ({x ∈ X : |u (x) − s_n (x)| > \epsilon}) = 0$$ for each $\epsilon >0$, and $$\lim\limits_{n,l\to\infty}\int_X|s_n − s_l | d\mu = 0.(*)$$ It may be shown that for every $E \in M$ the limit $\lim\limits_{n\to\infty}\int_E s_n d\mu$ exists in $\mathbb{R}$ and does not depend on the particular sequence ${s_n}$. The integral of $u$ over the measurable set E is defined by $$\int_E u d\mu := \lim\limits_{n\to\infty}\int_E s_n d\mu.$$
Now I think the fact that $\lim\limits_{n\to\infty}\int_E s_n d\mu$ exists is because by $(*)$ $\{\int_X s_nd\mu\}_{n\ge 1}$ is a Cauchy sequence and $\mathbb{R}$ is complete (is that correct?). However I am not sure how to show that the limit does not depend on the particular sequence. If I take two sequences $s_n$ and $\hat{s}_n$, then $$\int_E (s_n-\hat{s}_n)d\mu = \sum\limits_{i=1}^I c_i \mu(E_i)-\sum\limits_{j=1}^J \hat{c}_j\mu(\hat{E}_j)$$ and then I don't know how to conclude.