Dominated convergence theorem for finitely additive measures. Let $\Omega\subset\mathbb{R}^N$ be a bounded domain and $\mu$ a finitely additive, finite signed measure, which is absolutely continuous with respect to the Lebesgue measure in $\Omega$.
Let $\phi_n\in L^\infty(\Omega)$, $\phi\in L^\infty(\Omega)$. Assume that $\phi_n(x)\to \phi(x)$ for a.e. (Lebesgue measure) $x\in \overline{\Omega}$, $\|\phi_n\|_\infty <M$. May I conclude that $$\int_\Omega \phi_nd\mu\to\int_\Omega \phi d\mu. \tag{1}$$
Once $\mu$ is only finitely additive, the Dominated Convergence Theorem does not work, however, as we have that $\mu$ is absolutely continuous with respect to the Lebesgue measure in $\Omega$, maybe we can use this fact to prove the result, or maybe there is also a counter example.
If we can prove that $\phi_n\to \phi$ in $\mu$-measure, we are done, however I could not prove it also.
This problem comes from my answer here. In that answer, I used Dominated Convergence Theorem (which is not immediately true) to prove the limit $$\int_\Omega \frac{\partial (u\gamma_n-u)}{\partial x_i}d\mu_i\to 0.$$
I think that this limit is really zero and that I can save the answer.
Remark: If necessary, we can assume that the support of $\phi_n$ is contained in the set $$\{x\in \Omega:\ \operatorname{dist}(x,\partial\Omega)<1/n\}.$$
 A: No, this is false. I will give a counterexample for the case $\Omega = (0,1)$, but this should easily generalize to more general $\Omega$.
To see this, note that Zorn's Lemma implies (see my answer here (non)equivalence of definition of non-atomic measure for finitely additive measure) the existence of a finitely additive $0-1$ valued measure $\mu$ on $(1/3,2/3] \subset \Omega$ such that $\mu$ is absolutely continuous w.r.t. Lebesgue measure $\lambda$ (in the sense that $\lambda(A) = 0$ implies $\mu(A) = 0$).
For each $n \in \Bbb{N}$, we have
$$
(\frac{1}{3}, \frac{2}{3}] = \biguplus_{k=1}^{n} (\frac{1}{3} + \frac{k-1}{n}, \frac{1}{3} + \frac{k}{n}].
$$
Hence, there is a (unique) $k_n \in \{1, \dots, n\}$ such that
$$
A_n := (\frac{1}{3} + \frac{k_n-1}{3n}, \frac{1}{3} + \frac{k_n}{3n}]
$$
satisfies $\mu(A_n) = 1$.
But it is easy to see that there is a subsequence $(A_{n_k})_k$ such that $\chi_{A_{n_k}} \to 0$ Lebesgue almost everywhere. To see this, note for example that
$$
\int_0^1 \sum_{n=1}^\infty \chi_{A_{2^n}}(x) \, dx = \sum_{n=1}^\infty \int \chi_{A_{2^n}} \, dx = \sum_{n=1}^\infty \frac{1}{3\cdot 2^n} < \infty,
$$
so that the series in the integral converges for almost every $x \in (0,1)$.
But $$\int \chi_{A_{2^n}} \, d\mu = \mu(A_{2^n}) = 1 \not \to 0 = \int 0 \, d\mu ,$$
although $\Vert \chi_{A_{2^n}} \Vert_\infty \leq 1$ for all $n \in \Bbb{N}$.
Note that because of $A_n \subset (\frac{1}{3}, \frac{2}{3}]$, the supports of $\chi_{A_n}$ even stay away from the boundary of $\Omega = (0,1)$.
Finally observe that if we mean absolute continuity in the sense that for each $\epsilon > 0$ there is some $\delta > 0$ so that $\lambda(A) < \delta$ implies $\mu(A) < \varepsilon$, then the claim is true, as can be seen by invoking Egoroff's theorem.
