Context: I'm trying to prove the following:
Let $\mu$ be a finite additive function with real valure defined on algebra $\mathcal{A}$. Than $\mu$ is $\sigma$-additive iff for any decreasing sequence $A_n \searrow \emptyset $, we have $\mu(A_n)\to 0$
I was wondering, is it possible for an additive set function to have both positive negative values? Also, is the statement true if $\mu$ has both positive and negative values? Currently my attempt to prove it highly rely on $\mu$ being non-negative (or equivalently non-positive).
Yes, an additive set function can attain (strictly) positive as well as (strictly) negative values. For a simple example, let, for Borel sets $A$,
$$\mu(A) = \int_A \frac{\sin x}{1+x^2}\,d\lambda$$
with $\lambda$ the Lebesgue measure on $\mathbb{R}$. Then $\mu$ is a Borel measure on $\mathbb{R}$, hence additive (even $\sigma$-additive), and
$$\mu([\pi,2\pi]) < 0 < \mu([0,\pi]).$$
But that doesn't have any bearing on the criterion for $\sigma$-additivity. For that, only the finiteness assumption is needed. In the one direction, if $A \in \mathcal{A}$ and
$$A = \bigcup_{n = 0}^{\infty} A_n$$
for a disjoint family $\{ A_n : n \in \mathbb{N}\}$ in $\mathcal{A}$, then we have
$$A = B_N \cup \bigcup_{n = 0}^{N-1} A_n$$
where
$$B_N = \bigcup_{n = N}^{\infty} A_n$$
for every $N\in\mathbb{N}$. Further, $B_N \in \mathcal{A}$ and $B_N \searrow \varnothing$.
Thus, by additivity
$$\Biggl\lvert \mu(A) - \sum_{n = 0}^{N-1} \mu(A_n)\Biggr\rvert = \lvert \mu(B_N)\rvert.$$
Then $\mu(B_N) \to 0$ implies the $\sigma$-additivity
$$\mu(A) = \sum_{n = 0}^{\infty} \mu(A_n)$$
by the definition of the sum of an infinite series. For this direction, we do not need that $\mu$ only attains finite values.
For the other direction, note that for $A_n \searrow \varnothing$ and any $N \in \mathbb{N}$ we have
$$A_0 = A_N \cup \bigcup_{n = 0}^{N-1} \bigl(A_n \setminus A_{n+1}\bigr),$$
and since $\mathcal{A}$ is an algebra, $A_n \setminus A_{n+1} \in \mathcal{A}$. Thus
$$\mu(A_N) = \mu(A_0) - \sum_{n = 0}^{N-1} \mu(A_n \setminus A_{n+1}).$$
Since by $\sigma$-additivity
$$\mu(A_0) = \mu \Biggl(\bigcup_{n = 0}^{\infty} \bigl(A_n\setminus A_{n+1}\bigr)\Biggr) = \sum_{n = 0}^{\infty} \mu\bigl(A_n\setminus A_{n+1}\bigr),$$
the result $\mu(A_N) \to 0$ again follows by definition of convergence of a series.