Sets of finite measure satisfy $|\mu (A) −\mu (B)| \le \mu(A \triangle B)$ 
Let $(Ω,A,μ)$ be a measure space. If $A,B ∈ A$ are arbitrary sets satisfying $μ(A) < ∞$ or $μ(B) < ∞$, then $|μ(A) − μ(B)| ≤ μ(A △ B)$.

It is relatively easy to show if $A⊂B$. But since I can't assume that I am stuck. Please help :)
 A: The answer by Prahlad Vaidyanathan in the comments is completely sufficient and elementary. Here is another approach, based on the observation $|1_A-1_B|=1_{A\Delta B}$. It follows that
$$|\mu(A)-\mu(B)|=\left|\int 1_A d\mu-\int 1_B\, d\mu\right| = \left|\int (1_A-1_B) d\mu\right|\leq\int 1_{A\Delta B}d\mu=\mu(A\Delta B),$$
where the integral makes sense because either $\mu(A)$ or $\mu(B)$ is finite. The inequality is the triangle inequality.
A: Since $A\subseteq B\cup \left( A\triangle B \right) $
and $B\subseteq A\cup \left( A\triangle B \right)$ and then by sub-aditive we have $${ m }^{ \ast  }\left( A \right) \le { m }^{ \ast  }\left( B \right) { +m }^{ \ast  }\left( A\triangle B \right) $$
$${ m }^{ \ast  }\left( B \right) \le { m }^{ \ast  }\left( A \right) { +m }^{ \ast  }\left( A\triangle B \right) $$ by first inequality
$${ m }^{ \ast  }\left( A \right) -{ m }^{ \ast  }\left( B \right) \le { m }^{ \ast  }\left( A\triangle B \right)  $$ and by second inequality we have $$-{ m }^{ \ast  }\left( A\triangle B \right) \le { m }^{ \ast  }\left( A \right) -{ m }^{ \ast  }\left( B \right) $$
