For any measure right? Let $ A , B \subset \mathbb{R}$. Does  $m^* (A \cup B) + m^* (A \cap B) \le m^* (A) +m^* (B)$ hold  for every measure? Example, for outer measure, Lebesgue measure, etc.
 A: It is true. Let $m^*$ be an outer measure. (As pointed out by @Potato, an outer measure may not be a measure. However, (1) and (2) below can be shown to remain true for measures, too, when the domain is restricted to the corresponding $\sigma$-algebra.) By definition,
(1) $m^*(E)\leq m^*(F)$ if $E\subseteq F$; and
(2) $m^*(\bigcup_{j=1}^{\infty}E_j)\leq\sum_{j=1}^{\infty}m^*(E_j)$. (This is obviously true for finite unions and sums, too, as we can take $E_j=\varnothing$ for $j$ sufficiently large and greater, and we also know that $m^*(\varnothing)=0$ by the definition of outer measures or actual measures.)

Edited on April 30, 2016. As Vim pointed out, the inequalities (3) and (4) below are invalid in general. Parts of the text I have correspondingly retracted are struck out throughout my answer. (3) and (4) actually hold as equalities if $m^*$ is a measure and $A$ and $B$ are measurable sets, but they do not necessarily hold if $m^*$ is a generic outer measure and the sets $A$ and $B$ are not measurable. I do not have an immediate idea on how to fix this proof or construct a counterexample for general outer measures, so further input is welcome.

Now, note that
(3) $m^*(A)=m^*\left((A\setminus B)\bigcup(A\bigcap B)\right)\geq m^*(A\setminus B)+m^*(A\bigcap B)$ by (2), and similarly:
(4) $m^*(B)=m^*\left((B\setminus A)\bigcup(A\bigcap B)\right)\geq m^*(B\setminus A)+m^*(A\bigcap B)$.
But note also that $(A\setminus B)\bigcup(B\setminus A)\bigcup(A\bigcap B)=A\bigcup B$. Therefore, (2) implies that
(5) $m^*(A\setminus B)+m^*(B\setminus A)+m^*(A\bigcap B)\geq m^*\left((A\setminus B)\bigcup(B\setminus A)\bigcup(A\bigcap B)\right)=m^*(A\bigcup B)$.
Finally, combining (3)-(5) yields
$m^*(A)+m^*(B)\geq m^*(A\setminus B)+m^*(B\setminus A)+m^*(A\bigcap B)+m^*(A\bigcap B)\geq m^*(A\bigcup B)+m^*(A\bigcap B),$
which is the desired result. This is true for both measures and outer measures, since, as pointed out above, (1) and (2) hold for both measures and outer measures.
