Inequality with outer measure Let $ \mu : \mathcal{H} \to \mathbb{R}  $ be a measure on a seminring $ \mathcal{H} $ over a set $ X $ and $ \mu^{*} = \inf \{ \sum \mu(A_{i}) : A_{i} \in \mathcal{H}, \; A \subset \bigcup A_{i} \} $ the associated outer measure. 
I want to show that for every $ A,B \subset X $ the following inequality holds: $$ \mu^{*}(A \cup B) + \mu^{*}(A \cap B) \le \mu^{*}(A) + \mu^{*}(B) $$
The equality is true if $A$ or $B$ are $\mu^{*}$- measurable.
By definition of $\mu^{*}$, we know that for $ \epsilon \gt 0$ there are $A_1, A_2, ...,B_1,B_2,... \in \mathcal{H} $ with $ A \subset \bigcup_i A_i$ and $ B \subset \bigcup_i B_i $, so that:
$$ \mu^{*}(A) + \epsilon = \sum_{i=1}^{\infty}\mu(A_i) $$
$$ \mu^{*}(B) + \epsilon = \sum_{i=1}^{\infty}\mu(B_i) $$
$$ \mu^{*}(A \cup B) \le \sum_{i=1}^{\infty}\mu(A_i \cup B_i) $$
$$ \mu^{*}(A \cap B) \le \sum_{i=1}^{\infty}\mu(A_i \cap B_i) $$
that would implicate:
$$ \mu^{*}(A \cup B) + \mu^{*}(A \cap B) \le \sum_{i=1}^{\infty} ( \mu(A_i \cup B_i) + \mu(A_i \cap B_i)) = \sum_{i=1}^{\infty}(\mu(A_i) + \mu(B_i)) = \mu^{*}(A) + \mu^{*}(B) + 2 \epsilon $$
The problem with that, as you see, is that $ \mu^{*}(A \cap B) \le \sum_{i=1}^{\infty}\mu(A_i \cap B_i) $ is not always true.
Questions: Is there a way to fix this or is there a simpler solution? Does this inequality have a name?
 A: Here is an easier proof:
First, show that for every subset $A\subset X$, there is an $\mu^\ast$ measurable set $A*\ast$ with $A \subset A^\ast$ and $\mu*\ast (A) = \mu(A^\ast)$. This essentially follows because every set in $\mathcal{H}$ is measurable and by definition of the outer measure. 
This property is very useful in general, try to remember it!
Now, (why exactly?)
$$
\mu^\ast (A\cup B) +\mu^\ast(A\cap B) \leq \mu^\ast (A\ast \cup B^\ast) + \mu^\ast (A^\ast \cap B^\ast) \leq \mu^\ast (A^\ast) +\mu^\ast (B^\ast ) =\mu^\ast (A)+\mu^\ast (B). 
$$
Thus, we have circumvented the problem in your proof. 
EDIT: Let us show that equality holds if $A$ or $B$ is measurable. For concreteness, assume that $B$ is measurable. We only need to prove "$\geq$", because we have already established "$\leq$".
We have
$$
\mu^\ast (A) + \mu^\ast (B) = \mu^\ast (A\cap B) + \mu^\ast (A \setminus B) + \mu^\ast (B),
$$
because $B$ is measurable. We want to show that the RHS is $\leq \mu^\ast (A\cap B) + \mu^\ast (A \cup B)$, so that it suffices to show that $\mu^\ast (A \setminus B) + \mu^\ast (B) \leq \mu^\ast (A \cup B)$.
But using the measurability of $B$ again, we have
$$
\mu^\ast (A \cup B) = \mu^\ast ((A\cup B) \cap B) + \mu^\ast ((A \cup B) \setminus B) \geq \mu^\ast (B) + \mu^\ast (A \setminus B),
$$
where the last step simply used $(A \cup B) \cap B \supset B$ and $(A \cup B) \setminus B \supset A \setminus B$.
