Checking Caratheodory-measurability condition on sets of the semiring Let $\mathcal H$ be a semiring over the set $X$ and $\mu$ a pre-measure defined on $\mathcal H$. Then we associate an outer measure $\mu^\ast$ to $\mu$ (describe here: https://en.wikipedia.org/w/index.php?title=Outer_measure (method I))
A set $A$ is called Caratheodory-measurable by $\mu^\ast$ if
$\mu^\ast(Q) = \mu^\ast(Q \cap A) + \mu^\ast(Q \cap A^c)$ for all sets $Q \subset X$.
Now, would it be sufficient to check this condition against all sets $Q$ in the sigma-algebra generated by $\mathcal H$ or even only in the sets in the semiring $\mathcal H$?
 A: In fact, in the general case of an outer measure $\mu^*$ induced by a premeasure defined on a semi-ring $\mathcal H$, it is enough to check the condition against all sets $Q$ in the $\sigma$-algebra generated by $\mathcal H$, such that $\mu^*(Q)<\infty$. 
Proof: The condition for $A$ to be Caratheodory-measurable by $\mu^\ast$ is: 
$$\mu^\ast(Q) = \mu^\ast(Q \cap A) + \mu^\ast(Q \cap A^c)$$ 
for all sets $Q \subseteq X$. 
Note that this condition is trivially satisfied in $\mu^*(Q)=\infty$. In fact, since $\mu^*$ is sub-additive, we know that, for any $A\subseteq X$, for all sets $Q \subseteq X$, 
$$\mu^\ast(Q) \leqslant \mu^\ast(Q \cap A) + \mu^\ast(Q \cap A^c)$$
and, if $\mu^*(Q)=\infty$, we trivially have 
$$\mu^\ast(Q) \geqslant \mu^\ast(Q \cap A) + \mu^\ast(Q \cap A^c)$$ 
So, the condition for $A$ to be Caratheodory-measurable by $\mu^\ast$ can be equivalently re-stated as:
$$\mu^\ast(Q) = \mu^\ast(Q \cap A) + \mu^\ast(Q \cap A^c)$$ 
for all sets $Q \subseteq X$ such that $\mu^*(Q)<\infty$. 
Now, it is a known result in Measure Theory that, for any $Q \subseteq X$ such that $\mu^*(Q)<\infty$, there a set $B$ in the $\sigma$-algebra generated by $\mathcal H$, such that $Q\subseteq B$ and $\mu^*(B)=\mu^*(Q)$. So we have, for any $A\subseteq X$, 
$$ \mu^*(B)=\mu^*(Q) \leqslant \mu^\ast(Q \cap A) + \mu^\ast(Q \cap A^c) \leqslant \mu^\ast(B \cap A) + \mu^\ast(B \cap A^c) $$
where we used that $\mu^*$ is sub-additive and monotone (that is, if $C\subseteq D$ then $\mu^*(C)\leqslant\mu^*(D)$). 
It is then clear that, given any set $A\subseteq X$, $A$ satisfies the condition 

$$\mu^\ast(Q) = \mu^\ast(Q \cap A) + \mu^\ast(Q \cap A^c)$$ 
  for all sets $Q \subseteq X$ such that $\mu^*(Q)<\infty$, 

if and only if $A$ satisfies the condition 

$$\mu^\ast(B) = \mu^\ast(B \cap A) + \mu^\ast(B \cap A^c)$$ 
  for all $B$ in the $\sigma$-algebra generated by $\mathcal H$, such that $\mu^*(B)<\infty$. 

