# 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 $\mathcal H$?

• If there is a $Q_0$ in the $\sigma$-algebra generated by $\mathcal H$, such that $\mu^*(Q)<\infty$ and $A\subseteq Q_0$ then it is enough to check the condition against $Q_0$. In particular, if $\mu^*(X)<\infty$, it is enough to check the condition against $X$. – Ramiro Nov 23 '15 at 13:34

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 $\mu^*(B)<\infty$.