Proving Caratheodory measurability if and only if the measure of a set summed with the measure of its complement is the measure of the whole space.

Suppose we have a premeasure $\mu$ on a space $X$ such that $\mu(X) < \infty$. Prove that $E \subset X$ is Caratheodory measurable iff $ \mu^*(E)+ \mu^*(E^C) = \mu(X)$.

Going in the forward direction is fairly easy. Assuming that E is Caratheodory measurable, we can just substitute X into $\mu^*(A) = \mu^*(A \cap E) + \mu^*(A \cap E^C) $, and then we note that the outer measure and the premeasure of X themselves would have the same value.

The other direction is more difficult however. My primary idea of how to solve this part is to show that $\mu^*(A)$ and $\mu^*(A \cap E) + \mu^*(A \cap E^C) $ are both greater than and less than each other to show equality. However, I am not completely sure how to proceed with this. Can anybody provide any pointers as to how to prove the equality between these two expressions?

Solution 1:

Step 0: The inequality $\mu^{\ast}(A) \leqslant \mu^{\ast}(A\cap E) + \mu^{\ast}(A \cap E^C)$ for all $A \subset X$ follows directly from the subadditivity of outer measures.

Step 1: For a Carathéodory-measurable $B \subset X$ we have

$$\mu^{\ast}(E) = \mu^{\ast}(E \cap B) + \mu^{\ast}(E \cap B^C)\quad\text{and}\quad \mu^{\ast}(E^C) = \mu^{\ast}(E^C \cap B) + \mu^{\ast}(E^C \cap B^C),$$

whence

\begin{align} \mu(X) &= \mu^{\ast}(E) + \mu^{\ast}(E^C)\\ &= \mu^{\ast}(E\cap B) + \mu^{\ast}(E\cap B^C) + \mu^{\ast}(E^C\cap B) + \mu^{\ast}(E^C \cap B^C) \\ &= \bigl( \mu^{\ast}(E \cap B) + \mu^{\ast}(E^C \cap B)\bigr) + \bigl(\mu^{\ast}(E \cap B^C) + \mu^{\ast}(E^C \cap B^C)\bigr) \\ &\geqslant \mu^{\ast}(B) + \mu^{\ast}(B^C) \\ &= \mu(X), \end{align}

and so we must have

$$\mu^{\ast}(B) = \mu^{\ast}(B\cap E) + \mu^{\ast}(B \cap E^C)$$

for every Carathéodory-measurable $B$.

Step 2: Let $A\subset X$ arbitrary. For each $n \in \mathbb{N}$, there is a Carathéodory-measurable $B_n \supset A$ with $\mu^{\ast}(B_n) < \mu^{\ast}(A) + 2^{-n}$. Let

$$B = \bigcap_{n = 0}^\infty B_n.$$

Then $B \supset A$, and $B$ is Carathéodory-measurable with $\mu^{\ast}(B) = \mu^{\ast}(A)$, therefore

\begin{align} \mu^{\ast}(A) &= \mu^{\ast}(B) \\ &= \mu^{\ast}(B \cap E) + \mu^{\ast}(B \cap E^C) \tag{Step 1}\\ &\geqslant \mu^{\ast}(A\cap E) + \mu^{\ast}(A \cap E^C). \tag{$A \subset B$} \end{align}

Thus $E$ is Carathéodory-measurable.