If $E_{1}$ and $E_{2}$ satisfy the condition of caratheodory, then the intersection also satisfies

Question

The condition of caratheodory tells us the following:

A set $E\subset { R }^{ n }$ satisfies the condition of caratheodory if for any set $S\subset { R }^{ n }$ , ${ m }_{ e }\left( S\cap E \right) +{ m }_{ e }\left( S-E \right) ={ m }_{ e }\left( S \right) $.

I must prove that ${ m }_{ e }\left( S\cap \left( { E }_{ 1 }\cap E_{ 2 } \right) \right) +{ m }_{ e }\left( S-\left( { E }_{ 1 }\cap { E }_{ 2 } \right) \right) ={ m }_{ e }\left( S \right)$

But since $E_{1}$ and $E_{2}$ satisfy the condition we have the following

${ m }_{ e }\left( S\cap { E }_{ 1 } \right) +{ m }_{ e }\left( S-{ E }_{ 1 } \right) ={ m }_{ e }\left( S \right) $

${ m }_{ e }\left( S\cap { E }_{ 2 } \right) +{ m }_{ e }\left( S-{ E }_{ 2 } \right) ={ m }_{ e }\left( S \right) $

I could use the following equivalence but I can't get to the result

$S-{ \left( { E }_{ 1 }{ \cap E }_{ 2 } \right) = }\left( (S\cap { E }_{ 1 })-{ E }_{ 2 } \right) \cup \left( \left( S-{ E }_{ 1 } \right) \cap { E }_{ 2 } \right) \cup \left( \left( S-{ E }_{ 1 } \right) -{ E }_{ 2 } \right) $

Answer 1

The sets that satisfy the Caratheodory condition form an algebra (in fact a $\sigma$-algebra but that isn't necessary here). To see this, let $\mathcal{M}$ be the collection of all such sets. Since $S \setminus E = S \cap E^c$, the Caratheodory condition is symmetric wrt $E$ and $E^c$ so $\mathcal{M}$ is closed under complements. Now take any $E,F \in \mathcal{M}$ and any $S \subset \mathbb{R}^n.$ Then

\begin{align*} m(S)&=m(S \cap E)+m(S \cap E^c)\\ &=\left(m(S \cap E \cap F)+m(S \cap E \cap F^c) \right) + \left(m(S \cap E^c \cap F) + m(S \cap E^c \cap F^c))\right) \end{align*}

where the second equality holds because $S \cap E$ and $S \cap E^c$ are subsets of $\mathbb{R}^n$ and $F \in \mathcal{M}.$ Now observe that

$$E \cup F = (E \cap F) \cup (E \cap F^c) \cup (E^c \cap F),$$

and hence

$$S \cap (E \cup F) = (S \cap E \cap F) \cup (S \cap E \cap F^c) \cup (S \cap E^c \cap F).$$

So by subadditivity

$$m(S \cap (E \cup F)) \leq m(S \cap E \cap F) + m(S \cap E \cap F^c) +m(S \cap E^c \cap F).$$

And therefore,

\begin{align*} m(S)&=m(S \cap E \cap F)+m(S \cap E \cap F^c) + m(S \cap E^c \cap F) + m(S \cap E^c \cap F^c)\\ & \geq m(S \cap (E \cup F))+m(S \cap E^c \cap F^c)\\ &= m(S \cap (E \cup F))+m(S \cap (E \cup F)^c). \end{align*}

This shows $E \cup F \in \mathcal{M}$ since $m(S) \leq m(S \cap (E \cup F))+m(S \cap (E \cup F)^c)$ by subadditivity (note that $A=(A \cap B)\cup(A \cap B^c)$ for any sets $A,B$). So $\mathcal{M}$ is closed under unions. Now by DeMorgan's law,

$$E \cap F = (E^c \cup F^c)^c \in \mathcal{M}.$$