Outer measure inequality

Question

I came upon this statement in a proof I read today and I can't figure out, why the following inequality (at $(*)$) holds:

Let $\nu$ be outer measure on a set $X$, let $\mathcal M:=\{A\subset X|\nu(A)<\infty \}$, let $A,B\in\mathcal M$:

For $D\subset X$ arbitrary the following holds:

$$\nu(D)=\nu(D\cap A)+\nu(D\setminus A)=\nu(D\cap A)+\nu\bigl((D\setminus A)\cap B\bigr)+\nu\bigl((D\setminus A)\setminus B\bigr)\underset{(*)}\ge \nu\bigl(D\cap(A\cup B)\bigr)+\nu\bigl(D\setminus(A\cup B)\bigr)$$

Which means $A\cup B\in \mathcal M$.

Can someone tell me, why $(*)$ holds? As I see it, the sets used before the $\ge$ are equal to the sets on the LHS.

This might not even be a difficult question, but I don't see why.

(As context: This is part of a proof to show $\mathcal M$ is a $\sigma$-Algebra)

Answer 1

It follows by these:

• $D\cap (A\cup B)\,=\,(D\cap A)\cup(D\cap B)\,=\,(D\cap A)\cup ((D\setminus A)\cap B)$,
• and hence $ \nu(D\cap (A\cup B))\ \le\ \nu(D\cap A)+\nu((D\setminus A)\cap B)$
• $D\setminus(A\cup B)=(D\setminus A)\setminus B\,$.