Lebesgue Measure and Symetric Difference I need to know if this is true because I want to use it in a proof.
Let $\mu$ be the Lebesgue measure and $\Delta$ denote the symmetric difference. Is is true that $\mu(A\Delta B)-\mu(B\Delta D)-\mu(A\Delta C) \leq \mu(C\Delta D)$? I tried to apply the definition of $\Delta$ but got no result.
 A: I assume $A,B,C,D$ are measurable. Equivalently, you want to show that
$$\mu(A\Delta B)\leq \mu(B\Delta D)+\mu(A\Delta C)+\mu(C\Delta D).$$
Rewriting this using the definition of $\Delta$ gives us
$$\mu((A\setminus B)\cup (B\setminus A))\leq \mu((B\setminus D)\cup (D\setminus B))+\mu((A\setminus C)\cup (C\setminus A))+\mu((C\setminus D)\cup (D \setminus C))$$
and so by additivity it suffices to show that $A\setminus B$ and $B\setminus A$ are contained in
$$(B\setminus D)\cup (D\setminus B)\cup (A\setminus C)\cup (C\setminus A)\cup (C\setminus D)\cup (D \setminus C).$$
which follows from the facts that
$$A\setminus B\subseteq (A\setminus C)\cup (C\setminus D)\cup (D\setminus B)$$
$$B\setminus A\subseteq (B\setminus D)\cup (D\setminus C)\cup (C\setminus A)$$
A: Let me mention another way of viewing this: Given a finite measure space $(X, \mathcal{F}, m)$, call $A, B \in \mathcal{F}$ equivalent if $m(A \Delta B) = 0$ and let $[A]$ denote the equivalence class of $A$. Then the function $d([A], [B]) = m(A \Delta B)$ defines a metric and your inequality readily follows from triangle inequality.
