Given any sigma algebra sets $X, Y$ and $Z$ and a function, $f(X,Y):=\lambda((X-Y)\cup(Y-X))$ where $\lambda$ is a lebesgue measure, show that $f(X,Z)\leq f(X,Y) + f(Y,Z)$
So I have used the property that $f(X,Z)=\lambda((X-Z)\cup(Z-X))\leq \lambda(X-Z) + \lambda(Z-X)$, and similarly $f(X,Y) + f(Y,Z)\\=\lambda((X-Y)\cup(Y-X))+\lambda((Y-Z)\cup(Z-Y))\\\leq\lambda(X-Y) + \lambda(Y-X) +\lambda(Y-Z) + \lambda(Z-Y)$.
I then rearranged the right hand side to $\lambda(X-Y)+\lambda(Y-Z)+\lambda(Z-Y)+\lambda(Y-X)$.
I want to say that $\lambda(X-Y)+\lambda(Y-Z)=\lambda(X-Z) +\lambda(Z-X)$, but I know that is just nonsense. Any help is appreciated.