Let $(X,\mathcal{M})$ be a measurable space and let $\mu,\nu$ be measures on $(X,\mathcal{M})$ such that $\mu\geq \nu$. How to define a measure $\lambda$ such that $\mu=\nu+\lambda$?
I define $\lambda(E)=\sup\{\mu(F)-\nu(F):F\subset E,\nu(F)<\infty\}$ if $E\neq \emptyset$ and $\lambda(\emptyset)=0$. But I have failed to show it a measure. Can anyone please help?
Answer assuming that $\mu$ is sigma finite: since $\nu << \mu$ there exists $f$ such that $d\nu = fd \mu$. Note that $0 \leq f \leq 1$ a.e. $[\mu]$. Define $\lambda (E)=\int_E (1-f) d \mu$. Then $v(E)+\lambda (E)=\int_E f d \mu+ \int_E (1-f) d \mu=\mu (E)$.