Past exam of the real and harmonic analysis course of the University of Oxford / 2008 / 1 / a

The restriction of a measure to a measurable subset is $(\mu {|}_M)(A)=\mu (A\cap M)$ for every measurable $A$. Absolute continuity of measures, written $\mu \ll \kappa$, means $\kappa (A)=0⟹\mu (A)=0$. The measures are mutually singular when a measurable $S$ satisfies $\kappa (S)=0$ and $\mu (X\setminus S)=0$. Thus singularity means concentration on a null set of the other measure; it does not assert that every measurable set is null for one of them.

Suppose $\mu =\alpha +\sigma =\beta +\tau$ are two such decompositions. Choose measurable supports $S,T$ of the singular parts with $\kappa (S)=\kappa (T)=0$, and put $N=S\cup T$. Absolute continuity makes both $\alpha$ and $\beta$ vanish on $N$. On $X\setminus N$ the two singular parts vanish, so for every measurable $A$,Subtracting from the finite measure $\mu$ gives $\sigma =\tau$. This proves uniqueness of the Lebesgue decomposition without subtracting infinite masses.

For the specified $M$, the restriction to $M$ is singular because $\kappa (M)=0$, and the complementary restriction is absolutely continuous by assumption. Their sum is $\mu$. Uniqueness therefore gives

For the finite nonnegative measures there exist a unique nonnegative integrable density $h\in L^1(\kappa )$, up to $\kappa$-almost-everywhere equality, and a unique nonnegative measure ${\mu }^s\perp \kappa$, such thatThis combines the Lebesgue decomposition theorem with the Radon-Nikodym theorem. In particular ${\mu }^a=h\kappa$ and $h=d{\mu }^a/d\kappa$; if $\mu \ll \kappa$ the singular term is zero.