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

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.

Write $J(A)$ for the partition supremum. Empty cells are permitted, so $J\ge {\mu }_i$ for each $i$, and $J(A)\le {\sum }_i{\mu }_i(A)$ is finite. Also $J(\varnothing )=0$.

Let $A={⨆}_{j\ge 1}{B}_j$. Any measurable partition $A={⨆}_i{A}_i$ yieldsTaking its supremum gives one inequality. Conversely, for any $\epsilon >0$ choose for each $j$ a partition $B_j={⨆}_i{B}_{ji}$ whose sum exceeds $J(B_j)-\epsilon 2^{-j}$. The sets $A_i={⨆}_j{B}_{ji}$ partition $A$, and countable additivity of each ${\mu }_i$ gives $J(A)\ge {\sum }_{j}J(B_j)-\epsilon$. Let $\epsilon \downarrow 0$. Thus $J$ is a finite measure.

If another measure $\nu$ dominates every ${\mu }_i$, then for every such partition ${\sum }_i{\mu }_i(A_i)\le {\sum }_i\nu (A_i)=\nu (A)$. Consequently $J\le \nu$. Hence $J={\bigvee }_i{\mu }_i$ is the least dominating measure, the finite supremum of positive measures. It generally differs from the pointwise maximum of the numbers ${\mu }_i(A)$: the maximizing index may differ between pieces of $A$.

Choose measurable sets $S_i$ carrying ${\mu }_i^s$ with $\kappa (S_i)=0$, and put $S={\bigcup }_i{S}_i$. Every absolutely continuous component vanishes on $S$, and every singular component vanishes on its complement. The given restriction identity therefore givesThe first is singular with respect to $\kappa$. The second is absolutely continuous because it is bounded above by ${\sum }_i{\mu }_i^a$. They sum to $\mu$, so uniqueness identifies them as its Lebesgue components.

Let $h_i=d{\mu }_i^a/d\kappa$ and $h={\max }_i{h}_i$, using measurable representatives. Each partition sum is at most ${\int }_{A}hd\kappa$. For the reverse inequality, partition $A$ measurably by assigning a point to the smallest index attaining this finite maximum. On its assigned cell $h_i=h$, and the partition sum is exactly ${\int }_{A}hd\kappa$. ThusThe maximum of finitely many integrable nonnegative densities is integrable, since it is at most their sum.