Past exam of the real and harmonic analysis course of the University of Oxford / 2007 / 1 / b

For a signed measure define its total variation measure byThe value at the empty set is zero. For disjoint measurable sets $A_j$ with union $A$, any finite partition $E_r$ of $A$ givesTaking the supremum proves one inequality. In the reverse direction choose, for the first $k$ sets, finite partitions approaching their variation values. Combining these partitions and the remaining set $A\setminus {\bigcup }_{j\le k}{A}_j$ givesIf one term is infinite, its partition sums can instead be made arbitrarily large. Letting $\epsilon \downarrow 0$ and $k\to \infty$ proves countable additivity, including infinite values. Thus $|\mu |$ is a positive measure. This uses only countable additivity of $\mu$, without assuming a Hahn decomposition.

From now on $|\mu |(X)<\infty$, so all signed quantities below are finite. Since $|\mu |(E)\ge |\mu (E)|$, the set functionis nonnegative. Countable additivity follows from that of the two terms, so ${\mu }^{+}$ is a finite positive measure.

Set $s=\sup \{|\mu |(M):{\mu }^{+}(M)=0\}$. This is finite. Choose measurable $M_j$ with ${\mu }^{+}(M_j)=0$ and $|\mu |(M_j)\to s$. Thenis ${\mu }^{+}$-null by countable subadditivity, and monotonicity gives $|\mu |(M_{-})\ge s$. The definition of $s$ gives the reverse inequality. Hence $M_{-}$ attains the maximum, rather than merely approximating it.

Start with $R_0=A$. Having chosen disjoint $A_1,\dots ,A_i$, put $R_i=A\setminus {\bigcup }_{j\le i}{A}_j$ andThe empty set shows $s_i\ge 0$, and finite total variation bounds it above. If $s_i>0$, choose $A_{i+1}\subseteq R_i$ with $\mu (A_{i+1})\ge s_i/2>0$; if $s_i=0$, choose $A_{i+1}=\varnothing$. This is the required recursion, including the initial step $i=0$.

Disjointness and nonnegativity give ${\sum }_i\mu (A_i)\le |\mu |(A)<\infty$, hence $\mu (A_i)\to 0$. Write $R=A\setminus {\bigcup }_i{A}_i$. For every measurable $B\subseteq R$,Thus every measurable subset of $R$ has nonpositive measure. Each finite partition of $R$ consequently has variation sum exactly $-\mu (R)$, and therefore $|\mu |(R)=-\mu (R)$. It follows that ${\mu }^{+}(R)=0$. Finally countable additivity gives

If $S\subseteq M_{-}$, then ${\mu }^{+}(S)=0$, so $\mu (S)=-|\mu |(S)\le 0$. Suppose instead that $S\cap M_{-}=\varnothing$ but $\mu (S)<0$. The preceding construction gives $R\subseteq S$ with ${\mu }^{+}(R)=0$ and $|\mu |(R)=-\mu (R)>0$. Then $M_{-}\cup R$ is ${\mu }^{+}$-null with variation strictly larger than that of $M_{-}$, a contradiction. Hence $\mu (S)\ge 0$ on the complement of $M_{-}$. This is a Hahn decomposition.

On $M_{+}=X\setminus M_{-}$ every partition has nonnegative signed measures, so $|\mu |(E)=\mu (E)={\mu }^{+}(E)$ for $E\subseteq M_{+}$. On $M_{-}$, ${\mu }^{+}$ vanishes. Splitting an arbitrary measurable set between these two sets givesIn particular ${\mu }^{+}\ll |\mu |$, and the exhibited Radon-Nikodym derivative verifies the Radon-Nikodym theorem for this pair directly.