Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 33 / 1 / c

Write $\eta (A)={\int }_A{X}_{\infty }d\lambda$. This is a finite positive measure. For $A\in {\mathcal{F}}_n$, the conditional expectation inequality givesThe union $\mathcal{A}={\bigcup }_n{\mathcal{F}}_n$ is an algebra of sets generating $\mathcal{B}([0,1))$, since half-open dyadic intervals generate all open sets and hence all Borel sets. The class $\{A:\eta (A)\le \mu (A)\}$ is closed under increasing and decreasing limits, by continuity of these finite measures. The Monotone class theorem extends the inequality from $\mathcal{A}$ to every Borel set. Approximation by nonnegative simple functions and monotone convergence then giveThe difference $\nu =\mu -\eta$ is therefore a finite nonnegative measure, not merely a signed measure. For $A\in {\mathcal{F}}_n$,Consequently its restricted Radon-Nikodym derivative isChoose the cell-average versions of all the conditional expectations. Then $Y_n$ is constant on each level-$n$ cell and equals $2^n\nu (D_k^n)$ there, so these versions are nonnegative at every point.

Put $Z_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$. It is a conditional-expectation martingale. We justify why its limit is $X_{\infty }$, rather than just quoting convergence to an unspecified random variable.

First the family is uniformly integrable. For $A=\{Z_n>K\}\in {\mathcal{F}}_n$, nonnegativity and the conditional expectation identity giveThus its tail expectations are at most twice the integrable tail of $X_{\infty }$, uniformly in $n$. The uniformly integrable martingale convergence theorem gives a limit $Z$ almost surely and in $L^1$. For any $A\in {\mathcal{F}}_m$ and $n\ge m$,Pass to the $L^1$ limit and extend from $\mathcal{A}$ to all Borel sets by the Monotone class theorem. The equality of these integrals forces $Z=X_{\infty }$ almost surely. Combining this with the preceding convergence of $X_n$ proves

The event $A_{n,\epsilon }=\{Y_n\le \epsilon \}$ belongs to ${\mathcal{F}}_n$. Its $\nu$-measure can therefore be computed using the finite-level density:Let $H=\{{\mathrm{limsup}}_n{Y}_n=0\}$. Because the chosen versions satisfy $Y_n\ge 0$ everywhere, every point of $H$ eventually lies in all the $A_{n,\epsilon }$. For each $N$, the set ${\bigcap }_{n\ge N}{A}_{n,\epsilon }$ has $\nu$-measure at most $\epsilon$; these sets increase as $N$ increases. Continuity from below givesLetting $\epsilon \downarrow 0$ yields $\nu (H)=0$. In contrast, the preceding part says $\lambda (H)=1$. Thus $\nu$ is concentrated on the $\lambda$-null set $H^c$. This proves that $\nu$ and $\lambda$ are mutually singular measures, and completes the martingale construction of Lebesgue decomposition:The key step is that the density identity is available on each ${\mathcal{F}}_n$ even though $\nu$ need not have a density on the full Borel sigma-algebra.