Section 9.4

Martingale Theorems

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Problem 9.4.1

Fill in the details for the outline concerning the proof of (9.12).

Framework (from the section's opening discussion). Let Ω\Omega denote a nonempty set, F\mathcal{F} a σ\sigma-field of subsets of Ω\Omega, and μ\mu and η\eta arbitrary measures on F\mathcal{F}. Let {Fn}n=1∞\left\{ \mathcal{F}_{n}\right\}_{n=1}^{\infty } be a sequence of sub-σ\sigma-fields of F\mathcal{F} with ⋃n=1∞Fn⊆F\bigcup_{n=1}^{\infty } \mathcal{F}_{n} \subseteq \mathcal{F}. For each n∈Nn \in \mathbb {N}, let μn\mu_{n} and ηn\eta_{n} denote the restrictions of μ\mu and η\eta to Fn\mathcal{F}_{n}, assumed σ\sigma-finite on Fn\mathcal{F}_{n}, and let fnf_{n} denote a derivative of ηn\eta_{n} with respect to μn\mu_{n}. We write f‾=lim⁡‾nfn\underline{f}=\underline{\lim }_{n} f_{n} and fˉ=lim⁡‾nfn\bar{f}=\overline{\lim }_{n} f_{n}, that is, f‾=lim⁡ninf⁡{fn,fn+1,⋯ }\underline{f}=\lim_{n} \inf \left\{ f_{n}, f_{n+1}, \cdots \right\} and fˉ=lim⁡nsup⁡{fn,fn+1,⋯ }\bar{f}=\lim_{n} \sup \left\{ f_{n}, f_{n+1}, \cdots \right\}. In the situation treated by Claim 5, F1⊆F2⊆⋯\mathcal{F}_{1} \subseteq \mathcal{F}_{2} \subseteq \cdots; write F∞=σ(⋃n=1∞Fn)\mathcal{F}_{\infty }=\sigma \left(\bigcup_{n=1}^{\infty } \mathcal{F}_{n}\right), with μ∞\mu_{\infty } and η∞\eta_{\infty } denoting the restrictions of μ\mu and η\eta to F∞\mathcal{F}_{\infty }.

Claim 5 (restated in part; the two displayed inequalities to be established are numbered (9.11) and (9.12)). Suppose that F1⊆F2⊆⋯\mathcal{F}_{1} \subseteq \mathcal{F}_{2} \subseteq \cdots. Then f‾\underline{f} and fˉ\bar{f} are F∞/B∗\mathcal{F}_{\infty }/\mathcal{B}^{*}-measurable, and in addition we have the following: (a) f‾\underline{f} and fˉ\bar{f} are derivatives of η∞\eta_{\infty } with respect to μ∞\mu_{\infty }. (b) f‾=fˉ\underline{f}=\bar{f} μ\mu-a.e. on Ω\Omega, so that lim⁡nfn\lim_{n} f_{n} exists μ\mu-a.e. on Ω\Omega. (c) ∫Alim⁡nfn dμ=η∞,ac(A)\int_{A} \lim_{n} f_{n} \, \mathrm{d} \mu =\eta_{\infty , \mathrm{ac}}(A) for every A∈F∞A \in \mathcal{F}_{\infty }. Establishing (a) reduces (via the sets Lα,f‾={ω∈Ω:f‾(ω)⩽α}L_{\alpha , \underline{f}}=\left\{ \omega \in \Omega : \underline{f}(\omega ) \leqslant \alpha \right\} and Gα,fˉ={ω∈Ω:fˉ(ω)⩾α}G_{\alpha , \bar{f}}=\left\{ \omega \in \Omega : \bar{f}(\omega ) \geqslant \alpha \right\}, together with Claim 2) to proving

η(Lα,f‾∩A)⩽αμ(Lα,f‾∩A)for every A∈F∞ and α∈R(9.11) \eta \left(L_{\alpha , \underline{f}} \cap A\right) \leqslant \alpha \mu \left(L_{\alpha , \underline{f}} \cap A\right) \quad \text{for every } A \in \mathcal{F}_{\infty } \text{ and } \alpha \in \mathbb {R} \tag {9.11}

and

η(Gα,fˉ∩A)⩾αμ(Gα,fˉ∩A)for every A∈F∞ and α∈R.(9.12) \eta \left(G_{\alpha , \bar{f}} \cap A\right) \geqslant \alpha \mu \left(G_{\alpha , \bar{f}} \cap A\right) \quad \text{for every } A \in \mathcal{F}_{\infty } \text{ and } \alpha \in \mathbb {R}. \tag {9.12}

This should be straightforward if one understands the details concerning the proof of (9.11).

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Problem 9.4.2

Prove (1)-(3) and (1')-(3') in Claim 6.

Framework. Assume the same framework as in the preceding exercise, save that here F1⊇F2⊇⋯\mathcal{F}_{1} \supseteq \mathcal{F}_{2} \supseteq \cdots; write F0=⋂n=1∞Fn=lim⁡nFn\mathcal{F}_{0}=\bigcap_{n=1}^{\infty } \mathcal{F}_{n}=\lim_{n} \mathcal{F}_{n}, with μ0\mu_{0} and η0\eta_{0} denoting the restrictions of μ\mu and η\eta to F0\mathcal{F}_{0}.

Claim 6 (restated). Suppose that F1⊇F2⊇⋯\mathcal{F}_{1} \supseteq \mathcal{F}_{2} \supseteq \cdots. Then both f‾\underline{f} and fˉ\bar{f} are F0/B∗\mathcal{F}_{0}/\mathcal{B}^{*}-measurable, and we have the following: (a) Both f‾\underline{f} and fˉ\bar{f} are derivatives of η0\eta_{0} with respect to μ0\mu_{0}. (b) f‾=fˉ\underline{f}=\bar{f} μ\mu-a.e. on Ω\Omega, so that lim⁡nfn\lim_{n} f_{n} exists μ\mu-a.e. on Ω\Omega. (c) η0,ac(A)=∫Alim⁡nfn dμ\eta_{0, \mathrm{ac}}(A)=\int_{A} \lim_{n} f_{n} \, \mathrm{d} \mu for every A∈F0A \in \mathcal{F}_{0}.

As in the proof of Claim 5, establishing part (a) reduces to proving, for every α∈R\alpha \in \mathbb {R}, the two displays

η(Mα∩A)⩽αμ(Mα∩A)for every A∈F0(9.18) \eta \left(M_{\alpha } \cap A\right) \leqslant \alpha \mu \left(M_{\alpha } \cap A\right) \quad \text{for every } A \in \mathcal{F}_{0} \tag {9.18}

and

η(Hα∩A)⩾αμ(Hα∩A)for every A∈F0,(9.19) \eta \left(H_{\alpha } \cap A\right) \geqslant \alpha \mu \left(H_{\alpha } \cap A\right) \quad \text{for every } A \in \mathcal{F}_{0}, \tag {9.19}

where Mα={ω∈Ω:inf⁡{f1(ω),f2(ω),⋯ }<α}M_{\alpha }=\left\{ \omega \in \Omega : \inf \left\{ f_{1}(\omega ), f_{2}(\omega ), \cdots \right\} <\alpha \right\} and Hα={ω∈Ω:sup⁡{f1(ω),f2(ω),⋯ }>α}H_{\alpha }=\left\{ \omega \in \Omega : \sup \left\{ f_{1}(\omega ), f_{2}(\omega ), \cdots \right\} >\alpha \right\}.

To prove (9.18), fix α∈R\alpha \in \mathbb {R} and, for each n∈Nn \in \mathbb {N}, define

Jn={ω∈Ω:min⁡{f1(ω),⋯ ,fn(ω)}<α}, J_{n}=\left\{ \omega \in \Omega : \min \left\{ f_{1}(\omega ), \cdots , f_{n}(\omega )\right\} <\alpha \right\} , Jn,p={ω∈Ω:fp(ω)<α,fp+1(ω)⩾α,⋯ ,fn(ω)⩾α},p=1,⋯ ,n−1, J_{n, p}=\left\{ \omega \in \Omega : f_{p}(\omega )<\alpha , f_{p+1}(\omega ) \geqslant \alpha , \cdots , f_{n}(\omega ) \geqslant \alpha \right\} , \quad p=1, \cdots , n-1,

and Jn,n={ω∈Ω:fn(ω)<α}J_{n, n}=\left\{ \omega \in \Omega : f_{n}(\omega )<\alpha \right\}. The claims to be proved are:

  1. Jn,p∈FpJ_{n, p} \in \mathcal{F}_{p} for every n∈Nn \in \mathbb {N} and p∈{1,⋯ ,n}p \in \left\{ 1, \cdots , n\right\}.

  2. For each n∈Nn \in \mathbb {N} the finite sequence {Jn,p}p=1n\left\{ J_{n, p}\right\}_{p=1}^{n} is disjoint with union JnJ_{n}.

  3. J1⊆J2⊆⋯J_{1} \subseteq J_{2} \subseteq \cdots and lim⁡nJn=⋃n=1∞Jn=Mα\lim_{n} J_{n}=\bigcup_{n=1}^{\infty } J_{n}=M_{\alpha }. To prove (9.19), fix α∈R\alpha \in \mathbb {R} and, for each n∈Nn \in \mathbb {N}, define

Jn={ω∈Ω:max⁡{f1(ω),⋯ ,fn(ω)}>α}, J_{n}=\left\{ \omega \in \Omega : \max \left\{ f_{1}(\omega ), \cdots , f_{n}(\omega )\right\} >\alpha \right\} , Jn,p={ω∈Ω:fp(ω)>α,fp+1(ω)⩽α,⋯ ,fn(ω)⩽α},p=1,⋯ ,n−1, J_{n, p}=\left\{ \omega \in \Omega : f_{p}(\omega )>\alpha , f_{p+1}(\omega ) \leqslant \alpha , \cdots , f_{n}(\omega ) \leqslant \alpha \right\} , \quad p=1, \cdots , n-1,

and Jn,n={ω∈Ω:fn(ω)>α}J_{n, n}=\left\{ \omega \in \Omega : f_{n}(\omega )>\alpha \right\}. The claims to be proved are:

  1. For every n∈Nn \in \mathbb {N} and p=1,⋯ ,np=1, \cdots , n we have Jn,p∈FpJ_{n, p} \in \mathcal{F}_{p}.

  2. For each n∈Nn \in \mathbb {N} the finite sequence {Jn,p}p=1n\left\{ J_{n, p}\right\}_{p=1}^{n} is disjoint with union JnJ_{n}.

  3. J1⊆J2⊆⋯J_{1} \subseteq J_{2} \subseteq \cdots and lim⁡nJn=⋃n=1∞Jn=Hα\lim_{n} J_{n}=\bigcup_{n=1}^{\infty } J_{n}=H_{\alpha }.

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(a)
(b)
(c)
Problem 9.4.3

Assume the framework of Claim 5. If ηn≪μn\eta_{n} \ll \mu_{n} for every n∈Nn \in \mathbb {N}, and if for every EE in the field ⋃k=1∞Fk\bigcup_{k=1}^{\infty } \mathcal{F}_{k} we have lim⁡n∫Efn dμ=∫Ef∞ dμ\lim_{n} \int_{E} f_{n} \, \mathrm{d} \mu =\int_{E} f_{\infty } \, \mathrm{d} \mu, then η∞≪μ∞\eta_{\infty } \ll \mu_{\infty }. Here f∞f_{\infty } denotes a derivative of η∞\eta_{\infty } with respect to μ∞\mu_{\infty }.

Framework of Claim 5 (restated). Let Ω\Omega denote a nonempty set, F\mathcal{F} a σ\sigma-field of subsets of Ω\Omega, and μ\mu and η\eta arbitrary measures on F\mathcal{F}. Let F1⊆F2⊆⋯\mathcal{F}_{1} \subseteq \mathcal{F}_{2} \subseteq \cdots be a nondecreasing sequence of sub-σ\sigma-fields of F\mathcal{F}. For each n∈Nn \in \mathbb {N}, let μn\mu_{n} and ηn\eta_{n} denote the restrictions of μ\mu and η\eta to Fn\mathcal{F}_{n}, assumed σ\sigma-finite on Fn\mathcal{F}_{n}, and let fnf_{n} denote a derivative of ηn\eta_{n} with respect to μn\mu_{n}. Write F∞=σ(⋃n=1∞Fn)\mathcal{F}_{\infty }=\sigma \left(\bigcup_{n=1}^{\infty } \mathcal{F}_{n}\right), with μ∞\mu_{\infty } and η∞\eta_{\infty } denoting the restrictions of μ\mu and η\eta to F∞\mathcal{F}_{\infty }.

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Remark.
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If ηk≪μk\eta_{k} \ll \mu_{k} for each k∈Nk \in \mathbb {N}, we have ηk=ηk,ac\eta_{k}=\eta_{k, \mathrm{ac}} and ηk,s=0\eta_{k, \mathrm{s}}=0. Fix k∈Nk \in \mathbb {N} and E∈FkE \in \mathcal{F}_{k}; we have

η(E)=ηk(E)=ηk,ac(E)=∫Efk dμk=∫Efk dμ. \eta (E)=\eta _{k}(E)=\eta _{k, \mathrm{ac}}(E)=\int _{E} f_{k} \, \mathrm{d} \mu _{k}=\int _{E} f_{k} \, \mathrm{d} \mu .

For every n⩾kn \geqslant k, EE is also in Fn\mathcal{F}_{n}, so we may also write

η(E)=ηn(E)=ηn,ac(E)=∫Efn dμn=∫Efn dμ. \eta (E)=\eta _{n}(E)=\eta _{n, \mathrm{ac}}(E)=\int _{E} f_{n} \, \mathrm{d} \mu _{n}=\int _{E} f_{n} \, \mathrm{d} \mu .

Let n→∞n \rightarrow \infty above to obtain

η(E)=lim⁡n∫Efn dμ=∫Ef∞ dμ=∫Ef∞ dμ∞=η∞,ac(E); \eta (E)=\lim _{n} \int _{E} f_{n} \, \mathrm{d} \mu =\int _{E} f_{\infty } \, \mathrm{d} \mu =\int _{E} f_{\infty } \, \mathrm{d} \mu _{\infty }=\eta _{\infty , \mathrm{ac}}(E);

the last equality follows because f∞f_{\infty } is a derivative of η∞\eta_{\infty } with respect to μ∞\mu_{\infty }. Therefore η(E)=η∞(E)=η∞,ac(E)\eta (E)=\eta_{\infty }(E)=\eta_{\infty , \mathrm{ac}}(E) for every E∈FkE \in \mathcal{F}_{k}. Since kk is a generic positive integer, it follows that η∞=η∞,ac\eta_{\infty }=\eta_{\infty , \mathrm{ac}} on the field ⋃k=1∞Fk\bigcup_{k=1}^{\infty } \mathcal{F}_{k}.

Uniqueness Theorem (Section 3.5, restated): let μ1\mu_{1} and μ2\mu_{2} denote measures with domain σ(P)\sigma (\mathcal{P}), where P\mathcal{P} denotes a π\pi-system. If μ1\mu_{1} is σ\sigma-finite with respect to P\mathcal{P} and if μ1=μ2\mu_{1}=\mu_{2} on P\mathcal{P}, then μ2\mu_{2} is σ\sigma-finite on P\mathcal{P} and μ1=μ2\mu_{1}=\mu_{2} on σ(P)\sigma (\mathcal{P}).

Now use the Uniqueness Theorem in Section 3.5 (with ⋃k=1∞Fk\bigcup_{k=1}^{\infty } \mathcal{F}_{k} playing the role of the π\pi-system P\mathcal{P}) to obtain η∞=η∞,ac\eta_{\infty }=\eta_{\infty , \mathrm{ac}} on F∞\mathcal{F}_{\infty }.