Proof of Dyadic Sigma-Algebras on a Compact Interval and Conditional Expectation as Dyadic Averaging

Throughout fix $m\ge1$ and write $s_{p}=Tp\,2^{-m}$ for $p=0,1,\dots,2^{m}$, so that $0=s_{0}<s_{1}<\dots<s_{2^{m}}=T$, $I_{m,p}=[s_{p-1},s_{p})$ for $p<2^{m}$, and $I_{m,2^{m}}=[s_{2^{m}-1},T]$. Write $\lambda=\lambda_{[0,T]}$ and $\mathcal{B}=\mathcal{B}_{[0,T]}$.

Claim 1. Partition. Let $t\in[0,T)$. Then $0\le T^{-1}t<1$, so there is exactly one $p\in\{1,\dots,2^{m}\}$ with $s_{p-1}\le t<s_{p}$, namely the least element $p$ of $\{1,\dots,2^{m}\}$ with $t<s_{p}$, a set that is nonempty because $t<T=s_{2^{m}}$; indeed $s_{p-1}\le t$ holds for that $p$, when $p=1$ because $s_{0}=0\le t$ and when $p>1$ by minimality, and the strict increase of $s_{0}<s_{1}<\dots<s_{2^{m}}$ makes $p$ unique. Hence $t$ lies in $I_{m,p}$ for exactly one $p$ (noting that for $p=2^{m}$ the condition $s_{2^{m}-1}\le t<s_{2^{m}}$ does place $t$ in $I_{m,2^{m}}=[s_{2^{m}-1},T]$). The remaining point $t=T$ lies in $I_{m,2^{m}}$ and, since $s_{p}<T$ for $p<2^{m}$, in no other atom. So the atoms are pairwise disjoint with union $[0,T]$.

Measurability and measure. Each atom is the intersection with $[0,T]$ of an interval of $\mathbb{R}$, and intervals are Borel sets; so each atom lies in $\mathcal{B}$ by the description $\mathcal{B}=\{S\cap[0,T]:S\text{ Borel}\}$ in claim 1 of the restricted Lebesgue toolkit. Since Lebesgue measure assigns to an interval its length, $\lambda(I_{m,p})=s_{p}-s_{p-1}=T2^{-m}$, whence $P_{T}(I_{m,p})=T^{-1}\cdot T2^{-m}=2^{-m}$.

$\mathcal{G}_{m}$ is a $\sigma$-algebra. By definition $\mathcal{G}_{m}=\bigl\{\bigcup_{p\in S}I_{m,p}:S\subseteq\{1,\dots,2^{m}\}\bigr\}$. It contains $\emptyset$ (take $S=\emptyset$) and $[0,T]$ (take $S$ to be everything). Because the atoms partition $[0,T]$, the complement of $\bigcup_{p\in S}I_{m,p}$ in $[0,T]$ is $\bigcup_{p\notin S}I_{m,p}$, again a member. A union of countably many members corresponds to the union of the associated index sets, hence is a member. So $\mathcal{G}_{m}$ is a $\sigma$-algebra on $[0,T]$, and it is contained in $\mathcal{B}$ because each atom is and $\mathcal{B}$ is closed under finite unions.

Refinement. Let $1\le p<2^{m}$. Then $I_{m+1,2p-1}=[T(2p-2)2^{-(m+1)},T(2p-1)2^{-(m+1)})$ and $I_{m+1,2p}=[T(2p-1)2^{-(m+1)},T(2p)2^{-(m+1)})$, both half-open because $2p<2^{m+1}$; their union is $[T(2p-2)2^{-(m+1)},T(2p)2^{-(m+1)})=[s_{p-1},s_{p})=I_{m,p}$. For $p=2^{m}$, $I_{m+1,2^{m+1}-1}=[T(2^{m+1}-2)2^{-(m+1)},T(2^{m+1}-1)2^{-(m+1)})$ and $I_{m+1,2^{m+1}}=[T(2^{m+1}-1)2^{-(m+1)},T]$, whose union is $[T(2^{m}-1)2^{-m},T]=I_{m,2^{m}}$. Hence every level-$m$ atom is a union of level-$(m+1)$ atoms, so every union of level-$m$ atoms is one of level-$(m+1)$ atoms, giving $\mathcal{G}_{m}\subseteq\mathcal{G}_{m+1}$. Iterating, for $m'\ge m$ every level-$m$ atom is a union of level-$m'$ atoms; since the level-$m'$ atoms are pairwise disjoint with union $[0,T]$ and every level-$m'$ atom is nonempty, each level-$m'$ atom meets, hence is contained in, exactly one level-$m$ atom.

Claim 2. Write $\mathcal{G}_{\infty}=\sigma\bigl(\bigcup_{m\ge1}\mathcal{G}_{m}\bigr)$.

The inclusion $\mathcal{G}_{\infty}\subseteq\mathcal{B}$. By claim 1 each $\mathcal{G}_{m}$ is contained in $\mathcal{B}$, so $\mathcal{B}$ is a $\sigma$-algebra on $[0,T]$ containing $\bigcup_{m}\mathcal{G}_{m}$; by the minimality in the definition of the generated $\sigma$-algebra, $\mathcal{G}_{\infty}\subseteq\mathcal{B}$.

The inclusion $\mathcal{B}\subseteq\mathcal{G}_{\infty}$. Put $\mathcal{H}=\{S\subseteq\mathbb{R}:S\cap[0,T]\in\mathcal{G}_{\infty}\}$. This is a $\sigma$-algebra on $\mathbb{R}$: $\mathbb{R}\cap[0,T]=[0,T]\in\mathcal{G}_{1}\subseteq\mathcal{G}_{\infty}$; if $S\in\mathcal{H}$ then $(\mathbb{R}\setminus S)\cap[0,T]=[0,T]\setminus(S\cap[0,T])\in\mathcal{G}_{\infty}$; and if $S_{1},S_{2},\dots\in\mathcal{H}$ then $\bigl(\bigcup_{n}S_{n}\bigr)\cap[0,T]=\bigcup_{n}\bigl(S_{n}\cap[0,T]\bigr)\in\mathcal{G}_{\infty}$.

Let $U\subseteq\mathbb{R}$ be open for the metric $(s,t)\mapsto|s-t|$. For $m\ge1$ let $V_{m}$ be the union of those level-$m$ atoms that are contained in $U$; then $V_{m}\in\mathcal{G}_{m}\subseteq\mathcal{G}_{\infty}$. We show $U\cap[0,T]=\bigcup_{m\ge1}V_{m}$. Each $V_{m}$ is contained in $U$ and in $[0,T]$, which gives one inclusion. For the other, let $t\in U\cap[0,T]$ and choose a real $\varepsilon>0$ with $\{s\in\mathbb{R}:|s-t|<\varepsilon\}\subseteq U$. Since $2^{m}\ge m$ for every natural $m\ge1$ and the real numbers are Archimedean, there is $m\ge1$ with $T2^{-m}<\varepsilon$. By claim 1 there is a (unique) $p$ with $t\in I_{m,p}$, and every $s\in I_{m,p}$ satisfies $|s-t|\le T2^{-m}<\varepsilon$, because $I_{m,p}$ is contained in an interval of length $T2^{-m}$ containing $t$. Hence $I_{m,p}\subseteq U$, so $I_{m,p}$ is one of the atoms forming $V_{m}$ and $t\in V_{m}$. Therefore $U\cap[0,T]\in\mathcal{G}_{\infty}$, that is $U\in\mathcal{H}$.

So $\mathcal{H}$ is a $\sigma$-algebra on $\mathbb{R}$ containing every open set. Since the Borel $\sigma$-algebra is the $\sigma$-algebra generated by the open sets, minimality gives $\mathcal{B}(\mathbb{R})\subseteq\mathcal{H}$; that is, $S\cap[0,T]\in\mathcal{G}_{\infty}$ for every Borel $S$. As $\mathcal{B}=\{S\cap[0,T]:S\text{ Borel}\}$, we conclude $\mathcal{B}\subseteq\mathcal{G}_{\infty}$, and with the previous inclusion, $\mathcal{G}_{\infty}=\mathcal{B}$.

Claim 3. Each $\mathbf{1}_{I_{m,p}}$ is a bounded $\mathcal{B}$-measurable function, hence a square-integrable random variable on $([0,T],\mathcal{B},P_{T})$; by the definition of square-integrability, the product $X\mathbf{1}_{I_{m,p}}$ of two square-integrable random variables is integrable, so $a_{p}:=2^{m}\,\mathbb{E}[X\mathbf{1}_{I_{m,p}}]$ is a well-defined real number and $A_{m}X=\sum_{p=1}^{2^{m}}a_{p}\mathbf{1}_{I_{m,p}}$.

(i) $\mathcal{G}_{m}$-measurability. $A_{m}X$ takes the value $a_{p}$ on $I_{m,p}$ and, the atoms partitioning $[0,T]$, no other values. For a Borel set $B$, $(A_{m}X)^{-1}(B)=\bigcup\{I_{m,p}:a_{p}\in B\}\in\mathcal{G}_{m}$.

(ii) Square-integrability. With $C=\max_{p}|a_{p}|$, a maximum over a finite set, $|A_{m}X|\le C$ everywhere, so $(A_{m}X)^{2}\le C^{2}$ and $\mathbb{E}[(A_{m}X)^{2}]\le C^{2}<\infty$ by monotonicity of the integral and $P_{T}([0,T])=1$. Thus $A_{m}X$ is bounded and square-integrable.

(iii) The defining identity. Let $A\in\mathcal{G}_{m}$, say $A=\bigcup_{p\in S}I_{m,p}$ with $S\subseteq\{1,\dots,2^{m}\}$. By disjointness of the atoms, $\mathbf{1}_{A}=\sum_{p\in S}\mathbf{1}_{I_{m,p}}$ pointwise, and likewise $A_{m}X\,\mathbf{1}_{A}=\sum_{p\in S}a_{p}\mathbf{1}_{I_{m,p}}$. By linearity of the integral and $\mathbb{E}[\mathbf{1}_{I_{m,p}}]=P_{T}(I_{m,p})=2^{-m}$ from claim 1,

$$\mathbb{E}\bigl[A_{m}X\,\mathbf{1}_{A}\bigr]=\sum_{p\in S}a_{p}2^{-m}=\sum_{p\in S}\mathbb{E}\bigl[X\mathbf{1}_{I_{m,p}}\bigr]=\mathbb{E}\Bigl[X\sum_{p\in S}\mathbf{1}_{I_{m,p}}\Bigr]=\mathbb{E}\bigl[X\mathbf{1}_{A}\bigr].$$

Conditions (i), (ii) and (iii) of the definition of conditional expectation are therefore satisfied, so $A_{m}X$ is a conditional expectation of $X$ given $\mathcal{G}_{m}$.