Martingale Inequalities and Convergence

Put $Y_n=a+(X_n-a{)}^{+}$. The function $x\mapsto a+(x-a{)}^{+}$ is convex and $1$-Lipschitz up to an additive constant, so $Y$ is integrable and is a submartingale by F2. Moreover $X_s\le a$ iff $Y_s=a$, and $X_t\ge b$ iff $Y_t\ge b$; therefore $U_N[a,b](Y)=U_N[a,b](X)$.

Define $H_j\in \{0,1\}$ before the increment $Y_j-Y_{j-1}$: it switches from $0$ to $1$ after the first observation at level $a$, remains $1$ until an observation at least $b$, then repeats. This rule depends only on $Y_0,\dots ,Y_{j-1}$, so $H$ is predictable. Let $K_j=1-H_j$, also nonnegative and predictable.

Each completed holding interval contributes at least $b-a$ to $(H\cdot Y{)}_N={\sum }_{j=1}^N{H}_j(Y_j-Y_{j-1})$. Any final incomplete holding starts at $Y_s=a$ and contributes $Y_N-a\ge 0$. Hence, pathwise, $(b-a)U_N[a,b](Y)\le (H\cdot Y{)}_N\mathrm{.}$ No maximizing tuple from F1 was selected.

Since $H+K=1$, $Y_N-Y_0=(H\cdot Y{)}_N+(K\cdot Y{)}_N\mathrm{.}$ F3 gives $\mathbb{E}(K\cdot Y{)}_N\ge 0$, so step 2.1 yields $(b-a)\mathbb{E}{U}_N[a,b](X)\le \mathbb{E}(Y_N-Y_0)=\mathbb{E}(X_N-a{)}^{+}-\mathbb{E}(X_0-a{)}^{+}\mathrm{.}$ The last subtracted term is nonnegative, proving the second inequality. AC is used exactly as stated in F4.

Fix rationals $a<b$. By F1, $(b-a)\mathbb{E}{U}_N[a,b]\le \mathbb{E}(X_N-a{)}^{+}\le C+\mathrm{\mid }a\mathrm{\mid }\mathrm{.}$ As $U_N\uparrow U_{\mathrm{\infty }}$, F2 gives $\mathbb{E}{U}_{\mathrm{\infty }}[a,b]<\mathrm{\infty }$. Hence $U_{\mathrm{\infty }}[a,b]<\mathrm{\infty }$ almost surely.

Intersect these probability-one events over the countable set $\{(a,b)\in {\mathbb{Q}}^2:a<b\}$. F4 makes the intersection have probability one. On it, if $\mathrm{lim\hspace{0.17em}inf}{X}_n<\mathrm{lim\hspace{0.17em}sup}{X}_n$, density supplies rationals $a<b$ strictly between them; the path then completes infinitely many upcrossings of $[a,b]$, contradicting step 1.1. Thus $X_n$ has an extended-real limit.

The submartingale property gives $\mathbb{E}{X}_n\ge \mathbb{E}{X}_0$. Therefore $\mathbb{E}{X}_n^{-}=\mathbb{E}{X}_n^{+}-\mathbb{E}{X}_n\le C-\mathbb{E}{X}_0\mathrm{.}$ Fatou applied separately to $X_n^{+}$ and $X_n^{-}$ shows that neither $+\mathrm{\infty }$ nor $-\mathrm{\infty }$ can occur on a positive-measure set and that $\mathbb{E}\mathrm{\mid }{X}_{\mathrm{\infty }}\mathrm{\mid }\le C+(C-\mathbb{E}{X}_0)<\mathrm{\infty }\mathrm{.}$ Taking the limit on the full-measure event and defining it arbitrarily on its null complement yields the claimed finite integrable random variable. AC has only the inherited use in F5.

Define the disjoint first-crossing events $A_k=\{X_0<\lambda ,\dots ,X_{k-1}<\lambda ,X_k\ge \lambda \},0\le k\le N,$ with the preceding string empty for $k=0$. Each $A_k\in {\mathcal{F}}_k$, and their union is $A=\{{\max }_{j\le N}{X}_j\ge \lambda \}$.

On $A_k$, $X_k\ge \lambda$. By F2 and the conditional-expectation identity, $\lambda \mathbb{P}(A_k)\le \mathbb{E}[X_k{1}_{A_k}]\le \mathbb{E}[X_N{1}_{A_k}]\mathrm{.}$

Sum over the finite disjoint partition to get $\lambda \mathbb{P}(A)\le \mathbb{E}[X_N{1}_A]\mathrm{.}$ Since $X_N\ge 0$, the latter is at most $\mathbb{E}{X}_N$. This finite-time proof does not presuppose stopping-time or optional-sampling results. AC has exactly the inherited use in F4.

Put $X_N^{\ast }={\max }_{k\le N}{X}_k$ and fix $r>0$. Layer cake and F1 give $\begin{array}{rl}{\displaystyle \mathbb{E}(X_N^{\ast }\wedge r{)}^p}& {\displaystyle ={\int }_0^{r}p{\lambda }^{p-1}\mathbb{P}(X_N^{\ast }\ge \lambda )d\lambda }\\ {\displaystyle }& {\displaystyle \le {\int }_0^{r}p{\lambda }^{p-2}\mathbb{E}[X_N{1}_{\{X_N^{\ast }\ge \lambda \}}]d\lambda }\\ {\displaystyle }& {\displaystyle =q\mathbb{E}\text{\hspace{0.17em}⁣}[X_N(X_N^{\ast }\wedge r{)}^{p-1}]\mathrm{.}}\end{array}$ The last identity follows by integrating $p{\lambda }^{p-2}$ up to $X_N^{\ast }\wedge r$.

Hölder bounds the last expression by $q\mathrm{\parallel }{X}_N{\mathrm{\parallel }}_p\mathrm{\parallel }{X}_N^{\ast }\wedge r{\mathrm{\parallel }}_p^{p-1}\mathrm{.}$ If the truncated norm is nonzero, divide by its $(p-1)$st power; if it is zero, the desired inequality is immediate. Thus $\mathrm{\parallel }{X}_N^{\ast }\wedge r{\mathrm{\parallel }}_p\le q\mathrm{\parallel }{X}_N{\mathrm{\parallel }}_p$.

Let $r\uparrow \mathrm{\infty }$. F4 yields $\mathrm{\parallel }{X}_N^{\ast }{\mathrm{\parallel }}_p\le q\mathrm{\parallel }{X}_N{\mathrm{\parallel }}_p$, including the case where a priori the maximal moment might be infinite.

If $M$ is a martingale, $\mathrm{\mid }{M}_k\mathrm{\mid }$ is a nonnegative submartingale by F5 and its terminal value is $\mathrm{\mid }{M}_N\mathrm{\mid }$. Applying step 3.1 proves the second display. AC is exactly the inherited dependence in F6.

Since the underlying measure is a probability measure, Hölder gives ${\sup }_n\mathbb{E}\mathrm{\mid }{M}_n\mathrm{\mid }\le C^{1\mathrm{/}p}$. In particular ${\sup }_n\mathbb{E}(M_n{)}^{+}\le C^{1\mathrm{/}p}$. The martingale $M$ is also a submartingale, so F1 applies directly to $M$ and gives $M_n\to M_{\mathrm{\infty }}$ almost surely for a finite integrable $M_{\mathrm{\infty }}$.

For each $N$, F2 gives ${\parallel {\max }_{k\le N}\mathrm{\mid }{M}_k\mathrm{\mid }\parallel }_p\le q{\sup }_n\mathrm{\parallel }{M}_n{\mathrm{\parallel }}_p,q=\frac{p}{p-1}\mathrm{.}$ The maxima increase to $M^{\ast }={\sup }_n\mathrm{\mid }{M}_n\mathrm{\mid }$, so F3 yields the displayed infinite-horizon bound and $M^{\ast }\in L^p$. In particular $\mathrm{\mid }{M}_{\mathrm{\infty }}\mathrm{\mid }\le M^{\ast }$ almost surely, hence $M_{\mathrm{\infty }}\in L^p$.

We have $\mathrm{\mid }{M}_n-M_{\mathrm{\infty }}{\mathrm{\mid }}^p\le (2M^{\ast }{)}^p$ and pointwise convergence to zero. Dominated convergence gives $\mathrm{\parallel }{M}_n-M_{\mathrm{\infty }}{\mathrm{\parallel }}_p\to 0$.

Fix $n$ and take $m\ge n$. F4 gives $M_n=\mathbb{E}[M_m\mid {\mathcal{F}}_n]$. Conditional $L^p$ contraction and step 2.1 imply $\mathrm{\parallel }\mathbb{E}[M_m-M_{\mathrm{\infty }}\mid {\mathcal{F}}_n]{\mathrm{\parallel }}_p\le \mathrm{\parallel }{M}_m-M_{\mathrm{\infty }}{\mathrm{\parallel }}_p⟶0.$ The left conditional expectations therefore converge to zero while $M_n$ is fixed, proving $M_n=\mathbb{E}[M_{\mathrm{\infty }}\mid {\mathcal{F}}_n]$ almost surely. AC has only the inherited role in F5.

Uniform integrability implies ${\sup }_n\mathbb{E}\mathrm{\mid }{M}_n\mathrm{\mid }<\mathrm{\infty }$ by F1, hence ${\sup }_n\mathbb{E}{M}_n^{+}<\mathrm{\infty }$. Apply F2 to obtain a finite integrable $M_{\mathrm{\infty }}$ with $M_n\to M_{\mathrm{\infty }}$ almost surely.

Almost-sure convergence implies convergence in probability. The family $\{M_n:n\ge 0\}$ is uniformly integrable by hypothesis, so F3 yields $\mathbb{E}\mathrm{\mid }{M}_n-M_{\mathrm{\infty }}\mathrm{\mid }\to 0$. Mere $L^1$ boundedness was not substituted for uniform integrability. AC is used exactly as described in F4.

Assume (1). F1 supplies $M_{\mathrm{\infty }}\in L^1$ with convergence both almost surely and in $L^1$, proving (2) and the final convergence assertion.

Assume (2) and fix $n$. For every $m\ge n$, F2 gives $M_n=\mathbb{E}[M_m\mid {\mathcal{F}}_n]$. By F3, $\mathrm{\parallel }\mathbb{E}[M_m\mid {\mathcal{F}}_n]-\mathbb{E}[M_{\mathrm{\infty }}\mid {\mathcal{F}}_n]{\mathrm{\parallel }}_1\le \mathrm{\parallel }{M}_m-M_{\mathrm{\infty }}{\mathrm{\parallel }}_1\to 0.$ Consequently $M_n=\mathbb{E}[M_{\mathrm{\infty }}\mid {\mathcal{F}}_n]$ almost surely. Thus (3) holds with $X=M_{\mathrm{\infty }}$.

Assume (3). F4 applied to the single variable $X$ makes $\{\mathbb{E}[X\mid {\mathcal{F}}_n]:n\ge 0\}$ uniformly integrable. These variables are the $M_n$, so (1) follows.

Steps 1.1, 1.2, and 1.3 prove every direction, including the claimed choice of terminal variable. AC is used exactly through F5; no stronger limiting assertion is made for a merely $L^1$-bounded martingale.

Fix $N$. Read $X_N,X_{N-1},\dots ,X_0$ in that order with filtration ${\mathcal{G}}_N\subseteq \cdots \subseteq {\mathcal{G}}_0$; F1 makes this a finite ordinary martingale. An upcrossing of $X_0,\dots ,X_N$ becomes a downcrossing of the reversed list, hence an upcrossing of its negative through $[-b,-a]$. F3 bounds its expectation by endpoint positive parts, uniformly in $N$, because F2 gives uniform $L^1$ bounds. The same argument directly bounds downcrossings.

For every rational $a<b$, the total upcrossing and downcrossing counts are finite almost surely. Intersecting these countably many full-measure events, the usual rational-interval argument gives a finite or extended-real limit. Uniform integrability bounds the positive and negative tails uniformly, so Fatou excludes both infinite values. Denote the finite almost-sure limit by $X_{\mathrm{\infty }}$.

The almost-sure convergence from step 2.1 gives convergence in probability by F4. Using the uniformly integrable family from F2, F4 then upgrades it to $X_n\to X_{\mathrm{\infty }}$ in $L^1$.

Fix $r$. For all $n\ge r$, $X_n$ is ${\mathcal{G}}_n$-measurable and ${\mathcal{G}}_n\subseteq {\mathcal{G}}_r$, so F5 makes $X_{\mathrm{\infty }}$ measurable for ${\mathcal{G}}_r$. This holds for every $r$, hence $X_{\mathrm{\infty }}$ is ${\mathcal{G}}_{\mathrm{\infty }}$-measurable. If $A\in {\mathcal{G}}_{\mathrm{\infty }}$, then $A\in {\mathcal{G}}_n$ for every $n$ and F1 gives ${\int }_A{X}_{n}dP={\int }_A{X}_{0}dP\mathrm{.}$ The $L^1$ limit passes through the left integral, so F5 identifies $X_{\mathrm{\infty }}=\mathbb{E}[X_0\mid {\mathcal{G}}_{\mathrm{\infty }}]$. AC is used exactly as recorded in F6.

By F1, F2, F3 there is $Y\in L^1$ such that $M_n\to Y$ almost surely and in $L^1$. As an almost-sure limit of ${\mathcal{F}}_{\mathrm{\infty }}$-measurable variables, $Y$ has an ${\mathcal{F}}_{\mathrm{\infty }}$-measurable version.

The union $\mathcal{A}={\bigcup }_n{\mathcal{F}}_n$ is an algebra because the filtration is increasing. If $A\in \mathcal{A}$, then $A\in {\mathcal{F}}_N$ for some $N$, and for every $n\ge N$, ${\int }_A{M}_{n}dP={\int }_{A}XdP\mathrm{.}$ Passing to the $L^1$ limit yields ${\int }_{A}Y={\int }_{A}X$.

Let $\mathcal{D}=\{A\in {\mathcal{F}}_{\mathrm{\infty }}:{\int }_{A}Y={\int }_{A}X\}$. Integrability makes $\mathcal{D}$ a monotone class, and step 2.1 gives $\mathcal{A}\subseteq \mathcal{D}$. F4 yields $\sigma (\mathcal{A})={\mathcal{F}}_{\mathrm{\infty }}\subseteq \mathcal{D}$. Thus $Y$ has exactly the defining event integrals of $\mathbb{E}[X\mid {\mathcal{F}}_{\mathrm{\infty }}]$, proving the result. AC has the role stated in F5.

Put $X_n=\mathbb{E}[X\mid {\mathcal{G}}_n]$. If $m\le n$, then ${\mathcal{G}}_n\subseteq {\mathcal{G}}_m$, and F1 gives $\mathbb{E}[X_m\mid {\mathcal{G}}_n]=\mathbb{E}[\mathbb{E}[X\mid {\mathcal{G}}_m]\mid {\mathcal{G}}_n]=\mathbb{E}[X\mid {\mathcal{G}}_n]=X_n\mathrm{.}$ Thus $(X_n,{\mathcal{G}}_n)$ is a reverse martingale.

F2 gives convergence almost surely and in $L^1$ to $\mathbb{E}[X_0\mid {\mathcal{G}}_{\mathrm{\infty }}]$. Since $X_0=\mathbb{E}[X\mid {\mathcal{G}}_0]$ and ${\mathcal{G}}_{\mathrm{\infty }}\subseteq {\mathcal{G}}_0$, another tower identity identifies this with $\mathbb{E}[X\mid {\mathcal{G}}_{\mathrm{\infty }}]$. AC is used exactly through F3.

Fix $A\in \mathcal{T}$ and let ${\mathcal{H}}_n$ be generated by the first $n$ random elements. F2, F3 give $\mathbb{E}[1_A\mid {\mathcal{H}}_n]=\mathbb{P}(A)\text{a.s. for every }n\mathrm{.}$

The increasing union of the ${\mathcal{H}}_n$ generates the $\sigma$-algebra of the entire sequence. By F1, $A$ belongs to that $\sigma$-algebra. F4 therefore says the constants in step 1.1 converge almost surely to $\mathbb{E}[1_A\mid \sigma ({\bigcup }_n{\mathcal{H}}_n)]=1_A,$ where the last identity is F3's known-variable clause. Hence $1_A=\mathbb{P}(A)$ almost surely.

An indicator that is almost surely the constant $p=\mathbb{P}(A)$ can take only $0$ or $1$ on a nonnull set, so $p\in \{0,1\}$. This proof does not consume the already-published zero-one theorem that it recovers. AC has only the dependence in F5.

The inequalities and the deterministic width give $D-c\le A\le D\le B\le D+c$, so $A,B\in L^1$. Conditional order and F2 yield $A=\mathbb{E}[A\mid \mathcal{G}]\le 0\le \mathbb{E}[B\mid \mathcal{G}]=B\mathrm{.}$ Thus $\mathrm{\mid }D\mathrm{\mid }\le B-A\le c$, making $e^{\lambda D}$ bounded and its conditional expectation well-defined.

On $\{A=B\}$, step 1.1 forces $D=A=B=0$, so the result is equality. On $\{A<B\}$ write $w=B-A$ and $\theta =(D-A)\mathrm{/}w\in [0,1]$. F3 gives $e^{\lambda D}\le \frac{B-D}{w}{e}^{\lambda A}+\frac{D-A}{w}{e}^{\lambda B}\mathrm{.}$ The ratios $B\mathrm{/}w$ and $-A\mathrm{/}w$ lie in $[0,1]$, while the $\mathcal{G}$-measurable divided difference $K=(e^{\lambda B}-e^{\lambda A})\mathrm{/}w$ is bounded by $\mathrm{\mid }\lambda \mathrm{\mid }{e}^{\mathrm{\mid }\lambda \mathrm{\mid }c}$. Thus the right side is the sum of the two displayed bounded endpoint terms and $KD$; conditional linearity, the known-variable rule, F5, and $\mathbb{E}[D\mid \mathcal{G}]=0$ turn its conditional expectation into $\frac{B}{w}{e}^{\lambda A}+\frac{-A}{w}{e}^{\lambda B}\mathrm{.}$ This justifies pulling out the potentially small-width coefficient rather than formally dividing inside a conditional expectation.

Put $u=-A\mathrm{/}w\in [0,1]$ and $z=\lambda w$. The last expression is $h_u(z)=(1-u)e^{-uz}+u{e}^{(1-u)z}\mathrm{.}$ For $g_u=\log h_u$, direct differentiation gives $g_u(0)=g_u^{\mathrm{\prime }}(0)=0$ and $0\le g_u^{\mathrm{\prime }\mathrm{\prime }}(z)=p_z(1-p_z)\le \frac{1}{4},$ where $p_z=u{e}^{(1-u)z}\mathrm{/}{h}_u(z)\in [0,1]$. Integrating the second-derivative bound from $0$ to $z$ (with reversed limits when $z<0$) gives $g_u(z)\le z^2\mathrm{/}8$. Hence $h_u(z)\le e^{z^2\mathrm{/}8}\le e^{{\lambda }^2{c}^2\mathrm{/}8}$.

Combining the two measurable cases proves the conditional inequality. Random endpoints cause no hidden selection; only their deterministic width enters the final bound. AC is used exactly as recorded in F4.

Let $D_k=M_k-M_{k-1}$ and $V={\sum }_{k=1}^n{c}_k^2$. For $\lambda >0$, F1, F2, and F5 give $\begin{array}{rl}{\displaystyle \mathbb{E}{e}^{\lambda \sum _{k=1}^n{D}_k}}& {\displaystyle =\mathbb{E}\text{\hspace{0.17em}⁣}[e^{\lambda \sum _{k<n}{D}_k}\mathbb{E}(e^{\lambda D_n}\mid {\mathcal{F}}_{n-1})]}\\ {\displaystyle }& {\displaystyle \le e^{{\lambda }^2{c}_n^2\mathrm{/}8}\mathbb{E}{e}^{\lambda \sum _{k<n}{D}_k}\le e^{{\lambda }^{2}V\mathrm{/}8}\mathrm{.}}\end{array}$ The final inequality follows by finite induction, with the empty sum at time $0$.

Markov applied to $e^{\lambda (M_n-M_0)}$ yields $\mathbb{P}(M_n-M_0\ge t)\le \exp (-\lambda t+{\lambda }^{2}V\mathrm{/}8)\mathrm{.}$ If $V>0$, the quadratic is minimized at $\lambda =4t\mathrm{/}V$, giving $\exp (-2t^2\mathrm{/}V)$.

If $V=0$, every $c_k=0$. F1's hypotheses then force every $D_k=0$ almost surely, so the event is empty for $t>0$, agreeing with the stated convention. Apply step 1.1, step 2.1 to $-M$, whose endpoints are $-B_k,-A_k$, to obtain the lower-tail bound. AC has exactly the inherited role in F4.

Apply F1 with $A_k=-c_k$ and $B_k=c_k$. Their width is $2c_k$, so each one-sided probability is at most $\exp \text{\hspace{0.17em}⁣}(-\frac{2t^2}{{\sum }_k(2c_k{)}^2})=\exp \text{\hspace{0.17em}⁣}(-\frac{t^2}{2{\sum }_k{c}_k^2})\mathrm{.}$

The event $\{\mathrm{\mid }{M}_n-M_0\mathrm{\mid }\ge t\}$ is the union of the upper and lower tail events. F2 gives twice the bound in step 1.1. If all $c_k=0$, all increments vanish almost surely and the event is empty. Zero-width individual terms otherwise simply contribute zero to the sum. AC is used exactly as stated in F3.

Write $v_{n,k}=\mathbb{E}[Z_{n,k}^2\mid {\mathcal{F}}_{n,k-1}]$. For every $\epsilon >0$, $v_{n,k}\le {\epsilon }^2+\mathbb{E}[Z_{n,k}^2{1}_{\{\mathrm{\mid }{Z}_{n,k}\mathrm{\mid }>\epsilon \}}\mid {\mathcal{F}}_{n,k-1}]\mathrm{.}$ Taking the supremum in $k$, bounding it by the sum of the nonnegative tail terms, and taking expectations gives $\mathbb{E}{\sup }_k{v}_{n,k}\le {\epsilon }^2+L_n(\epsilon )\mathrm{.}$ Consequently $\mathbb{E}{\sup }_k{v}_{n,k}\to 0$ after first taking $n\to \mathrm{\infty }$ and then $\epsilon \downarrow 0$.

First suppose ${\mathrm{\Gamma }}_{n,\mathrm{\infty }}\le c$ almost surely for one deterministic $c$. Fix $t\in \mathbb{R}$ and put $Q_{n,m}=\exp (it{M}_{n,m})\exp (t^2{\mathrm{\Gamma }}_{n,m}\mathrm{/}2)\mathrm{.}$ The exact telescoping identity is $Q_{n,m}-Q_{n,m-1}=e^{it{M}_{n,m-1}}{e}^{t^2{\mathrm{\Gamma }}_{n,m}\mathrm{/}2}(e^{it{Z}_{n,m}}-e^{-t^2{v}_{n,m}\mathrm{/}2})\mathrm{.}$ Here the prefactor before the parentheses is ${\mathcal{F}}_{n,m-1}$-measurable and has modulus at most $e^{t^{2}c\mathrm{/}2}$.

For any bounded ${\mathcal{F}}_{n,m-1}$-measurable complex $H$ and integrable complex $Y$, the defining conditional-expectation event integrals in F3 give $\mathbb{E}[HY]=\mathbb{E}[H\mathbb{E}[Y\mid {\mathcal{F}}_{n,m-1}]]$: prove it first for simple real $H$, approximate bounded real and imaginary parts by bounded simple functions, and apply F7 to each integrable product. The prefactor in step 1.2 is such an $H$, so this pull-out identity licenses conditioning the telescoping increment. Conditional centering, F2, and a split at $\mathrm{\mid }{Z}_{n,m}\mathrm{\mid }=\epsilon$ give $\mid \mathbb{E}[e^{it{Z}_{n,m}}\mid {\mathcal{F}}_{n,m-1}]-(1-t^2{v}_{n,m}\mathrm{/}2)\mid \le C_t\epsilon v_{n,m}+t^2\mathbb{E}[Z_{n,m}^2{1}_{\{\mathrm{\mid }{Z}_{n,m}\mathrm{\mid }>\epsilon \}}\mid {\mathcal{F}}_{n,m-1}],$ where $C_t$ is deterministic. The elementary exponential remainder also gives $\mathrm{\mid }{e}^{-t^2{v}_{n,m}\mathrm{/}2}-(1-t^2{v}_{n,m}\mathrm{/}2)\mathrm{\mid }\le C_{t,c}{v}_{n,m}{\sup }_j{v}_{n,j}\mathrm{.}$ Sum the expected telescoping errors from step 1.2. Since ${\sum }_m{v}_{n,m}={\mathrm{\Gamma }}_{n,\mathrm{\infty }}\le c$, step 1.1 and the Lindeberg hypothesis yield $\mathrm{\mid }\mathbb{E}{Q}_{n,\mathrm{\infty }}-1\mathrm{\mid }\le C_{t,c}(\epsilon +L_n(\epsilon )+\mathbb{E}{\sup }_j{v}_{n,j})⟶0$ after $n\to \mathrm{\infty }$ and then $\epsilon \downarrow 0$. The infinite telescoping limit is legitimate because $M_{n,m},{\mathrm{\Gamma }}_{n,m}$ converge almost surely and $\mathrm{\mid }{Q}_{n,m}\mathrm{\mid }\le e^{t^{2}c\mathrm{/}2}$; the displayed summable error bound controls passage of expectation through the partial telescopes.