Predictably weighted bounded martingale differences have a convergent series

Put $Y_n=B_n{D}_n$. The square-sum bound implies $\mathrm{\mid }{B}_n\mathrm{\mid }\le \sqrt{K}$ almost surely, so $Y_n$ is square integrable. Since $B_n$ is ${\mathcal{F}}_{n-1}$-measurable, [F2] and [F5] give $E[Y_n\mid {\mathcal{F}}_{n-1}]=0$. For $i<j$, [F1] gives ${\mathcal{F}}_i\subseteq {\mathcal{F}}_{j-1}$, and [F4] gives $E[Y_j\mid {\mathcal{F}}_i]=0$. Taking out the bounded ${\mathcal{F}}_i$-measurable factor $Y_i$ with [F5], then using expectation preservation [F6], gives $E[Y_i{Y}_j]=0$. Expanding a finite square therefore yields $E\text{\hspace{0.17em}⁣}\left[{\left({\sum }_{n=N}^m{Y}_n\right)}^2\right]={\sum }_{n=N}^{m}E[Y_n^2]\le C^{2}E\text{\hspace{0.17em}⁣}\left[{\sum }_{n=N}^m{B}_n^2\right]\mathrm{.}$

Let $R_N={\sum }_{n=N}^{\mathrm{\infty }}{B}_n^2$. By [F8], $R_N$ is measurable; the hypothesis gives $0\le R_N\le K$ and $R_N\downarrow 0$ almost surely. The constant $K$ is integrable by [F7], so [F9] gives $E[R_N]\to 0$.

For finite $m\ge N$, put $S_j={\sum }_{n=N}^j{Y}_n$ for $N\le j\le m$, with $S_{N-1}=0$. Each $S_j$ is adapted and integrable; [F2] and [F6] give $E[S_{j+1}\mid {\mathcal{F}}_j]=S_j+E[Y_{j+1}\mid {\mathcal{F}}_j]=S_j$, so $(S_j)$ is a martingale. Since $S_j$ is ${\mathcal{F}}_j$-measurable, [F5] takes it out of the conditional expectation; the MDS identity from step 1.1 gives $E[2S_j{Y}_{j+1}\mid {\mathcal{F}}_j]=2S_{j}E[Y_{j+1}\mid {\mathcal{F}}_j]=0$. The conditional square expansion therefore gives $E[S_{j+1}^2\mid {\mathcal{F}}_j]=S_j^2+E[Y_{j+1}^2\mid {\mathcal{F}}_j]\ge S_j^2,$ so $(S_j^2)$ is a nonnegative submartingale by [F3]. Let $A_j$ be the event that $j$ is the first index in $[N,m]$ with $\mathrm{\mid }{S}_j\mathrm{\mid }\ge \epsilon$. These events are disjoint and $A_j\in {\mathcal{F}}_j$. Repeated use of the submartingale inequality and [F4] gives $E[S_m^2\mid {\mathcal{F}}_j]\ge S_j^2$, hence $E[{\mathbf{1}}_{A_j}{S}_m^2]\ge E[{\mathbf{1}}_{A_j}{S}_j^2]\ge {\epsilon }^{2}P(A_j)\mathrm{.}$ Summing over $j$ and using $S_m^2\ge 0$ gives $P\text{\hspace{0.17em}⁣}({\max }_{N\le j\le m}\mathrm{\mid }{S}_j\mathrm{\mid }\ge \epsilon )\le \frac{E[S_m^2]}{{\epsilon }^2}\le \frac{C^{2}E[R_N]}{{\epsilon }^2}\mathrm{.}$

Since $C^{2}E[R_N]\to 0$, for each integer $r\ge 1$ choose the least $N_r>N_{r-1}$ (with $N_0=0$) such that $C^{2}E[R_{N_r}]\le 2^{-4r}$; it exists by [F13]. For finite $m\ge N_r$, step 2.1 with $\epsilon =2^{-r}$ bounds the event ${\max }_{N_r\le j\le m}\mathrm{\mid }{\sum }_{n=N_r}^j{Y}_n\mathrm{\mid }\ge 2^{-r}$ by $2^{-2r}$. These events increase with $m$, so [F10] gives the same bound for their union $A_r$ over all $m$. For each $R$, [F11] gives $P({\bigcup }_{r\ge R}{A}_r)\le {\sum }_{r\ge R}{2}^{-2r}$, which tends to zero; hence almost surely only finitely many $A_r$ occur. Outside that null event, for every sufficiently large $r$ and all $p\le q$ with $p,q\ge N_r$, the tail sum satisfies $\mathrm{\mid }{\sum }_{n=p}^q{Y}_n\mathrm{\mid }\le 2^{1-r}$. Thus the partial sums are Cauchy.

Completeness [F12] makes the Cauchy sequence of partial sums converge in $\mathbb{R}$. Since $Y_n=B_n{D}_n$, this is the asserted convergence of ${\sum }_{n\ge 1}{B}_n{D}_n$.