[proofplan] The proof is a one-step martingale increment calculation. Boundedness of $H_i$ and integrability of the martingale increments make each summand integrable, while predictability ensures that $H_{n+1}$ is known at time $n$. Conditioning the next increment $Y_{n+1}-Y_n$ on $\mathcal F_n$ then factors out $H_{n+1}$ and leaves the martingale increment $\mathbb E[M_{n+1}-M_n\mid\mathcal F_n]=0$. [/proofplan] [step:Verify adaptedness and integrability of the transform] For $n=0$, $Y_0=0$ is $\mathcal F_0$-measurable and integrable. Fix $n\geq 1$. For each $1\leq i\leq n$, predictability gives $H_i$ is $\mathcal F_{i-1}$-measurable, hence $\mathcal F_i$-measurable. Since $M_i$ and $M_{i-1}$ are $\mathcal F_i$-measurable, the product $H_i(M_i-M_{i-1})$ is $\mathcal F_i$-measurable and therefore $\mathcal F_n$-measurable. Thus $Y_n$, a finite sum of $\mathcal F_n$-measurable random variables, is $\mathcal F_n$-measurable. Let $C<\infty$ be such that $|H_i|\leq C$ almost surely for every $i\geq 1$. Since $M_i$ and $M_{i-1}$ are integrable, \begin{align*} \mathbb E[|H_i(M_i-M_{i-1})|] &\leq C\,\mathbb E[|M_i-M_{i-1}|] \\ &\leq C\left(\mathbb E[|M_i|]+\mathbb E[|M_{i-1}|]\right) <\infty. \end{align*} Therefore each finite sum $Y_n$ is integrable. [/step] [step:Compute the conditional expectation of the next increment] For $n\geq 0$, the increment of the transform is \begin{align*} Y_{n+1}-Y_n &= H_{n+1}(M_{n+1}-M_n). \end{align*} Because $H_{n+1}$ is bounded and $\mathcal F_n$-measurable, [Taking Out What is Known](/theorems/1151) gives \begin{align*} \mathbb E[Y_{n+1}-Y_n\mid\mathcal F_n] &= \mathbb E[H_{n+1}(M_{n+1}-M_n)\mid\mathcal F_n] \\ &= H_{n+1}\,\mathbb E[M_{n+1}-M_n\mid\mathcal F_n]. \end{align*} The martingale property of $(M_n)_{n\geq 0}$ gives \begin{align*} \mathbb E[M_{n+1}-M_n\mid\mathcal F_n] &= \mathbb E[M_{n+1}\mid\mathcal F_n]-M_n \\ &= M_n-M_n \\ &= 0. \end{align*} Hence \begin{align*} \mathbb E[Y_{n+1}-Y_n\mid\mathcal F_n] &= 0. \end{align*} [/step] [step:Conclude the martingale identity for $Y$] Since $Y_n$ is $\mathcal F_n$-measurable and integrable, the previous increment computation gives \begin{align*} \mathbb E[Y_{n+1}\mid\mathcal F_n] &= \mathbb E[Y_n+(Y_{n+1}-Y_n)\mid\mathcal F_n] \\ &= Y_n+\mathbb E[Y_{n+1}-Y_n\mid\mathcal F_n] \\ &= Y_n. \end{align*} Thus $(Y_n)_{n\geq 0}$ is a martingale with respect to $(\mathcal F_n)_{n\geq 0}$. [/step]