[proofplan] For simple predictable integrands, the martingale property follows by writing the increment from $s$ to $t$ as a finite sum of future Brownian increments with predictable coefficients and taking conditional expectations. For a general square-integrable predictable integrand, approximate by simple predictable processes and use the [Itô isometry](/theorems/2092) to pass the martingale identity to the $L^2$ limit. Square integrability follows directly from the isometry. [/proofplan] [step:Verify the martingale property for simple predictable integrands] Let $H$ be a bounded simple predictable process on $[0,T]$. For each $0\leq t\leq T$, the integral \begin{align*} M_t&=\int_0^t H_u\,dW_u \end{align*} is a finite sum of elementary terms: there are an integer $p\geq 1$, deterministic times $0\leq a_i<b_i\leq T$, and bounded $\mathcal F_{a_i}$-measurable real-valued random variables $C_i$ for $1\leq i\leq p$ such that \begin{align*} M_t&=\sum_{i=1}^{p} C_i\left(W_{b_i\wedge t}-W_{a_i\wedge t}\right). \end{align*} Thus $M_t$ is $\mathcal F_t$-measurable and integrable. Fix $0\leq s\leq t\leq T$. Refining the time partition if necessary, write the increment $M_t-M_s$ as \begin{align*} M_t-M_s &= \sum_{j=1}^{r_0} K_j\left(W_{v_j}-W_{u_j}\right), \end{align*} where $r_0\geq 1$ is an integer, $s\leq u_j<v_j\leq t$ are deterministic times, and each coefficient $K_j$ is a bounded $\mathcal F_{u_j}$-measurable real-valued random variable. Since $\mathcal F_s\subseteq\mathcal F_{u_j}$, the tower property and independence of Brownian increments give \begin{align*} \mathbb E\left[K_j(W_{v_j}-W_{u_j})\mid\mathcal F_s\right] &= \mathbb E\left[\mathbb E\left[K_j(W_{v_j}-W_{u_j})\mid\mathcal F_{u_j}\right]\mid\mathcal F_s\right] \\ &= \mathbb E\left[K_j\,\mathbb E[W_{v_j}-W_{u_j}\mid\mathcal F_{u_j}]\mid\mathcal F_s\right] \\ &=0. \end{align*} Summing over $j$ gives \begin{align*} \mathbb E[M_t-M_s\mid\mathcal F_s]&=0, \end{align*} and hence $\mathbb E[M_t\mid\mathcal F_s]=M_s$. [/step] [step:Approximate a general integrand by simple predictable integrands] Let $\mathcal H_T$ be the predictable square-integrable space with norm \begin{align*} \|K\|_{\mathcal H_T}^2&=\mathbb E\left[\int_0^{\,T} K_u^2\,d\mathcal L^1(u)\right]. \end{align*} By the [Density of Simple Processes](/theorems/2091), there are bounded simple predictable processes $(H_r)_{r\geq 1}$ such that \begin{align*} \|H_r-H\|_{\mathcal H_T}&\to 0. \end{align*} For each $r$, define the stochastic integral process $M_{r,\cdot}=(M_{r,t})_{0\leq t\leq T}$ by \begin{align*} M_{r,t} &= \int_0^t H_r(u)\,dW_u. \end{align*} The first step shows that $(M_{r,t})_{0\leq t\leq T}$ is a martingale. The [Itô Isometry](/theorems/2092), applied on $[0,t]$, gives \begin{align*} \mathbb E\left[|M_{r,t}-M_t|^2\right] &= \mathbb E\left[\int_0^t (H_r(u)-H_u)^2\,d\mathcal L^1(u)\right] \\ &\leq \|H_r-H\|_{\mathcal H_T}^2\to 0. \end{align*} Thus $M_{r,t}\to M_t$ in $L^2(\Omega,\mathcal F,\mathbb P)$ for every $t\in[0,T]$. [/step] [step:Pass the conditional expectation identity to the limit] Fix $0\leq s\leq t\leq T$ and $A\in\mathcal F_s$. Since $M_{r,t}\to M_t$ and $M_{r,s}\to M_s$ in $L^2$, they also converge in $L^1$. Therefore \begin{align*} \mathbb E[\mathbb 1_A M_t] &= \lim_{r\to\infty}\mathbb E[\mathbb 1_A M_{r,t}] \\ &= \lim_{r\to\infty}\mathbb E[\mathbb 1_A M_{r,s}] \\ &= \mathbb E[\mathbb 1_A M_s], \end{align*} where the middle equality uses the martingale property of $M_{r,\cdot}$. Since this holds for every $A\in\mathcal F_s$ and $M_s$ is $\mathcal F_s$-measurable, the defining property of conditional expectation gives \begin{align*} \mathbb E[M_t\mid\mathcal F_s]&=M_s. \end{align*} Finally, the [Itô isometry](/theorems/3544) gives \begin{align*} \mathbb E[M_t^2] &= \mathbb E\left[\int_0^t H_u^2\,d\mathcal L^1(u)\right] \leq \mathbb E\left[\int_0^{\,T} H_u^2\,d\mathcal L^1(u)\right]<\infty. \end{align*} Thus $(M_t)_{0\leq t\leq T}$ is a square-integrable martingale. [/step]