Assume now that each $f\left(t_{k}, \omega\right)$ is continuous, and non-anticipating , and that $\mathbb{E}\left[f\left(t_{k}, \omega\right)^{2}\right]$ is finite. The discrete stochastic integral is $$ I_{n}(f) \stackrel{\text { def }}{=} \sum_{k=0}^{n-1} f\left(t_{k}, \omega\right)\left[B\left(t_{k+1}\right)-B\left(t_{k}\right)\right] $$ Show that, $$\operatorname{Var}\left[I_{n}(f)\right]=\mathbb{E}\left[I_{n}(f)^{2}\right]=\sum_{k=0}^{n-1} \mathbb{E}\left[f\left(t_{k}, \omega\right)^{2}\right] \Delta t$$ and $I_{n}(f)$ is a discrete martingale.
I know that $\mathbb{E}[\left.I_{n}(f)\right]=0$. The expression for the variance then reduces to $$ \begin{aligned} \mathbb{E}[&\left.I_{n}(f)^{2}\right] \\ &=\mathbb{E}\left\{\sum_{k=0}^{n-1} f\left(t_{k}, \omega\right)\left[B\left(t_{k+1}\right)-B\left(t_{k}\right)\right]\right.\\ &\left.\times \sum_{m=0}^{n-1} f\left(t_{m}, \omega\right)\left[B\left(t_{m+1}\right)-B\left(t_{m}\right)\right]\right\} \end{aligned} $$
Now, I am stuck. Any help will be appreciated.
$$E(I_{n+1} | \mathcal{F}_n) = E\big( \sum_{k=0}^{n} f\left(t_{k}, \omega\right)\left[B\left(t_{k+1}\right)-B\left(t_{k}\right)\right] | \mathcal{F}_n \big) $$ $$= \sum_{k=0}^{n-1} f\left(t_{k}, \omega\right)\left[B\left(t_{k+1}\right)-B\left(t_{k}\right)\right] + E\big( f\left(t_{n}, \omega\right)\left[B\left(t_{n+1}\right)-B\left(t_{n}\right)\right] | \mathcal{F}_n \big)$$ $$ = I_n + f\left(t_{n}, \omega\right) E\big(\left[B\left(t_{n+1}\right)-B\left(t_{n}\right)\right] | \mathcal{F}_n \big) = I_n + 0.$$ Hence $I_n$ is martingale. Put $I_0 = 0$.
$$Var(I_n) = Var(\sum_{k=1}^{n-1} (I_{k+1}-I_{k})) = cov \big(\sum_{k=1}^{n-1} (I_{k+1}-I_{k}), \sum_{j=1}^{n-1} (I_{j+1}-I_{j}) \big)$$ $$ = \sum_{i,j} cov \big(I_{k+1}-I_{k}, I_{j+1}-I_{j} \big) = \sum_{j=i} cov \big(I_{k+1}-I_{k}, I_{j+1}-I_{j} \big)$$ because every square-integrable martingale has uncorrelated increments ( see, e.g. math.stackexchange.com/questions/1257086/uncorrelated-successive-differences-of-martingale ). Thus $ Var(I_n) = \sum_{j=i} D (I_{k+1}-I_{k})$. But $$ D (I_{k+1}-I_{k}) = E(I_{k+1}-I_{k})^2 = E \xi_k$$ where $$ \xi_k = E\big( (I_{k+1}-I_{k})^2 | \mathcal{F}_k\big) $$ $$ = E\big( f^2\left(t_{k}, \omega\right)\left[B\left(t_{k+1}\right)-B\left(t_{k}\right)\right]^2| \mathcal{F}_k\big) = f^2\left(t_{k}, \omega\right) \big(E\left[B\left(t_{k+1}\right)-B\left(t_{k}\right)\right]^2| \mathcal{F}_k\big) $$ $$ = f^2\left(t_{k}, \omega\right) E\left[B\left(t_{k+1}\right)-B\left(t_{k}\right)\right]^2 = f^2\left(t_{k}, \omega\right) \Delta_k t$$ Thus $ D (I_{k+1}-I_{k}) = E \xi_k = Ef^2\left(t_{k}, \omega\right) \Delta_k $ and
$$\operatorname{Var}\left[I_{n}(f)\right]=\sum_{k=1}^{n-1} D (I_{k+1}-I_{k}) =\sum_{k=1}^{n-1} \mathbb{E}\left[f\left(t_{k}, \omega\right)^{2}\right] \Delta_k t.$$