variance of stochastic integral for non-anticipating random step function 
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.
 A: $$E(I_{n+1} | \mathcal{F}_n) = E\big( \sum_{k=0}^{n} f\left(t_{k}, \omega\right)\left[B\left(t_{k+1}\right)-B\left(t_{k}\right)\right] | \mathcal{F}_n \big) $$
$$= \sum_{k=0}^{n-1} f\left(t_{k}, \omega\right)\left[B\left(t_{k+1}\right)-B\left(t_{k}\right)\right]  + E\big( f\left(t_{n}, \omega\right)\left[B\left(t_{n+1}\right)-B\left(t_{n}\right)\right] | \mathcal{F}_n \big)$$
$$ = I_n +  f\left(t_{n}, \omega\right) E\big(\left[B\left(t_{n+1}\right)-B\left(t_{n}\right)\right] | \mathcal{F}_n \big) = I_n + 0.$$
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
$$\operatorname{Var}\left[I_{n}(f)\right]=\sum_{k=1}^{n-1}  D (I_{k+1}-I_{k})  =\sum_{k=1}^{n-1}  \mathbb{E}\left[f\left(t_{k}, \omega\right)^{2}\right] \Delta_k t.$$
