 Solution

The correct quadratic variation of a stochastic integral identity isThe brackets around the final integrator are missing in the PDF. Without them the assertion is false: for $H=1$ and $M$ Brownian motion, the left side is $t$ while the printed right side is $B_t$.

For elementary predictable $H$, the integral is obtained by scaling each martingale increment by its known coefficient. Quadratic variation on each interval is therefore scaled by the square of that coefficient, giving the displayed formula. To pass to a general $L^2(M)$ integrand on $[0,T]$, take elementary $H_n\to H$ in the integrand norm and put $N_n=H_n\cdot M$, $N=H\cdot M$. The Itô isometry gives $\mathbb{E}(N_n-N{)}_T^2\to 0$, hence $\mathbb{E}[N_n-N{]}_T\to 0$. The bracket Cauchy-Schwarz boundand polarization show that $[N_n]\to [N]$ uniformly in probability. On the other side, Cauchy-Schwarz for the quadratic-variation measure givesTaking limits proves the formula in the square-integrable case. Stopping and pasting as in part b proves it for locally bounded predictable integrands and continuous local martingales.