 Solution

The polarization identity for quadratic covariation, together with $\mathbb{E}[M\pm N{]}_s=\mathbb{E}(M_s\pm N_s{)}^2$, yieldsThe bracket identity for each zero-starting square-integrable martingale is the Itô isometry for integration of the constant integrand against that martingale. Polarizing both identities gives the displayed covariance identity for continuous square-integrable martingales. The bracket is integrable: its total variation is at most $\sqrt{[M{]}_s[N{]}_s}$ by the Kunita-Watanabe inequality, whose expectation is finite by the Cauchy-Schwarz inequality. Therefore condition (iii) is equivalent to (ii).