Square minus predictable quadratic variation is a martingale

Put $D_n=M_n-M_{n-1}$ for $n\ge 1$. It is in $L^2$ by $D_n^2\le 2M_n^2+2M_{n-1}^2$, and its conditional mean given ${\mathcal{F}}_{n-1}$ is zero. Cauchy–Schwarz gives $E\mathrm{\mid }{M}_{n-1}{D}_n\mathrm{\mid }\le (E{M}_{n-1}^2{)}^{1\mathrm{/}2}(E{D}_n^2{)}^{1\mathrm{/}2}<\mathrm{\infty }$. The factor $M_{n-1}$ is finite and known at time $n-1$, so taking-out is legitimate and gives $E[M_{n-1}{D}_n\mid {\mathcal{F}}_{n-1}]=M_{n-1}E[D_n\mid {\mathcal{F}}_{n-1}]=0$.

For the finite integral and conditional linearity used here, augment every finite disjoint nonnegative-simple display by its zero-coefficient complement. Intersections of two augmented displays partition the space and carry equal coefficients on nonempty cells, so finite additivity and $0\cdot (+\mathrm{\infty })=0$ prove representation independence. Common refinements give simple addition and monotonicity; scalar zero is direct and positive scalars are termwise. Supremum over simple minorants, increasing simple approximation and the sets $\{f_j\ge cs\}$ for $0<c<1$ give monotone convergence and nonnegative additivity. Positive/negative and real/imaginary decompositions give finite $L^1$ linearity; substituting these facts at the base validates the event-integral construction and algebra of the cited conditional expectations. The square and bracket are adapted and integrable, hence so is $Z$. Expand $M_n^2-M_{n-1}^2=2M_{n-1}{D}_n+D_n^2$. Every term is integrable. Conditional linearity and step 1.1 give $E[M_n^2-M_{n-1}^2\mid {\mathcal{F}}_{n-1}]=E[D_n^2\mid {\mathcal{F}}_{n-1}]=⟨M{⟩}_n-⟨M{⟩}_{n-1}$. The last difference is known at time $n-1$. Subtracting it and conditioning the known $Z_{n-1}$ proves $E[Z_n\mid {\mathcal{F}}_{n-1}]=Z_{n-1}$. The initial bracket is zero, so $Z_0=M_0^2$.

Set $N_n=M_n-M_0$. Since $M_0$ is ${\mathcal{F}}_0$-measurable, it is known for every ${\mathcal{F}}_n$. Linearity gives $E[N_{n+1}\mid {\mathcal{F}}_n]=M_n-M_0=N_n$. Also $N_n^2\le 2M_n^2+2M_0^2$ is integrable, so $N$ is a square-integrable martingale with $N_0=0$. Its increments equal $D_n$, hence $⟨N⟩=⟨M⟩$ as classes. Apply the already proved step 2.1 to $N$ to conclude that $W=N^2-⟨M⟩$ is a martingale with $W_0=0$. AC is inherited from CE and bracket version construction; no unproved assertion about the product $M_0{M}_n$ is used.