A nonadapted step integrand breaks the Ito isometry

On the event $A$ one has $B_T^2>T$ by definition of $A$, while off $A$ the integrand $H=1_A$ vanishes; hence $H^2{B}_T^2=1_A{B}_T^2\ge 1_{A}T$, and the inequality is strict on the nonempty event $A$ (where $B_T^2>T$ strictly), so $E[H^2{B}_T^2]>TP(A)$.

The integrand $H=1_A$ is not admissible: $A\in \sigma (B_T)\subseteq {\mathcal{F}}_T$ and $P(A)\in (0,1)$ by [F1]; an event in ${\mathcal{F}}_0$ differs from a germ event in ${\mathcal{F}}_{0+}^0$ only by a null set, so by [F2] every set in ${\mathcal{F}}_0$ has probability $0$ or $1$; hence $A\mathrm{\notin }{\mathcal{F}}_0$, so a block coefficient equal to $1_A$ on $(0,T]$ violates the left-endpoint measurability requirement at time $0$.

Since $H^2=1_A$, one has $E[H^2]=P(A)$ and ${\int }_0^T{H}^{2}ds=H^{2}T=1_{A}T$, so $TE[H^2]=TP(A)<E[H^2{B}_T^2]$ by step 1.1; this is the asserted violation of the isometry-type equality, and $P(A)\in (0,1)$ by [F1], so the strict inequality is between finite positive numbers.

The witness therefore separates the two hypotheses: with the left-endpoint measurability clause enforced, the elementary isometry of item 7 computes $E[H^2{B}_T^2]=TE[H^2]$ for the admissible case; dropping that clause and using the future information contained in $B_T$ admits the strict violation above. Note that the symmetry of $B_T$ does not rescue the mean: it is the second moment, not the first, that the isometry governs, and the failure exhibited is a second-moment failure. AC enters only through [F3].