Integrable stopping time alone does not suffice for arbitrary martingale increments

On $([0,1],\mathcal{B},\lambda )$ put $A_n=(0,2^{-n}]$, ${\mathcal{F}}_0$ trivial, ${\mathcal{F}}_n=\sigma (A_1,\dots ,A_n)$, $M_0=1$, and $M_n=2^n{1}_{A_n}$ for $n\ge 1$. On the atom $A_n$, $M_{n+1}$ equals $2^{n+1}$ on a half-measure subatom and zero on the other half, so its conditional average is $2^n$; off $A_n$ both variables vanish. Thus F1 proves directly that $M$ is a nonnegative martingale with $\mathbb{E}{M}_n=1$.

Let $\tau =\inf \{n\ge 1:M_n=0\}$. For $n\ge 1$, $\{\tau >n\}=A_n\in {\mathcal{F}}_n$, so F2 makes $\tau$ a stopping time. Its tail sum is $\mathbb{E}\tau ={\sum }_{n\ge 0}P(\tau >n)=1+{\sum }_{n\ge 1}{2}^{-n}=2.$

The intersection of the $A_n$ is empty, so every path eventually leaves and $M_{\tau }=0$. Hence $M_{\tau }$ is integrable but $\mathbb{E}{M}_{\tau }=0\mathrm{\ne }1=\mathbb{E}{M}_0\mathrm{.}$ On $A_n\setminus A_{n+1}$ the next increment has magnitude $2^n$, so no deterministic increment bound exists; this is exactly the missing hypothesis flagged by F4. The proof reconstructs its martingale locally and does not depend on a B-page supplier. AC has only the role in F5.