 Solution

There is an indexing error at the printed initial time. The dyadic sigma-algebra ${\mathcal{F}}_0$ is trivial, whereas the stated formula givesThis is not ${\mathcal{F}}_0$-measurable. Moreover $\mathbb{E}{X}_0=1/2$ and $\mathbb{E}{X}_1=1$, so the process cannot be a martingale on all of ${\mathbb{N}}_0$ as printed.

For every $n\ge 1$, its support $I_n=(1/2-2^{-n},1/2]$ is one atom of ${\mathcal{F}}_n$, so $X_n$ is adapted and integrable with expected value one. At the next dyadic level, $I_n$ splits into two equal-measure children; $X_{n+1}$ is zero on its left child and $2^{n+1}$ on its right child $I_{n+1}$. Its conditional average on the parent is $2^n$, and on every other parent it is zero. HenceThus the printed process is a martingale from time one onward. The intended full-index repair is to define $X_0\equiv 1$ and leave all $n\ge 1$ terms unchanged, since then $\mathbb{E}(X_1\mid {\mathcal{F}}_0)=1$.

If $\omega \ne 1/2$, it eventually lies outside the shrinking intervals, so $X_n(\omega )\to 0$. At the single null set $\omega =1/2$, the values tend to infinity. ThereforeThe repaired martingale is not uniformly integrable and does not converge in $L^1$. This is the time-zero correction for a half-point dyadic spike martingale.