= Solution
{c}
There is an indexing error at the printed initial time. The dyadic <sigma-algebra> $\mathcal F_0$ is trivial, whereas the stated formula gives
$$
X_0=\mathbf1_{(0,1/2]}.
$$
This is not $\mathcal F_0$-measurable. Moreover $\mathbb EX_0=1/2$ and $\mathbb EX_1=1$, so the process cannot be a <martingale> on all of $\mathbb N_0$ as printed.
For every $n\geq1$, 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. Hence
$$
\mathbb E(X_{n+1}\mid\mathcal F_n)=X_n
\qquad(n\geq1).
$$
Thus \b[the printed process is a <martingale> from time one onward]. The intended full-index repair is to define $X_0\equiv1$ and leave all $n\geq1$ terms unchanged, since then $\mathbb E(X_1\mid\mathcal F_0)=1$.
If $\omega\neq1/2$, it eventually lies outside the shrinking intervals, so $X_n(\omega)\to0$. At the single <null set> $\omega=1/2$, the values tend to infinity. Therefore
$$
\boxed{X_n\to0\text{ almost surely},\qquad
\lim_n\mathbb EX_n=1.}
$$
The 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>.
\Image[/past-exam-of-the-martingales-and-financial-mathematics-course-of-the-university-of-oxford/2007-dyadic-spike.png]
{title=Shrinking dyadic support preserves <expected value> while the values tend to zero almost everywhere}
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