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The Martingale convergence theorem in its $L^1$-bounded form says that a martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$ has an integrable limit $M_{\infty }$ andThe norm bound on the limit follows from Fatou lemma. Boundedness in $L^1$ by itself does not imply convergence in L1. The uniformly integrable martingale convergence theorem gives the stronger conclusion: if $(M_n)$ is uniformly integrable, then $M_n\to M_{\infty }$ both almost surely and in L1 norm, and $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$. Conversely, convergence in L1 implies uniform integrability.

For the distinction, the fair-coin doubling martingale $M_n=2^n{1}_{\{\text{first }n\text{ tosses are heads}\}}$ has expectation one for every $n$ but converges almost surely to zero. Its L1 norm remains one, so its convergence is not in L1 norm. These two formulations specify exactly which hypothesis is needed in part (d).

Let ${\xi }_1,{\xi }_2,\dots$ be independent and identically distributed random variables with the fair Bernoulli distribution, and putRelative to ${\mathcal{F}}_n=\sigma ({\xi }_1,\dots ,{\xi }_n)$, this fair-coin doubling martingale satisfies $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$: while alive it doubles with probability $1/2$ and becomes zero otherwise. Its expected value is $\mathbb{E}{M}_n=2^n{2}^{-n}=1$.

The probability of an infinite run of ones is ${\lim }_n{2}^{-n}=0$, so $M_n$ is eventually zero almost surely. However,Thus the martingale lacks convergence in L1 to its limit from almost sure convergence. Nor can it have convergence in L1 to any other limit: convergence in L1 implies convergence in probability, whose limit is unique up to almost sure equality.