= Solution
Let $\xi_1,\xi_2,\ldots$ be <independent and identically distributed random variables> with the fair <Bernoulli distribution>, and put
$$
M_0=1,\qquad M_n=2^n\mathbf1_{\{\xi_1=\cdots=\xi_n=1\}}.
$$
Relative to $\mathcal F_n=\sigma(\xi_1,\ldots,\xi_n)$, 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 $\mathbb E M_n=2^n2^{-n}=1$.
The probability of an infinite run of ones is $\lim_n2^{-n}=0$, so $M_n$ is eventually zero <almost surely>. However,
$$
\boxed{M_n\to0\ \text{almost surely},\qquad \mathbb E|M_n-0|=1\ \text{for every }n.}
$$
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>.
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