= Solution
The <Martingale convergence theorem> says that a discrete-time <martingale> $(M_n,\mathcal F_n)$ satisfying
$$
\sup_n\mathbb E|M_n|<\infty
$$
has an <almost sure convergence> limit $M_\infty$, finite <almost surely> and in $L^1$. The theorem asserts that the limit is integrable; it does not assert <convergence in L1>. More generally, the <almost sure submartingale convergence theorem> applies to a <submartingale> with $\sup_n\mathbb E M_n^+<\infty$. <Uniform integrability> is the additional condition that upgrades a <martingale>'s convergence to <convergence in L1>.
For the requested distinction, let $(\xi_k)$ be independent fair Bernoulli variables and use their <natural filtration>. The <coin-doubling martingale>
$$
M_0=1,\qquad M_n=2^n\mathbf1_{\{\xi_1=\cdots=\xi_n=1\}}
$$
is a <nonnegative martingale>: conditionally on $\mathcal F_n$, the next factor is $2\xi_{n+1}$ with mean one, so $\mathbb E(M_{n+1}\mid\mathcal F_n)=M_n$. Also $\mathbb E|M_n|=\mathbb E M_n=1$ for every $n$, giving the required uniform $L^1$ bound. The probability that all the Bernoulli variables equal one is $\lim_n2^{-n}=0$. Therefore a zero is eventually encountered <almost surely>, after which $M_n$ stays zero. Thus
$$
\boxed{M_n\longrightarrow0\text{ almost surely},\qquad
\mathbb E|M_n-0|=1\text{ for every }n.}
$$
There can be no other $L^1$ limit, since <convergence in L1> implies <convergence in probability>, whose limit must agree with the almost sure limit. \b[This <martingale> satisfies the almost sure theorem but does not converge in $L^1$.]
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