= Solution
For each fixed horizon $t$, the stipulated property is the <Class D process> condition on $[0,t]$, and imposing it for every finite horizon gives a <Class DL process>.
First suppose $M$ is a true <martingale>. For every <stopping time> $T\le t$, the bounded-time <optional sampling theorem> gives
$$
M_T=\mathbb E(M_t\mid\mathcal F_T).
$$
The <conditional expectations> of one integrable variable form a <uniformly integrable> family. To see the uniformity explicitly, put $V=\mathbb E(U\mid\mathcal H)$. For any $r,K>0$, conditional Jensen and Markov inequalities give
$$
\mathbb E\bigl[|V|\mathbf1_{\{|V|>K\}}\bigr]
\le\mathbb E\bigl[|U|\mathbf1_{\{|U|>r\}}\bigr]+\frac{r\mathbb E|U|}{K}.
$$
Choose $r$ large and then $K$ large. The bound does not depend on the conditioning <sigma-algebra>. This proves the required <uniform integrability>. Bounded-time optional sampling here can also be justified directly: approximate $T$ from above by finite-grid <stopping times>, use discrete optional sampling with terminal value $M_t$, and use this same conditional-expectation <uniform integrability> and right continuity to pass to the limit.
Conversely, take a <localizing sequence> $\tau_n$ for the <local martingale>. For fixed $s\le t$, both families $M_{t\wedge\tau_n}$ and $M_{s\wedge\tau_n}$ are <uniformly integrable>, because their times are bounded by $t$. They converge almost surely to $M_t$ and $M_s$, and therefore converge in $L^1$. The stopped <martingale> identity
$$
\mathbb E(M_{t\wedge\tau_n}\mid\mathcal F_s)=M_{s\wedge\tau_n}
$$
passes to the limit by $L^1$ contractivity of <conditional expectation>. Thus $\mathbb E(M_t\mid\mathcal F_s)=M_s$, and all time values are integrable. Hence \b[a <local martingale> is a true <martingale> exactly when it is of <Class DL>]. This is a finite-horizon criterion, not a claim of <uniform integrability> over all times simultaneously.
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