EDIT 2: With the correct definition, I think I have a proof.
Want to show $\lim_{M\to\infty} \sup_t E[|X_{t\wedge n}|; |X_{t\wedge n}|\ge M]=0$. Fix $n$. Note that $\sup_t E[|X_{t\wedge n}; |X_{t\wedge n}|\ge M]=\sup_{t\in [0,n]} E[|X_{t}|; |X_{t}|\ge M]<\infty$ (left-limits existing implies bounded on compact intervals [need to verify; know it's true if left and right limits exist]). Then DCT gives the rest, since for $\epsilon>0$ given and each $t$, there exists $M(t)>0$ so that $E[|X_{t\wedge n}|;|X_{t\wedge n}|\ge M(t)]<\epsilon$. Let $K=\sup_{t\in[0,n]} M(t)$. Then for $M\ge K$, $\sup_t E[|X_{t\wedge n}|; |X_{t\wedge n}|\ge M]<\epsilon$.
EDIT 1: I've just noticed that I misread the definition of local martingale - $n$ should be fixed, and $X_{t\wedge n}$ should be uniformly integrable in $t$, not in $n$. Will update ASAP with new attempt. My attempts below use the wrong definition, so nothing to see there.
Preface: I've seen the duplicate question (Demonstrate that every martingale is a local martingale.), but I'm also hoping for a proof-specific hint about how to use the dominated convergence theorem to show uniform integrability.
Claim: A cadlag (right continuous with left-limits existing) adapted martingale $X=\{X_t\}_{t\ge0}$ is a local martingale (i.e. there exists a sequence of increasing stopping times $T_n$ with $\lim_{n\to\infty} T_n=\infty$ so that $X_{t \wedge T_n}1_{T_n>0}$ is a uniformly integrable martingale for each $n$). We are given the hint that $T_n=n$ suffices. [Notation: $a\wedge b=\min(a,b)$.]
My progress: I've shown that $X$ is an adapted martingale, so it remains to show uniform integrability. I've observed that $X_{t\wedge n}\to X_t$ as $n\to\infty$ a.s., $E|X_t|<\infty$ for every $t$ fixed, and for $Y=\max(|X_t|,|X_n|)$, $|X_{t\wedge n}|\le Y$, and $E|Y|<\infty$. So the conditions for DCT are met, and we have $E[X_{t\wedge n}]\to E[X_t]$, and hence $E|X_t|-E|X_{t\wedge n}|\to 0$ as $n\to\infty$.
We want to show $\lim_{M\to\infty} \sup_n E[|X_{t\wedge n}|; |X_{t\wedge n}|\ge M]=0$. Since $E|X_t|<\infty$ for all $t$, we know $E[|X_t|;|X_t|\ge M]\to 0$ as $M\to\infty$. It seems like $\{|X_{t\wedge n}|\ge M\}\to \{|X_t|\ge M\}$ for $n$ large enough, so we should have something like $E[|X_{t\wedge n}|; |X_{t\wedge n}|\ge M]\to E[|X_t|;|X_t|\ge M]$, in which case we'd be done. Is this idea justified formally, or should I be taking a different approach?
Thanks in advance!
Your reasoning doesn't work. If each sample path $[0,n] \ni t \mapsto X_t(\omega)$, this does not imply
$$\sup_{t \in [0,n]} \mathbb{E}(|X_t|)<\infty,$$
in fact, $\mathbb{E}(|X_t|)$ might be infinite. The trouble is simply that the upper bound $$C(\omega) := \sup_{t \in [0,n]} |X_t(\omega)|$$
depends on $\omega$.
However, since $(X_t)_{t \geq 0}$ is a martingale, we know from Jensen's inequality that $(|X_t|)_{t \geq 0}$ is a submartingale and this implies
$$\sup_{t \in [0,n]} \mathbb{E}(|X_t|) \leq \mathbb{E}(|X_n|)<\infty.$$
In particular,
$$\sup_{t\in [0,n]} E[|X_{t}|; |X_{t}|\ge M]<\infty.$$