Claim: Let $(X_n)$ be a bounded sequence of random variables, i.e. $|X_n| \leq M$ for some $M > 0$. In this case, $X_n \overset{P}{\longrightarrow} 0$ implies $X_n \overset{L^1}{\longrightarrow} 0$. (Or at least it seems that way; see my proof below.)
I've never seen this fact mentioned anywhere (even when actively googling around), which makes me mistrust it. I'd be grateful if someone would check my proof.
Proof:
Fix $\delta > 0$.
$$ \begin{aligned} E|X_n| &= \int_{\{|X_n| \leq \delta \}} \underbrace{|X(\omega)|}_{\leq \delta} P(d\omega) + \int_{\{|X_n| > \delta \}} \underbrace{|X(\omega)|}_{\leq M} P(d\omega) \\ &\leq \delta \cdot \underbrace{P(|X_n| \leq \delta)}_{\to 1} + M \cdot \underbrace{ P(|X_n| > \delta) }_{\to 0} \to \delta \quad \text{as $n \to \infty$.} \end{aligned} $$ so $$ \limsup_{n \to \infty} E|X_n| \leq \delta. $$ Allowing $\delta \downarrow 0$ then yields $\lim_{n \to \infty} E|X_n| = 0. \quad \square$