If $\{X_{n}\}$ is a uniformly integrable submartingale, then clearly, so is $\{X_{n}^{+}\}$. Is the converse true?
My attempt: For a submartingale. $\textbf{E}(|X_{n}|)=\textbf{E}(X_{n}^{+})+\textbf{E}(X_{n}^{-})=2\textbf{E}(X_{n}^{+})-\textbf{E}(X_{n})\leq2\textbf{E}(X_{n}^{+})-\textbf{E}(X_{0})$. Hence, $L^{1}-$boundedness of $\{X_{n}^{+}\}$ is equivalent to that of $\{X_{n}\}$. But I couldn't proceed further. Any help would be appreciated.