$\lim_{n\rightarrow \infty} E[\min(|X_n-X|,1)]=0\quad\Rightarrow\quad\lim_{n\rightarrow \infty} E[|X_n-X|]=0$

Question

Consider a sequence of measurable functions $X_n$ and $X$ measurable on some probability space $(\Omega,\mathcal{F},P)$.

I want to show, that following holds

$$\lim_{n\rightarrow \infty} E[\min(|X_n-X|,1)]=0\quad\Rightarrow\quad\lim_{n\rightarrow \infty} E[|X_n-X|]=0$$

This statement is intuitively clear, but I fail to find the actual argument. Thanks in advance!

Answer 1

This is not true. Consider $(0,1)$ with Lebesgue measure and let $A_n=(0, 1 -\frac 1 n)$. Let $X_n=\frac 1 n$ on $A_n$ and $n$ on $A_n^{c}$. Then $E X_n \wedge 1 =\frac 1 n P(A_n)+P(A_n^{c}) \leq \frac 2 n \to 0$ but $EX_n \geq nP(A_n^{c}) =1 $ for all $n$.