Suppose $X_n$ is a sequence of real value random variable and $E(\|X_n\|)<\infty$. Then $X_n=O_p(1)$.
My thinking is to use Markov inequality.
$$
P(\|X\|\geq a)\leq \frac{E\|X\|}{a}
$$
Now choose $a = \frac{E\|X\|}{\epsilon}$. Then
$$
(\forall \epsilon>0)(\exists a<\infty)(P(\|X\|<a)>1-\frac{E\|X\|}{E\|X\|/\epsilon}=1-\epsilon
$$
as required.
Is this correct? Any alternative proof?
Definition:
A sequence of random variables $\{X_n\}_{n\geq 1}$ is said to be $O_p(1)$ if and only if
$$
(\forall \epsilon>0)(\exists K<\infty )(P(\|X_n\|<K)>1-\epsilon\ \ \forall n \in N)
$$
Best Answer
With your assumption the claim is false and $X_n=n$ is a counter-example. If you assume that $\sup_n E\|X_n\|<\infty$ then you can prove it as follows:
Let $C=\sup_n E\|X_n\|$. Then $P(\|X_n\|\geq M)\leq \frac {E\|X_n\|} M\leq \frac C M <\epsilon$ for $M >\frac C {\epsilon}$. Hence, $P(\|X_n\|<M)>1-\epsilon$ if $M >\frac C {\epsilon}$.