Prove $\frac{Card\{X_1,\cdots,X_n\}}{\sqrt{n}}\rightarrow0$ in probability with i.i.d. $X_i\in\mathbb{N}, \ \mathbb{E}[X_1] < \infty$ Let $\{X_n\}$ be a sequence of independent, identically distributed random variables taking values in $\mathbb{N}$, the set of positive integers. Define
$$R_n=Card\{X_1,\cdots,X_n\}$$
i.e., $R_n$ is the number of distinct integers in $\{X_1,\cdots,X_n\}$. Suppose that $\mathbb{E}[X_1] < \infty$. Prove that $\frac{R_n}{\sqrt{n}}\rightarrow0$ in probability.
$$$$
I suppose that the $\mathbb{E}[X_1] < \infty$ condition is used in law of large numbers, but the answer seems not corresponding, for the $\frac{R_n}{\sqrt{n}}$ it looks like CLE, but the answer is $0$ other than normal distribution. So what could possibly be the way of approaching this problem? Thanks!
Edit: I've a possible idea of approaching, assume that $\mathbb{E}[X_1]=T>0$, then for any $\epsilon>0$ we have $\mathbb{P}[X_1\ge\epsilon\sqrt{n}]\le\frac{T}{\epsilon\sqrt{n}}$, hence we have (in intuition, approximately)$Card\{X_1,\cdots,X_n\}\le\epsilon\sqrt{n}+\frac{T}{\epsilon\sqrt{n}}\cdot n=\sqrt{n}(\epsilon+\frac{T}{\epsilon})$(we pick all integers $\le\epsilon\sqrt n$,and $X_i$ cannot be too large), so $\frac{R_n}{\sqrt{n}}$ definitely have an upper-bound $2T$, a more accurate estimation can bound this upper-bound to $T$, and I think this idea is not far from the solution. The problem is, how to reduce this upper-bound to $0$?
 A: Continuing the idea in your edit...
We want to show $P(R_n>\newcommand\e{\varepsilon}\e\sqrt n)\to 0$ for all $\e>0$.
Fix a particular $n$. Call an number $x$ large if $x>\frac12\e\sqrt n$, and small otherwise. The set $\{X_1,\dots,X_n\}$ certainly contains at most $\frac12\e\sqrt n$ small numbers, since the only possibilities for a small number are $\{1,\dots,\frac12\e\sqrt n\}$. This means that in order for the event $\{R_n>\e \sqrt n\}$ to occur, there need to be at least $\frac12 \e \sqrt n$ large numbers appearing in the sample. Let $p_n$ be the probability that $X_i$ is large, that is,
$$p_n=P(X_1>\tfrac12 \e \sqrt n).$$
Each $X_i$ is large with probability $p_n$, independently of the other variables. This means that the number of large numbers is binomially distributed ($n$ numbers, each independently large with probability $p_n$). The number of distinct large numbers less than or equal to the number of large numbers, because there may be repeats.
Putting this altogether,
$$
P(R_n>\e\sqrt n)\le P(\text{Bin}(n,p_n)>\tfrac12\e\sqrt n)\stackrel{\text{Markov}}{\le} \frac{\Bbb E[\text{Bin}(n,p_n)]}{\tfrac12\e\sqrt{n}}=\tfrac 2\e \cdot {p_n\sqrt n}
$$
The first inequality follows since $R_n>\frac12 \e\sqrt n$ can only occur if at least $\frac12 \e \sqrt n$ of the $X_i$ in are large, and the second inequality is the Markov inequality $P(Y>a)\le E[Y]/a$, where $Y$ is the number of indices for which $X_i$ is large.
It therefore suffices to show $p_n\sqrt n \to 0$ as $n\to\infty$. This can be proved with the dominated convergence theorem, the details of which I leave to you, dear reader.
Further small hint:

 In Durrett's probability text, one of the early exercises asks you to prove $aP(X>a)\to 0$ as $a\to+\infty$ when $E|X|<\infty$.

