Show that $\sum_{i=1}^n X_i / \sqrt{n} \Rightarrow W$ implies $EX_1^2 <\infty$ for an i.i.d. sequence $(X_i)$. Let $X_i$ be an i.i.d sequence of random variables such that $\sum_{i=1}^n X_i / \sqrt{n} $ converges in distribution to something. I want to show that $EX_1 ^2 <\infty$. (This is problem 3.4.3 of Durrett.)
The hint in the book suggests that if $EX_1 ^2 =\infty$, we define an independent copy of the sequence, say $X_i'$, and put $Y_i := X_i - X_i'$. Then put $U_i:=Y_i \mathbb 1_{|Y_i |\leq A}$ and $V_i:=Y_i \mathbb 1_{|Y_i |> A}$ for some $A>0$. Finally, for a fixed $K>0$,
$$P(\sum_{i=1}^n Y_i \geq K\sqrt n)\geq P(\sum_{i=1}^n U_i \geq K\sqrt n , \sum_{i=1}^n V_i \geq 0 ) \geq 0.5 P(\sum_{i=1}^n U_i \geq K\sqrt n ) \geq 0.2$$ and  he argues that this is a contradiction by taking $n$ to infinity.
I don't understand the second and third inequality above and the final limit part of his argument. Durrett suggests the inequality $P(\sum_{i=1}^n U_i \geq K\sqrt n , \sum_{i=1}^n V_i \geq 0 ) \geq P(\sum_{i=1}^n U_i \geq K\sqrt n , \sum_{i=1}^n V_i \leq 0 )$ follows from symmetry, but I don't see why (and this does imply the second inequality). Also, he uses the usual CLT in the third inequality, but we don't know whether $EU_1=0$. Likewise, we don't know if $EY_i=0$ hence the final limit process does not seem to work.
Any help is appreciated.

For future visitors, the last argument is as follows;
by independence, $\lim_{n\to \infty}\phi_{\sum_{i=1}^n Y_i /\sqrt{n}} (t)= \phi_{W} (t) \phi_{W} (-t) $ which has value $1$ and continuous at $t=0$, so we know that $\sum_{i=1}^n Y_i /\sqrt{n}$ converges in distribution to something. Finally take $K \to \infty$ to derive a contradiction.
 A: 1st Proof. By symmetry, $ Y'_i = U_i - V_i$ has the same distribution as $Y_i$. Moreover,
$$ Y'_i \mathbf{1}_{\{ |Y'_i| \leq A \}} = U_i \qquad\text{and}\qquad Y'_i \mathbf{1}_{\{ |Y'_i| > A \}} = -V_i. $$
So by replacing $Y_i$'s by $Y'_i$'s, we establish:
$$ \mathbb{P} \left( \sum_{i=1}^{n} U_i \geq K\sqrt{n} \text{ and } \sum_{i=1}^{n} V_i \geq 0 \right)
= \mathbb{P} \left( \sum_{i=1}^{n} U_i \geq K\sqrt{n} \text{ and } \sum_{i=1}^{n} V_i \leq 0 \right). $$

2nd Proof. Write $[n] = \{1, \dots, n\}$ for simplicity, and let $\mathcal{I}$ be the random set of indices defined by
$$\mathcal{I} = \{ i \in [n] : |Y_i| \leq A\}.$$
Then we find that

*

*$\sum_{i=1}^{n} U_i = \sum_{i \in \mathcal{I}} Y_i$ and $\sum_{i=1}^{n} V_i = \sum_{i \in [n]\setminus \mathcal{I}} Y_i$ are conditionally independent given $\mathcal{I}$, and


*The conditional distribution of $\sum_{i=1}^{n} V_i$ given $\mathcal{I}$ is symmetric.
So by the law of iterated expectation,
\begin{align*}
&\mathbb{P} \left( \sum_{i=1}^{n} U_i \geq K\sqrt{n} \text{ and } \sum_{i=1}^{n} V_i \geq 0 \right) \\
&= \mathbb{E} \left[
  \mathbb{P} \left( \sum_{i=1}^{n} U_i \geq K\sqrt{n} \text{ and } \sum_{i=1}^{n} V_i \geq 0 \,\middle|\, \mathcal{I} \right)
  \right] \\
&= \mathbb{E} \left[
    \mathbb{P} \left( \sum_{i=1}^{n} U_i \geq K\sqrt{n} \,\middle|\, \mathcal{I} \right) 
    \mathbb{P} \left(  \sum_{i=1}^{n} V_i \geq 0 \,\middle|\, \mathcal{I} \right)
  \right] \\
&= \mathbb{E} \left[
    \mathbb{P} \left( \sum_{i=1}^{n} U_i \geq K\sqrt{n} \,\middle|\, \mathcal{I} \right) 
    \mathbb{P} \left(  \sum_{i=1}^{n} V_i \leq 0 \,\middle|\, \mathcal{I} \right)
  \right] \\
&= \mathbb{P} \left( \sum_{i=1}^{n} U_i \geq K\sqrt{n} \text{ and } \sum_{i=1}^{n} V_i \leq 0 \right).
\end{align*}
