Suppose $X_{k}$ and $X_{k'}$ are i.i.d random variables and that $n^{-1/2}\sum_{k=1}^{n}X_{k}\Rightarrow Z$ for a real-valued random variable $Z$.
I have showed that
$(1)$ $Y_{k}=X_{k}-X_{k}'$ satisfies $n^{-1/2}\sum_{k=1}^{n}Y_{k}\Rightarrow Z-Z'$, where $Z$ and $Z'$ are i.i.d.
$(2)$ For $m>0$, let $U_{k}:=Y_{k}\mathbb{1}_{|Y_{k}|\leq m}$ and $V_{k}=Y_{k}\mathbb{1}_{|Y_{k}|>m}$. Then, for every $u<\infty$ and all $n$, we have $$\mathbb{P}\Big(\sum_{k=1}^{n}Y_{k}\geq u\sqrt{n}\Big)\geq\mathbb{P}\Big(\sum_{k=1}^{n}U_{k}\geq u\sqrt{n}, \sum_{k=1}^{n}V_{k}\geq 0\Big)\geq\dfrac{1}{2}\mathbb{P}\Big(\sum_{k=1}^{n}U_{k}\geq u\sqrt{n}\Big).$$
Now, I want to apply CLT to prove that $$n^{-1/2}\sum_{k=1}^{n}U_{k}\Rightarrow \mathcal{N}(0,\mathbb{E}U_{1}^{2})$$
To do this, I set $N_{k}:=\dfrac{U_{k}}{\sqrt{\mathbb{E}U_{1}^{2}}}$, for each $1\leq k\leq n$, and to apply i.i.d CLT to $n^{-1/2}\sum_{k=1}^{n}N_{k}$. ($N_{k}$ are i.i.d since $Y_{k}$ are i.i.d)
Then if I showed $$n^{-1/2}\sum_{k=1}^{n}N_{k}\Rightarrow\mathcal{N}(0,1),$$ then I would have $$n^{-1/2}\sum_{k=1}^{n}U_{k}=\sqrt{\mathbb{E}U_{1}^{2}}n^{-1/2}\sum_{k=1}^{n}N_{k}\Rightarrow\sqrt{\mathbb{E}U_{1}^{2}}\mathcal{N}(0,1)$$
To use CLT, I need to prove $\mathbb{E}N_{k}=0$, $Var(N_{k})=1$ and $$(N_{1}+\cdots+N_{n})/\longrightarrow_{p} 0\ \text{in probability}.$$
The second and third are easy to prove, but I don't know how to show $\mathbb{E}N_{k}=0$, which is equivalently showing $\mathbb{E}U_{k}=0$, for which we need to show $\mathbb{E}Y_{k}=0$, I know that $\mathbb{E}Y_{k}=\mathbb{E}X_{k}-\mathbb{E}X_{k}'$, but this question does not assume the integrability of $X_{k}$ and $X_{k}'$, so if $\mathbb{E}X_{k}=\infty$, we cannot conclude $\mathbb{E}Y_{k}=0$, because I cannot say $\infty-\infty=0$.
What can I do here?
I think you are worrying too much: