I believe it is easier if I print the proof below:
Several other papers uses the same argument to bound $\lvert R_n(x)-E R_n(x) \rvert$ almost surely, so I believe it is correct. It should be clear that the aim of this result is to establish the (uniform) almost sure convergence of $R_n(x)$ to its expectation (but here, drop the uniformity part, for the sake of simplicity). It is not clear to me what "$O(a_n)$ almost surely" exactly means, but since the aim is show the rate that the almost surely convergence occurs, I believe that a suitable definition could be \begin{align} \lvert R_n(x)-E R_n(x) \rvert=O(a_n) \text{ almost surely}\\ \text{ iff } P(\lim\sup_n\{\lvert R_n(x)-E R_n(x) \rvert\leq Ca_n\})=1,\text{ for some }C>0 \end{align} this means that $\lim\sup_n\{\lvert R_n(x)-E R_n(x) \rvert\leq Ca_n\}$ holds $P$-almost everywhere. Please, correct me if you have further information.
As far as I understand, he is trying to give a bound from $\lvert R_n(x)-E R_n(x) \rvert\leq\lvert R_n(x) \rvert +\lvert E R_n(x) \rvert$.
As shown above, $\lvert E R_n(x) \rvert=O(a_n)$, or equivalentely, $\lvert E R_n(x) \rvert\leq C_1 a_n$ for some $C_1>0$ and all large $n$.
Further, he is saying that $\lvert R_n(x) \rvert=0$ almost everywhere (or, with probability one) when $n$ is large enough. This step is not clear to me, and I hope you can help me.
My attempt (Updated)
I'm on this problem for some days, but not suceeded. I'll try something different now.
I want to show that \begin{align} P(\lim\sup_n \{\lvert R_n-ER_n\rvert\leq C a_n\})=1 \end{align} for some $C>0$ with $a_n\to 0$. Note that \begin{align} &w\in\bigg\{\frac{1}{n}\sum_{i=1}^n Y_i1(\vert Y_i\rvert>\tau_n)>0\bigg\}\\ \implies &\exists i\in\{1,\dotsc,n\}:w\in\{\vert Y_i\rvert>\tau_n\}. \end{align} Using this result, the triangle inequality and the fact $\lvert E R_n\rvert=O(a_n)$, for any $n:n>n^*$, \begin{align} &P( \lvert R_n-ER_n\rvert>C_1 a_n) \\ \leq &P( \lvert R_n\rvert+C_1a_n>C_1 a_n)\\ =&P( \lvert R_n\rvert>0)\leq P(\lvert Y_i\rvert>\tau_n \text{ for some }i\in\{1,\dotsc,n\}) \end{align} If $Y_i$ were strictly stationary (he assumed $Y_i$ i.i.d.), then we could put $ P(\lvert Y_i\rvert>\tau_n \text{ for some }i\in\{1,\dotsc,n\}) = P(\lvert Y_n\rvert>\tau_n) $ and so \begin{align} \sum_{i=1}^\infty P( \lvert R_n-ER_n\rvert>C_1 a_n)&\leq n^*+\sum_{i=n^*}^\infty P( \lvert R_n-ER_n\rvert>C_1 a_n)\\ &\leq n^* +\sum_{i=n^*}^\infty P( \lvert Y_n\rvert>\tau_n)<\infty \end{align} The application of Borel-Cantelli's Lemma gives the result. Thanks in advance!
