Suppose that $Z,Z\_1,Z\_2,\ldots$ are iid random variables such that $\operatorname EZ=0$, $\operatorname EZ^2=1$ and $\operatorname E|Z|^s<\infty$ with some $s>2$. Let $\tilde Z\_t=Z\_tI\_{\{|Z\_t|\le n^{1/s} \}}$ with $1\le t\le n$ be the triangular array of the truncated random variables, where $I\_A$ is the indicator function of a set $A$. Denote the periodograms
$$
I\_{n,Z}(\lambda)=\Bigl|n^{-1/2}\sum\_{t=1}^n\exp(-i\lambda t)Z\_t\Bigr|^2
\quad\text{and}\quad
I\_{n,\tilde Z}(\lambda)=\Bigl|n^{-1/2}\sum\_{t=1}^n\exp(-i\lambda t)\tilde Z\_t\Bigr|^2
$$
for $\lambda\in[0,\pi]$.



> 
> I would like to show that
>  \begin{equation}\label{eq1}\tag{#}
> \max\_{1\le j\le q}I\_{n,Z}(\omega\_j)-\max\_{1\le j\le q}I\_{n,\tilde Z^{(n)}}(\omega\_j)\to0
> \end{equation}
>  a.s. as $n\to\infty$, where $\omega\_j=2\pi j/n$ with $j=1,\ldots q$ and $q=\lfloor(n-1)/2\rfloor$. Intuitively speaking, this means that the truncation does not affect the asymptotic behaviour of the maximum of a periodogram.
> 
> 
> 


I am trying to verify a proof of this statement, which is given [here](https://projecteuclid.org/euclid.aop/1022677270) (see Lemma 3.3). The proof establishes that
\begin{equation}\label{eq2}\tag{\*}
\sum\_{t=1}^n|Z\_t-\tilde Z\_t|=\sum\_{t=1}^n|Z\_t|I\_{\{|Z\_t|>n^{1/s}\}}
\end{equation}
converges to $0$ almost surely as $n\to\infty$ and then it is claimed that the periodograms of the sequences $Z\_1,\ldots,Z\_n$ and $\tilde Z\_1,\ldots,\tilde Z\_n$ have to be identical identical a.s. for all $n$ sufficiently large. However, I am struggling to see why \eqref{eq2} implies \eqref{eq1}. In my opinion, \eqref{eq2} implies that
$$
\max\_{1\le j\le q}\Bigl|n^{-1/2}\sum\_{t=1}^n\exp(-i\omega\_jt)Z\_t-n^{-1/2}\sum\_{t=1}^n\exp(-i\omega\_jt)\tilde Z\_t\Bigr|^2
\le\Bigl|n^{-1/2}\sum\_{t=1}^n|Z\_t-\tilde Z\_t|\Bigr|^2\to0
$$
a.s. as $n\to\infty$ and I do not see a way to conlcude that \eqref{eq1} holds. Maybe I am missing something trivial or maybe I do not understand the proof properly. So does \eqref{eq2} imply that \eqref{eq1} holds?


Any help is greatly appreciated!



Maximum of the periodogram of a truncated sequence
Let $A\_N$ be the event defined by 
$$
A\_N:=\bigcup\_{n=2^{N-1}+1}^{2^N}\left\{\max\_{1\le j\le q}I\_{n,Z}(\omega\_j)\neq \max\_{1\le j\le q}I\_{n,\tilde Z^{(n)}}(\omega\_j)\right\}.
$$
Then the following inclusion holds
$$
A\_N\subset \bigcup\_{n=2^{N-1}+1}^{2^N}\bigcup\_{t=1}^n\left\{ Z\_t\neq \widetilde{Z\_t}^{(n)} \right\}= \bigcup\_{n=2^{N-1}+1}^{2^N}\bigcup\_{t=1}^n\left\{ \left\lvert Z\_t\right\rvert\gt n^{1/s} \right\}=\bigcup\_{t=1}^{2^N}\left\{ \left\lvert Z\_t\right\rvert\gt \left(\max\left\{2^{N-1}+1,t\right\}\right)^{1/s} \right\}.
$$
Using the fact that the random variables $Z\_t,1\leqslant t\leqslant n$, have the same distribution, we infer that 
$$
\Pr\left(A\_N\right)\leqslant 2^N\Pr\left\{ \left\lvert Z\_1\right\rvert\gt 2^{\frac{N-1}s} \right\}
$$
and since $\mathbb E\left\lvert Z\right\rvert^s$ is finite, we derive finiteness of $\sum\_{N=1}^{+\infty}\Pr\left(A\_N\right)$ and the Borel-Cantelli lemma allows to conclude.


