Does $\lim_{t\to\infty}\int_t^{t+T}|\dot{x}(s)+a(s)x(s)|ds = 0$ imply $\lim_{t\to\infty}\int_t^{t+T}a(s)x(s)ds = 0$? Let $x:[0,\infty)\to[0,\infty)$ be an absolutely continuous function and $a:[0,\infty)\to[0,1]$ be a measurable function. If
\begin{equation}
    \lim_{t\to\infty}\int_t^{t+T}|\dot{x}(s)+a(s)x(s)|ds = 0
\end{equation}
for all $T>0$, then can we say that
\begin{equation}
\lim_{t\to\infty}\int_t^{t+T}a(s)x(s)ds = 0
\end{equation}
for all $T>0$?
 A: By  $ 0<a(s)<1$ and $x(s)>0$ is a simple exercise to proof that 
\begin{equation}
\lim_{t\to\infty}\int_t^{t+T}a(s)x(s)ds = 0\quad 
\mbox{ if, only if }\quad
\lim_{t\to\infty}x(t) = 0
\end{equation}
Once $x$ is a absolutely continuous function, hold $x(t+T)-x(t)=\int_{t}^{t+T}\dot{x}(s) d\xi>0 $. Then for all $T>0$  we too have 
\begin{equation}
\lim_{t\to\infty}\int_{t}^{t+T}\dot{x}(s) d s=0 
\end{equation}
and it's implies ( under the conditions  $T>0$, $x(t)>0$) 
$$ 
\lim_{t\to\infty}x(t)=0
\quad \mbox{ if, only if } \quad
\lim_{t\to\infty}\dot{x}(t)=0 \quad 
\mbox{ for } t\in \mathbb{R}-A
$$
here $A$ is a set whit measure zero. Therefore, applying the inverted triangular inequality 
$$
 \lim_{t\to\infty}\int_t^{t+T}|\dot{x}(s)+a(s)x(s)|ds 
\geq 
 \lim_{t\to\infty}\int_t^{t+T}|\dot{x}(s)| ds
- 
\lim_{t\to\infty}\int_t^{t+T}|a(s)x(s)|ds 
$$
and
$$
 \lim_{t\to\infty}\int_t^{t+T}|\dot{x}(s)+a(s)x(s)|ds 
+
 \lim_{t\to\infty}\int_t^{t+T}|\dot{x}(s)| ds
\geq 
\lim_{t\to\infty}\int_t^{t+T}|a(s)x(s)|ds 
$$
implies 
$$
 \lim_{t\to\infty}\int_t^{t+T}|a(s)x(s)|ds =0
$$
A: The answer is no. A counter example is as follows.
Define $x:[0,\infty)\to[0,\infty)$ and $a:[0,\infty)\to[0,1]$ as 
\begin{align}
x(t) &= \begin{cases} e^{-1}, & t\in[0,1)\cup [n^2+n+1,(n+1)^2) \\ e^{-1} + (1-e^{-1})(t-n^2)/n, & t\in [n^2,n^2+n) \\ e^{-(t-n^2-n)}, & t\in[n^2+n,n^2+n+1) \end{cases} \\
a(t) &= \begin{cases} 1, & t\in[n^2+n,n^2+n+1) \\ 0, & t\in [0,1)\cup[n^2,n^2+n)\cup[n^2+n+1,(n+1)^2) \end{cases}
\end{align}
for $n=1,2,\dots$. Then,
\begin{align}
\dot{x}(t) &= \begin{cases} 0, & t\in(0,1)\cup (n^2+n+1,(n+1)^2) \\ (1-e^{-1})/n, & t\in (n^2,n^2+n) \\ -e^{-(t-n^2-n)}, & t\in(n^2+n,n^2+n+1) \end{cases} \\
a(t)x(t) &= \begin{cases} e^{-(t-n^2-n)}, & t\in[n^2+n,n^2+n+1) \\ 0, & t\in [0,1)\cup[n^2,n^2+n)\cup[n^2+n+1,(n+1)^2) \end{cases}
\end{align}
, so
\begin{equation}
\dot{x}(t) + a(t)x(t) = \begin{cases} (1-e^{-1})/n, & t\in(n^2,n^2+n) \\ 0, & t\in(0,1)\cup(n^2+n,n^2+n+1)\cup(n^2+n+1,(n+1)^2) \end{cases}
\end{equation}
for $n=1,2,\dots$. It is obvious that $\lim_{t\to\infty}\int_t^{t+T}|\dot{x}(s)+a(s)x(s)|ds = 0$ for all $T>0$ but $\lim_{t\to\infty}\int_t^{t+1}a(s)x(s)ds \neq 0$.
