T_0$. Since $x^n$ satisfy $(\ref{cs})_1$ and $x^n\to x$ in $(C^1([0,t]))^{Nd}$, then $x$ satisfies $(\ref{cs})_1$ with $v=\lim_{n\to\infty}v^n$. Now for $\phi\in (C_c^\infty([0,t]))^d$, we have $$\begin{aligned}
\int_0^{t}v_i^n\dot{\phi} ds =-\int_0^{t}\dot{v_i^n}\phi ds=-\int_0^{t}\frac{1}{N}\sum_{k=1}^N(v_k^n-v_i^n)\psi_n(|x_i^n-x_k^n|)\phi ds\end{aligned}$$ and the left-hand side converges to $\int_0^{t} v\dot{\phi} ds$. Thus it remains to show that the right-hand side converges to $[-\int_0^{t}\dot{v}\phi ds]$, where $$\begin{aligned}
\label{podzial}
\dot{v}:=\frac{1}{N}\sum_{k=1}^N(v_k-v_i)\psi(|x_i-x_k|).\end{aligned}$$ To this end we require for example that $\dot{v}^n\rightharpoonup\dot{v}$ in $(L^1([0,t]))^{Nd}$, which follows from Lemma \[weakstrong\] applied to functions $f_n=v^n_k-v^n_i$, $f=v_k-v_i$, $g_n=\psi(|x^n_i-x^n_k|)$, $g=\psi(|x_i-x_k|)$.

After arriving at (\[podzial\]) we may apply a stronger argument that $\dot{v^n}\to \dot{v}$ in $(L^1([0,t]))^{Nd}$. Clearly $x^n\to x$ and $v^n\to v$ a.e. and thus also $\dot{v^n}\to\dot{v}$ a.e.. Moreover $\dot{v^n}\in (L^1([0,t]))^{Nd}$ for all $n$. Therefore if we show that functions $v^n$ are uniformly integrable, then by Lemma \[vit\] the proof will be finished. Since $\dot{v^n}$ are uniformly bounded in $[\delta,t]$ for all $\delta>0$ (as there are no collisions in $(0,T_0)$[^1]), then the set $\{0\leq s\leq t: |\dot{v^n}|>c\}$ is included in some interval $[0,s_c]$ with $s_c\to 0$ as $c\to\infty$. Therefore by Lemma \[equi\] we get $$\begin{aligned}
\int_0^{t}|\dot{v^n}|\chi_{\{s:|\dot{v^n}|>c\}} ds\leq \int_0^{s_c}|\dot{v^n}|ds\leq L|s_c|^{1-\alpha}\to 0\end{aligned}$$ as $c\to\infty$, which proves that $\dot{v^n}$ are uniformly integrable.

As our last effort in this section let us make an obvious remark involving properties 