Strong L1-convergence and continuous convergence We recall that a sequence of functions $(v_{\varepsilon})$ continuously converge on $(0,1)$ to a function $v_0$ as $\varepsilon\rightarrow 0$ if it holds that: for every sequence $(s_{\varepsilon})$ in $(0,1)$ such that $s_{\varepsilon}\rightarrow s_0$ we have $v_{\varepsilon}(s_{\varepsilon})\rightarrow v_0(s_0)$. Continuous convergence implies pointwise convergence. On the other hand, strong L1 convergence implies almost everywhere convergence (up to a subsequence). My question is: Is it possible to establish a result which provides (under some reasonable conditions on $v_{\varepsilon}$) that 
strong L1 convergence of $v_{\varepsilon}$ implies continuous convergence (up to a subsequence and/or up to a zero-measure set in $(0,1)$)? (We do not expect that L1 convergence implies continuous convergence for general sequence $(v_{\varepsilon})$). More precisely, I am interested in proving the following weaker and more specific result: 
Assume that a sequence of functions $v_{\varepsilon}$ in ${\rm H}^2(0,1)$ satisfies $|v_{\varepsilon}|_{{\rm L}^1(0,1)}\leq M\varepsilon$ (and, for convenience, $v_{\varepsilon}(0)=0$). Consider a sequence of strictly positive real numbers $\rho_{\varepsilon}$ such that  $\lim_{\varepsilon\rightarrow 0}\rho_{\varepsilon}=0$ and $\lim_{\varepsilon\rightarrow 0}\rho_{\varepsilon}^{-1}\varepsilon=0$. Then there exists $0<\delta<1$ and two sequences $(s_{\varepsilon}^{(1)})$ and $(s_{\varepsilon}^{(2)})$ with the properties: 
(1) $c_1\delta\rho_{\varepsilon}\leq |s_{\varepsilon}^{(1)}-s_{\varepsilon}^{(2)}|$ for every $\varepsilon\leq\varepsilon_0(\delta)$ and some $c_1>0$, 
(2) $\rho_{\varepsilon}^{-1}|v_{\varepsilon}(s_{\varepsilon}^{(i)})|\rightarrow 0$ as $\varepsilon\rightarrow 0$ for $i=1,2$. 
Remark: Here $v_0=0$. In the simplified version of the problem, by assumption $v_{\varepsilon}(0)=0$, we can choose $s_{\varepsilon}^{(2)}:=0$ (assumption $v_{\varepsilon}(0)=0$ is probably not essential), and the problem is reduced to finding the sequence $(s_{\varepsilon}^{(1)})$ which satisfies $\rho_{\varepsilon}^{-1}|v_{\varepsilon}(s_{\varepsilon}^{(1)})|\rightarrow 0$ as $\varepsilon\rightarrow 0$, which is much weaker property then the actual continuous convergence of $(v_{\varepsilon})$ to $v_0=0$ since we require only existence of a single sequence $(s_{\varepsilon}^{(1)})$ which satisfies (1) and  $\rho_{\varepsilon}^{-1}v_{\varepsilon}(s_{\varepsilon}^{(1)})\rightarrow 0$ (and thus $v_{\varepsilon}(s_{\varepsilon}^{(1)})\rightarrow 0$).  Hint: Note that, by assumption, we have $\rho_{\varepsilon}^{-1}|v_{\varepsilon}|_{{\rm L}^1(0,1)}\rightarrow 0$ as $\varepsilon\rightarrow 0$. Then, by the integral mean value theorem we can find $(s_{\varepsilon}^{(1)})$ which satisfies (2), but how to satisfy condition (1)?
 A: I will answer the second, specific part of the question. To find sequences $(s_{\varepsilon}^{(i)})$, $i=1,2$, it suffices to consider a subinterval in $(0,1)$ $J:=(a_{\varepsilon}, b_{\varepsilon})$, such that $|b_{\varepsilon}-a_{\varepsilon}|=c_2\delta$ and to divide $J$ into two parts: $J_1:=(a_{\varepsilon}, a_{\varepsilon}+{1\over{4}}|b_{\varepsilon}-a_{\varepsilon}|)$ and $J_2:=(a_{\varepsilon}+{3\over{4}}|b_{\varepsilon}-a_{\varepsilon}|,b_{\varepsilon})$. Then, since $\rho_{\varepsilon}^{-1}|v_{\varepsilon}|_{{\rm L}^1(J_i)}\leq M\rho_{\varepsilon}^{-1}\varepsilon\rightarrow 0$, by the integral mean value theorem we can find $s_{\varepsilon}^{(i)}\in J_i$ such that (2) holds, while by the construction we get 
$c_2\delta\geq|s_{\varepsilon}^{(1)}-s_{\varepsilon}^{(2)}|\geq{1\over{2}}c_2\delta$, and we can choose $c_1:={1\over{2}}c_2$ and $0<\delta<1$ to obtain (1). Hence, (1) and (2) are satisfied, and on top of that we have the upper bound in (2). This seems to suggest that there are many sequences $(s_{\varepsilon}^{(i)})$ in $(0,1)$ which satisfy (1) and (2). But then, I expect that the similar argument can probably be used to prove that strong L1 convergence implies (up to a subsequence) "almost everywhere" continuous convergence. Is this true?
