# Existence of a `right' sequence

Suppose that $$f_n: \mathbb{R} \times [0,\infty) \to \mathbb{R}$$ is a sequence of bounded variation functions such that there exists $$C>0$$: $$|f_n| < C$$ for every $$n$$ and, as $$n \to \infty$$, for $$t>0$$, $$\begin{equation} \tag{1} f_n(\cdot,t) \longrightarrow f(\cdot,t), \quad \text{in } L^1(\mathbb{R}). \end{equation}$$ Let $$x_n: [0,\infty) \to \mathbb{R}$$ be a sequence of absolutely continuous functions that satisfies $$\begin{equation} \tag{2} \frac{d}{dt} x_n(t) = f_n(x_n(t),t), \quad \text{a.e.} - t, \; x_n(0)=0, \end{equation}$$ and $$x_n(\cdot) \to x(\cdot)$$ uniformly in any bounded interval in $$[0,\infty)$$, for some $$x: [0,\infty) \to \mathbb{R}$$.

Is the following claim true?

Claim. For a.e.-$$t$$, there exists a sequence $$y_n$$ such that, $$y_n \ge x_n (t)$$, $$y_n \to x(t)$$ and $$f_n(y_n,t) \to f(x(t)+,t),$$ where $$f(x(t)+,t)$$ stands for the right limit in the nomal sense.

Remark. This is an attempt to solve a slightly stronger version of this problem.