# The continuous convergence given the a.e. convergence

Suppose that $$f_n: \mathbb{R} \times [0,\infty) \to \mathbb{R}$$ is a uniformly bounded sequence (i.e., there exists $$C>0$$: $$|f_n| < C$$ for every $$n$$) such that $$f_n \in C^2_x \times C^1_t,$$ and, as $$n \to \infty$$, $$$$\tag{1} f_n(x,t) \longrightarrow f(x,t), \quad \text{a.e.} -(x,t) \in \mathbb{R} \times [0,\infty).$$$$ Let $$x_n: [0,\infty) \to \mathbb{R}$$ be a sequence of absolutely continuous functions that satisfies $$$$\tag{2} \frac{d}{dt} x_n(t) = f_n(x_n(t),t), \quad \text{a.e.} - t, \; x_n(0)=0,$$$$ and $$$$\tag{3} x_n(\cdot) \to x(\cdot), \quad \text{uniformly in any bounded interval in } [0,\infty),$$$$ for some function $$x: [0,\infty) \to \mathbb{R}$$.

Question.

1. Is the following statement true: As $$n \to \infty$$, $$$$\tag{4} f_n(x_n(t),t) \longrightarrow f(x(t),t), \quad \text{a.e.} - t \in [0,\infty)?$$$$
2. Is (4) true for any $$x_n$$ not necessarily satisfies (2)?
3. If (4) is wrong, what are additional requirements needed on $$f_n$$ to make it true?

Remark.

• The limit function $$f$$ does not have to be $$C^1$$, even $$C^0$$.
• The conclusion (4) is true if (1) is true uniformly, without the assumption (2).

Updated. The second question is answered negatively thanks to the counterexample by Iosif Pinelis.

The answer is no. E.g., for all natural $$n$$ and all $$(u,t)\in\mathbb R\times[0,\infty)$$, let $$f_n(u,t)=e^{-nu^2}$$, $$f(u,t)=0$$, and $$x_n(t)=x(t)=0$$. Then $$f_n\to f$$ a.e. on $$\mathbb R\times[0,\infty)$$ and $$x_n\to x$$ uniformly on $$[0,\infty)$$, whereas for all $$t\in[0,\infty)$$ $$f_n(x_n(t),t)=1\to1\ne0=f(x(t),t).$$
• Thanks for your negative answer to my first question. So I shouldn't expect too much if I allow such a freedom on $x_n$. What if now $x_n$ is closely related to $f_n$, namely, for every $n$, $\frac{d}{dt} x_n(t) = f_n(x_n(t),t)$, a.e-$t$ with $x_n(0) = x(0)$? Would it be enough to draw a positive conclusion? Commented Aug 29, 2019 at 13:18