Let $f: \mathbb{R}^j \rightarrow \mathbb{R}^j$, a continuous function.
Now, if we assume that $(x_n)_n \subset L^{\infty}([0,1],\mathbb{R}^j)$ that converges to $x \in L^{\infty}([0,1],\mathbb{R}^j)$. We deduce that $f(x_n(t))\rightarrow f(x(t))$ (since $f$ is continuous).
But If we assume that $x_n,x \in L^1([0,1],\mathbb{R}^j)$, do we still have: $f(x_n(t))\rightarrow f(x(t))$ ?
My idea:
Probbaly no, but we have a convergent subsequence $(x_{\phi(n)})_n$ almost everywhere. Hence we have
$$f(x_{\phi(n)}(t))\rightarrow f(x(t)).$$
I want to know if either my statement or my idea are wrong.