$L^p$ convergence of composition Let $f \in L^{p}(U)$ and $(t_{n})$ be a sequence of functions from $U$ to $U$ such that $$t_{n} \rightarrow \operatorname{id},$$
in the space $L^{\infty}(U)$. 
Does the following hold:
$$f \circ t_{n} \rightarrow f,$$
in the space $L^{p}(U)$?
I tried using the dominated convergence theorem for $p_{n}(x)=\left|f(x) - f(t_{n}(x))\right|^{p}$ to show that the statement is true, but I couldn't find the "dominating" function. 
 A: For the convergence to happen, it is important to know how $t_n$ mixes the domain $U$. And there is a subtle technical issue associated with the definition of $L^p$. Let $U=[0,1]$ equipped with the Lebesgue measure and $t_n$ be
$$
t_n(x) = \frac{k}{2^n}
$$ for $x\in [\frac{k}{2^n},\frac{k+1}{2^n})$ for $k=0,1,\ldots,2^n$. We can easily see that $t_n \to \operatorname{id}$ in $L^\infty$. An issue here is that
$$
f\mapsto  f\circ t_n
$$ is not well-defined on $L^p([0,1])$. To see this, recall that $L^p([0,1])$ space is not a family of functions, but a family of equivalence classes of functions. Pick $0 = 1_\mathbb{Q} $ in $L^p([0,1])$. The composition with $t_n$ gives incompatible results that
$$
1_\mathbb{Q}\circ t_n (x) = 1,
$$ while
$$
0\circ t_n(x) = 0. 
$$ Therefore, we see that composition may not be well-defined in general. 
Instead of $L^p$ of equivalence classes, we may adopt the space 
$$
\mathcal{L}^p = \{f:U\to \mathbb{C}\;|\;f\text{ is measurable},\int_U|f|^p dx <\infty\}
$$ of actual functions. However, we can see that the convergence can also fail in this case by the example
$$
1_\mathbb{Q}\circ t_n = 1 \not\to 1_\mathbb{Q}.
$$ Thus, unfortunately, we conclude that mixing up the domain does not work properly in general. To properly define composition, some additional conditions such as preserving measure should be imposed on $t_n$.
