I am following a course of real analysis and my teacher, while proving the continuity of translation of functions in $L^p$, used the dominated convergence theorem (DCT) in a strange way. I write the first half of proof, for being clear.
Given $f \in L^p(E)$, with $E$ measurable and $1\le p< \infty$, then for every $\varepsilon >0$ exists $\delta=\delta(\varepsilon)$ such that $ \Vert T_hf-f\Vert_p<\varepsilon \ \ \text{ if}\ \ \vert h \vert < \delta $ Where the operator $T_hf$ is defined by $ T_hf(x)= \begin{cases} f(x+h) & \text{if}\ x+h \in E, \\ 0 & \text{if}\ x+h \in \mathbb{R}^N-E \end{cases} $
Proof.First we may suppose $f$ continuous and compactly supported (and at the end of the proof uses the density of such functions in $L^p$). For all $x \in E$ we have that $ \vert T_hf(x)- f(x)\vert^p=\vert f(x+h)-f(x) \vert^p \to 0 \ \ \text{if}\ \ \vert h \vert \to 0$ Now $f$ is continuous and compactly supported, so $\Vert f \Vert_\infty < \infty$, than we can write $ \vert T_hf(x)- f(x)\vert^p \le \left(\vert f(x+h)\vert+\vert f(x) \vert\right)^p \le 2^p \Vert f \Vert_\infty$ So, letting $K$ be the compact support of $f$ and $\vert h\vert < \delta$, we have that the function $\vert T_hf-f \vert$ vanishes outside the set $K_\delta=K+B(0,\delta)$ (the ball centered in $0$ with radius $\delta$), that is measurable. So we can dominate $ \vert T_hf-f \vert \le 2^p \Vert f \Vert_\infty \chi_{K_\delta}$ Now using the DCT we can say that $\Vert T_hf-f \Vert_p <\varepsilon$ for $|h|<\delta$. For $f \in L^p(E)$ generic we use the density of continuous and compactly supported functions.
The DCT is true for numerable families of functions $\{f_n \}$, but here is used for a non-countable family $\{f_t\}_{t \in I \subseteq \mathbb{R}}$, such $T_hf$. I managed as follow for giving a sense to the proof.
The DCT states that if we have a sequences of measurable function $f_n$ such that $f_n \to f$ pointwise, and all functions $f_n$ are "dominated" by a summable function $g$, i.e $|f_n(x)| \le |g(x)|, \ \forall n, \ \forall x$, so $\lim_{n \to \infty} \int_{E} f_{n} d \mu= \int_{E} f d \mu$ Now we have a subset $I$ of real numbers (we suppose for semplicity $I$ as interval), and a family of function $\{f_t\}_{t \in I}$ (dominated by a summable function $g$). Consider the function $ \begin{aligned} \phi \colon I &\longrightarrow \mathbb{R} \\ t &\longmapsto \int_{E}f_t d \mu \end{aligned} $ Now suppose we have the property that, fixed $\bar t \in I$, for every sequences $\{t_n\} \subseteq I$ such that $t_n \to \bar t$ as $n$ increase, result that $f_{t_n} \to f_{t}$ poinwise. By DCT we have that $\lim_{n\to \infty}\phi(t_n) = \phi(\bar t)$ for every sequence $t_n$ convergent to $\bar t$. So for a well known theorem of topology, we have $ \lim_{t \to \bar t} \phi(t)=\phi(\bar t) $
Choosing $\phi(h)=T_hf$ and $\bar t=0$ we have concluded the problem.
Because I don't have practice with $L^p$ space my question is if the above reasoning is right or not. My teacher use that fact easy, without long arguments, so there may be a faster reasoning way?
Thank you a lot.