A Question on Epsilon-Delta Proofs in Measure Theory

I am currently stuck on rewriting a proof for a proposition in Measure Theory. In particular, I am a little bit confused about the validity of my proof. I think the confusion is originated from the basic epsilon-delta constructions. I will present the statement of the proposition (Revised Version of Folland Proposition 8.5) here and also my attempt:

Proposition:

If $1 \leq p < \infty$, translation is continuous in the in the $L^p$ norm; that is, if $f \in L^p$ and $h \in \mathbb{R}^n$, then $\lim_{y \to 0} \| S_{h}f - f \|_p = 0$.

Note here we define $S_h := f(x + h)$.

Attempt:

Let $f \in L_p(\mathbb{R}^n, m)$, where $1 < p < \infty$. For $h \in \mathbb{R}^n$ define $S_hf(x) = f(x + h)$. We wish to prove that $$ \| S_hf - f \|_p \to 0 $$ as $|h| \to 0$. To show this, we let $\epsilon > 0$ be given. Note that since $C_c(\mathbb{R}^n)$ is dense in $L_p(\mathbb{R}^n, m)$(as $\mathbb{R}^n$ is locally compact Hausdorff and the Lebesgue measure $m$ on $\mathbb{R}^n$ is Radon), there exists a sequence $\{ g_n \}_n \subseteq C_c(\mathbb{R}^n)$ such that $\| f - g_n \|_p \leq \epsilon$ for $n \geq N$, $N \in \mathbb{N}$ large enough. In particular, this means that $\| f - g_N \|_p \leq \epsilon$. Now note that since shifting the function does not change the integral value, we can conclude that we have $\| S_hf - S_hg_N \|_p \leq \epsilon$ for $N \in \mathbb{N}$ large. Moreover, we claim that $\| S_h g_N - g_N \|_p \leq \epsilon$ whenever $|h|$ is small enough. To this end, we consider $$ \| S_h g_N - g_N \|_p^p = \int |S_h g_N - g_N|^p \,dm = \int|g_N(x + h) - g_N(x)|^p \,dm(x) $$ Now note that if $|h| < 1$ for example, we must be able to uniformly find a compact set $K$ such that $$sppt(g_N) + h, sppt(g_N) \subseteq K,$$ for all $h$ such that $|h| < 1$ as we have $sppt(g_N) + h, sppt(g_N)$ to both be compact in $\mathbb{R}^n$ and hence bounded for all $h$ such that $|h| < 1$. Note that this choice of $K$ is independent of $\epsilon > 0$. Moreover, since we have both $g_N \in C_c(\mathbb{R}^n)$, it has to be uniformly continuous. In particular, this means that we have $|S_h g_N - g_N|_\infty < \epsilon$ on $K \subseteq \mathbb{R}^n$ compact for $|h|$ small enough. Therefore, we have $$ \int|g_N(x + h) - g_N(x)|^p \,dm(x) \leq |S_h g_N - g_N|_\infty ^p m(K) \leq \epsilon^p m(K). $$ Note here that $m(K)$ is independent of $\epsilon > 0$ (i.e. as long as we choose $h$ such that $|h| < 1$, such an compact set $K$ exists). Therefore, we have $$ \| S_h f - f \|_p \leq \| S_h f - S_h g_N \|_p + \| S_h g_N - g_N \|_p + \| g_N - f \|_p \leq 2\epsilon + \epsilon m(K)^{\frac{1}{p}} $$ whenever $h$ such that $|h| < 1$ is small enough. Since $\epsilon > 0$ is arbitrary, we are done.

Question:

I am a little bit concerned about my use of ``choosing $K$ independently of $\epsilon$'' here. Intuitively I think it is correct. However, I am having trouble while actually try and write it down in term of $\epsilon-\delta$ definitions. Namely, fixing $\epsilon > 0$, we are always able to choose some compact $K$ so that the properties in the above proof holds true by choosing $|h| < 1$. Moreover, we know there exist $\delta' > 0$ such that for any $|h| < \delta'$, $|S_hg_N - g_N|_\infty < \frac{\epsilon}{m(K)}$. Now we should be able to choose $\delta = \min\{ \delta', 1 \}$ and if $|h| < \delta$, we have $|S_hg_N - g_N|_\infty < \frac{\epsilon}{m(K)}$. It seems like here we need to choose $\delta < 1$ first to guarantee the existence of $K$, and then we can find $\delta'$. However, we need to use $\delta'$ to define $\delta$? It doesn't seem possible to choose $\delta$ to be the minimum of $\delta'$ and $1$ at the same time. What is my logical fallacy here? What is a good way to think about similar proofs like this in the future, so I don't have to be wind up on this next time?

This is my first post here on Stack Exchange, so I am not entirely sure if I have the question format correct. I really hope my question makes sense, please let me know if any more clarification is needed.

Solution 1:

Let $A$ be the subspace in $L_p$ for which your condition holds. $A$ is closed since if $g_k \in A$ and $g_k \to g$ then $\int_X |g(x+h)-g(x)| d \mu \leq \int_X|g(x+h)-g_k(x+h)| + \int_X|g_k(x+h)-g_k(x)|+\int_X|g_k(x)-g(x)|$. The first and last terms are arbitrarily small and the middle term goes to zero by definition. $A$ also contains all the continuous functions with compact support, as you have shown. It follows that $A$ is closed and dense. Hence it must be all of $L_p$.

In general, a common argument in measure theory and analysis is to show that the set $A$ of things with property $p$ is closed and dense, hence it must be the whole space.