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.
 A: 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.
A: I realized that I was simply running into a logical fallacy: I was thinking that $m(K)$ could not have been found before setting the $\delta$. However, that is simply not true. Assuming $|h| < 1$ could always be done and hence finding the value of $m(K)$ is always possible regardless choosing that final $\delta$ or not. i.e. The value of $m(K)$ can be predetermined. Therefore, it is fine to choose $\delta = \{ \delta', 1 \}$ with $\delta' > 0$ chosen so that $|S_hg_N - g_N|_\infty < \frac{\epsilon}{m(K)}$ (due to uniform continuity), and it wouldn't be cyclical.
