Norm continuity in L^p space? I'm trying to solve the following problem in Stein and Shakarchi's Functional Analysis book:

Suppose $1 \leq p < \infty$ and that $\mathbb{R}^d$ is equipped with Lebesgue measure. Show the if $f \in L^p(\mathbb{R})$,then $$||f(x+h)-f(x)||_{L^p} \to 0$$ as $|h| \to 0$.

Since continuous functions with compact support are dense in $L^p(\mathbb{R})$, I was thinking of using continuous function $g$ to approximate, but I couldn't get anywhere. Could someone help me out? Thanks!
 A: Let us write that norm out, actually omitting the $p$-th root because it's irrelevant for "tends to 0":
$$\|f(x+h)-f(x)\|_p^p=\int_{\mathbb{R}^d}|f(x+h)-f(x)|^pdx.$$
First assume $f\in\mathcal{C}_c(\mathbb{R}^d)$. That integral then is only on the union of $\mathrm{supp}(f)$ and $\mathrm{supp}(f)-h$, for outside that the integrand is zero. If we assume $|h|<\delta$ for some $\delta$, we can enlarge that to an integral over $X_\delta:=\{x:d(x,\mathrm{supp}(f))<\delta$, for both pieces of the union are included in that. This is a compact set, and now it only depends on $\delta$, not directly on $h$, so we can try dominated convergence. By continuity, for all $\varepsilon$ we can find $\delta$ such that $|h|<\delta\implies|f(x+h)-f(x)|<\varepsilon$. So if $|h|<\delta$, the integrand is always less than $\varepsilon^p$, and the integral will be less than $\epsilon^p\lg(X_\delta)$, so:
$$\|f(x+h)-f(x)\|_p\leq\varepsilon\sqrt[p]{\lg(X_\delta)}.$$
For $\delta\to0$, the RHS goes to zero because $X_\delta$ tends to be a single point, so for $h\to0$ we can bound the LHS from above with the RHS that goes to zero, hence the LHS goes to zero, as we wanted. If $f$ is not continuous with compact support, by the density theorem we find $g_\epsilon$ such that $\|f-g_\epsilon\|_p<\epsilon$, for any $\epsilon$, and then:
$$\|f(x+h)-f(x)\|_p\leq\|f(x+h)-g_\epsilon(x+h)\|_p+\|g_\epsilon(x+h)-g_\epsilon(x)\|_p+\|g_\epsilon(x)-f(x)\|_p,$$
but the middle term tends to zero and the other two are both equal to $\|f-g_\epsilon\|$, so basically we have bounded our norm from above with any $3\epsilon$, any time $h$ is chosen such that $\|g_\epsilon(x+h)-g_\epsilon(x)\|_p\leq\epsilon$. Hence, the norm goes to zero, completing the proof.
