Let $f \in L^{p}(\mathbb{R})$, for $h>0$ define:
$f_{h}(x)=\frac{1}{h}\int_{x}^{x+h}f(t)dt$
Show that $f_{h}$ is continuous and that continuous functions are dense in $L^{p}(\mathbb{R})$, by showing $||f_{h}-f||_{p} \to 0$ as $h \to 0$.
I know this seems like a really easy question, but I don't know where to start. What would be the best way to approach this problem?
By the integral form of Minkowski's inequality $\|f_h-g_h||_p <\epsilon$ if $\|f-g\|_p <\epsilon$. Any $f$ in $L^{p}$ can be approximated in the $L^{p}$ norm by a bounded measurable function with compact support. [Just look at $fI_A$ where $A=\{x:|f(x)| \leq N,|x|\leq N\}$ with $N$ large enough]. Hence it suffices to prove the result when $f$ is a bounded measurable function with compact support. In that case we can use Lebesgue's Theorem (which tells you that $f_h \to f$ almost everywhere ) along with DCT to prove that $\|f_h-f||_p \to 0$ as $h \to 0$. Let me know if a more detailed answer is needed.