Problem: Let $f \in L^1(\mathbb{R},~\mu)$, where $\mu$ is the Lebesgue measure. For any $h \in \mathbb{R}$, define $f_h : \mathbb{R} \rightarrow \mathbb{R}$ by $f_h(x) = f(x - h)$. Prove that: $\lim_{h \rightarrow 0} \|f - f_h\|_{L^1} = 0.$
My attempt: So, I know that given $\epsilon > 0$, we can find a continuous function $g : \mathbb{R} \rightarrow \mathbb{R}$ with compact support such that $\int_{\mathbb{R}} |f - g|d\mu < \epsilon.$ We can then use the inequality $|f - f_h| \leq |f - g| + |g - g_h| + |g_h - f_h|$ to reduce the problem to the continuous case, so to speak, since the integral of the first and last terms will be $< \epsilon$. But now I'm stuck trying to show that $\lim_{h \rightarrow 0} \|g - g_h\|_{L^1} = 0.$ I tried taking a sequence $(h_n)_{n \in \mathbb{N}}$ converging to $0$ and considering $g_n := g_{h_n}$, but I don't have monotonicity and the convergence doesn't seem to be dominated either, so I don't know what to do.
Any help appreciated. Thanks.