Continuity of $L^1$ functions with respect to translation Let $f\in L^1$, consider the map $t\mapsto f_t=f(x-t)$, then how can one show that $t\mapsto f_t$ is continuous? More explicitly one wants to show that $\lim_{h\to 0}|f_{t+h}-f_t|_{L^1}=0$. I tried to use approximation by $C_0(\mathbb{R})$ functions $g^n$ to approximate $f$ in $L^1$ norm. Then one has $\lim_{h\to 0}|g^n_{t+h}-g^n_t|_{L^1}=0$, but then I came across the problem: how can one show that the two limits can exchange so that one has $$\lim_{h\to 0}|f_{t+h}-f_t|_{L^1}=\lim_{h\to 0}\lim_{n\to\infty}|g^n_{t+h}-g^n_t|_{L^1}=\lim_{n\to\infty}\lim_{h\to 0}|g^n_{t+h}-g^n_t|_{L^1}=0.$$
Can someone help me with some conditions on which two limits can be exchanged, or do you have a better way of proving the continuity? Thank you!
 A: Let $g_n$ be a sequence in $C_0(\mathbb{R})$ such that $g_n\to f$ in $L^1(\Omega)$. Note that $$|f(x-t)-f(x)|\leq |f(x-t)-g_n(x-t)|+|g_n(x-t)-g_n(x)|+|g_n(x)-f(x)|$$
Can you finish?
A: The translations $\tau_t$, where $\tau_t(f)(x) = f(x-t)$ are isometries of $L^1$ (more genral, any $L^p$).
Thus the family $T = \{\tau_t : t \in \mathbb{R}^n\}$ is equicontinuous.
For $f \in C_c(\mathbb{R}^n)$, we have $\lim\limits_{t \to 0} \tau_t(f) = f$ in $L^p$ by the uniform continuity of $f$. For  $1 \leqslant p < \infty$, $C_c(\mathbb{R}^n)$ is dense in $L^p$.
By the theorem of Ascoli (-Bourbaki), on a family $\mathcal{F}$ of equicontinuous functions $\varphi \colon X \to Y$, the topology of pointwise convergence on $X$, the topology of pointwise convergence on a (fixed) dense subset of $X$ and the topology of uniform convergence on compact subsets of $X$ coincide.
Hence not only $\lim\limits_{t\to 0} \tau_t(f) = f$ for all $f \in L^p$ [$p < \infty$, again], the convergence is even uniform on all compact subsets of $L^p$.

If you prefer a direct argument without citing topological theorems, let $\varepsilon > 0$. Find a $g \in C_c(\mathbb{R}^n)$ with $\lVert f - g\rVert_1 < \varepsilon/3$. By the uniform continuity of $g$, you have a $\delta > 0$ such that $\lVert \tau_h(g) - g\rVert_1 < \varepsilon/3$ whenever $\lVert h\rVert < \delta$. Then, for $\lVert h \rVert < \delta$, you have
$$\lVert \tau_h(f) - f\rVert_1 \leqslant \lVert \tau_h(f) - \tau_h(g)\rVert_1 + \lVert \tau_h(g) - g\rVert_1 + \lVert g - f \rVert_1 < \varepsilon.$$
