Is the $ L^1$ norm continuous in the following sense? Suppose I have $f \in L^1\cap L^\infty$, and I want to take the following limit
$$\lim_{h\to0} \int|f(x+h)-f(x)|dx$$
Does this follow from the dominated convergence theorem? Since $|f(x+h)-f(x)| \leq |f(x+h)|+|f(x)|$, and $\|f(x+h)\|_{L^1(\mathbb{R})}<\infty$ and $\|f(x)\|_{L^1(\mathbb{R})}<\infty$, we then have $$\lim_{h\to0} \int|f(x+h)-f(x)|dx =  \int\lim_{h\to0}|f(x+h)-f(x)|dx = 0$$?
I'm not convinced, and I think a better argument would be to involve the density of continuous function in $L^1$. If the approach in this post does not work, can anyone explain why?
 A: Let $g\in L^{1}$ be continuous such that $\left\|g-f\right\|_{L^{1}(\mathbb{R})}<\epsilon$. Let $N>0$ be sufficiently large so that
$$\max\left\{\int_{[-N,N]^{c}}\left|f\right|,\int_{[-N,N]^{c}}\left|g\right|\right\}<\epsilon$$
for $\epsilon>0$ given. For $0\leq x, h<N$, $\left|g(x+h)\right|\leq \sup_{x\in [-2N,2N]}\left|g(x)\right|=:M$. So $\left|g(x+h)-g(x)\right|\leq 2M$. By the DCT,
$$\lim_{h\rightarrow 0}\int_{-N}^{N}\left|g(x+h)-g(x)\right|dx=0$$
Observe that
$$\int_{[-N,N]^{c}}\left|g(x+h)-g(x)\right|dx\leq 2M\epsilon$$
This should get you started.
A: (I marked the question as a duplicate, but then realised that you were asking about the approach to a proof.)
You can use the DCT as part of the proof, but it doesn't follow directly.
To use the DCT, you need an integrable bound and a limit function.
In the above, you need to provide an integral upper bound that is 'independent' of $h$. And the limit doesn't necessarily exist unless the function is continuous ae. For example, $f(x)=1_{\mathbb{Q}^c}(x) e^{-|x|}$ is nowhere continuous.
An approach using compactly supported continuous approximations is fairly  standard.
