Let $f \in L_{\mathbb{R}^+}^1(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d),\lambda_d).$ Consider the sequence $f_k=\min(f,k)\min(1,\max(0;k+1-|x|))$ to show the continuity in the mean: $$\lim_{y \to 0}\int_{\mathbb{R}^d}|f(x+y)-f(x)|dx=0.$$ This result remains true if we take a Radon measure on $(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d))$?
Is the following attempt correct:
Let $\phi_k(x)=\min(1,\max(0;k+1-|x|)),$ $\phi_k$ is continuous with compact support. Let $(y_n)_n$ be a sequence converging to $0,$ so there exist $n_0$ such that for all $n \geq n_0,|y_n| \leq 1$
$$\int_{\mathbb{R}^d}|f(x+y_n)-f(x)|dx \leq2 \int_{\mathbb{R}^d}|f_k(x)-f(x)|dx+\int_{\mathbb{R}^d}|f_k(x+y_n)-f_k(x)|dx$$ So, $$\int_{\mathbb{R}^d}|f(x+y_n)-f(x)|dx \leq2 \int_{\mathbb{R}^d}|f_k(x)-f(x)|dx+k\int_{\mathbb{R}^d}|\phi_k(x+y_n)-\phi_k(x)|dx$$ taking $\limsup_n,$ we have $\int_{\mathbb{R}^d}|\phi_k(x+y_n)-\phi_k(x)|dx \to 0$ using the dominated convergence theorem (since $\phi_k$ are continuous with compact support),
then taking $k \to \infty$ and again using the dominated convergence theorem, we conclude that $$\lim_n \int_{\mathbb{R}^d}|f(x+y_n)-f(x)|dx=0$$
It'd help to know what level you're at to know what level of detail would be good to comment on this!
Looks good to me but here are a few suggestions. If you're allowed to its a lot quicker to just use density of continuous functions with compact support are dense in $L^1$. However you've essentially proved this using your $\phi$ functions. Secondly I'd just make a few points a bit clearer, your first integral inequality could have one more intermediate line and when you use dominated convergence at the end its not clear which limit you are doing first are you sending $k$ to infinity first and then $n$ or the other way round. One way leads to a horrible mess the other is fine.
Anyway good job looks good to me mathematically.