Let $\mu$ be a finite regular Borel measure on $\mathbb R^n$ and $A$ is a Borel set. I am trying to prove that $x \mapsto \mu(A+x)$ is continuous. Here $\mu$ is regular means it satisfies assumptions in this link.
In fact, there are several posts on this site similar to this question, but they all assume $\mu$ is ab. continuous w.r.t Lebesgue measure $m$. I found myself a proof without assuming this but couldn't find something wrong with it. Also, there is an answer by John Dawkins poitning out that if $\mu$ is not ab. continuous w.r.t $m$ then $x \mapsto \mu(A+x)$ is not continuous.
I hope you can help me find what's wrong in my proof. Here is it:
Since $\mu$ is finite and regular, $C_c(\mathbb R^n)$ is dense in $L^p(\mathbb R^n, \mu)$. To prove $x \mapsto \mu(A+x)$ is continuous, fix one $y \in \mathbb R^n$ I use
$$ |\mu(A+x) - \mu(A+y)| \leq \int |1_{A+y}(u+y-x)- 1_{A+y}(u)| \mu(du) $$
Let $g \in C_c(\mathbb R^n)$ be such that $ \| g - 1_{A+y} \|_{L^1(\mathbb R^n, \mu)} \leq \epsilon$ and since $g$ is uniformly continuous so when $h = y-x$ is going to 0, we have
\begin{align*} \int |1_{A+y}(u+h)- 1_{A+y}(u)| \mu(du) &\leq \| g - 1_{A+y} \|_{L^1(\mathbb R^n, \mu)} + \int |g(u+h)- g(x)| \mu(du)\\ &\qquad + \int |1_{A+y}(u+h)- g(u+h)| \mu(du) \\ &< 3 \epsilon. \end{align*}
So that $x \mapsto \mu(A+x)$ is continuous. In particular, I would like to chose $\mu = \delta_{0}$, the Dirac measure at 0, and $A$ is an open set so that $ x \mapsto \delta_{0}(A+x):= \phi(x)$ is continuous. However, $\phi^{-1}(-\infty, 1/2) = - \bar A$ is closed, which I suspect something is off but counldn't tell what went wrong
If the measure $\mu$ is not translation invariant, then the third term $\int |1_{A+y}(u+h)-g(u+h)|\,\mu(du)$ is not equal to $\|g-1_{A+y}\|_{L^1(d\mu)}$, and in particular, you can't say it's less than $\epsilon$. An example is $\mu = \delta_0$ and $A = (0,1]^n$. Let $e = \frac{1}{\sqrt n}(e_1+\dots+e_n)$. Then $\int 1_{A}(u+he)\,d\mu = \mu(A-h e) = 1$ for all $h > 0$ sufficiently small, but $\int 1_A(u)\,d\mu = 0$.