Continuity of $ T(f,y):=\int_A f(x-y)\, dx$

Question

Let us consider a measurable set $A\subset {\mathbb{R}}$ and, for all $f\in L^1(\mathbb{R})$ and $y\in\mathbb{R}$, $$ T(f,y):=\int_A f(x-y)\, dx.$$ How could I prove that the function $y \rightarrow T(f,y)$ is continuous on $\mathbb{R}$?

Answer 1

The function $B: y \in R \mapsto B(y) = (x \mapsto f(x-y)) \in L^1(R)$ is (uniformly) continuous (hint: density of compactly supported functions). Let $T_f$ denote the map $y \mapsto T(f,y)$. Then $T_f = L \circ B$, where $L: g \in L^1(R) \mapsto \int_A g(x)dx$ is obviously bounded. So $T_f$ is indeed uniformly continuous, not just continuous.