Continuity of an integral operator

Question

I'm stuck with this exercise:

Let $A \subset \mathbb{R}$ be a measurable set. For each $f \in L^1(\mathbb{R})$ and $y \in \mathbb{R}$, let:

$T(f, y) = \int_{A}{f(x-y)\mathrm{d}x}$.

I have to show that $\forall f \in L^1(\mathbb{R})$, the operator $y \mapsto T(f,y)$ is continuous.

What I've tried doing is to prove it by subsequences-continuity, but I get the integral of a difference quotient of $f$ that I don't know how to bound under the only hypothesis that it's in $L^1(\mathbb{R})$.

Any suggestions?

Thanks in advance

Answer 1

Given $y,z\in\mathbb{R}$ $$ |T(f,y)-T(f,z)|\le\int_A|f(x-y)-f(x-z)|\,dx\le\int_{A-y}|f(x)-f(x+y-z)|\,dx. $$ Call $h=y-z$, $B=A-y$. To show: $$ \lim_{h\to0}\int_B|f(x)-f(x+h)|\,dx=0. $$ This is done in 2 steps:

1. Prove it for nice functions (say continuous with compact support, using uniform continuity.)
2. Use a density argument.