Proving that $\int_{\mathbb R} |f(x-y)|d\lambda(x) = \int_{\mathbb R} |f(x)|d\lambda(x)$

Question

Bartle's book exercise 10.R is asking to use the following result which I couldn't prove: given $f$ a Lebesgue integrable function on the real line $\mathbb R$ with Borel sigma-algebra $\mathbb B$, denoting by $\lambda$ the Lebesgue measure on $\mathbb B$, it is a fact that $\int_{\mathbb R} |f(x-y)|d\lambda(x) = \int_{\mathbb R} |f(x)|d\lambda(x)$

How to prove this assertion? It seems that I could use some kind of change of variable theorem for the composition $f\circ \varphi_y = f(x-y)$ for $\varphi_y(x) = x-y$, but no change of variable result has been stated in the book yet. So I kinda feel like there must be a more elementary way to prove it. Maybe approximating $|f(x-y)|$ by simple functions and then using the fact that $\lambda(E+y) = \lambda(E)$ could help somehow, but I'm not sure wheter the limit of those "translated" simple functions would give me for instance $|f(x)|$.

Any help is much appreciated.

Answer 1

Hint

1.Prove the statement for functions of the form $1_A$ where $A$ is an interval.

2.Then you have it for every step function by linearity of the integral.

3.Then use the fact that step functions are dense in the class of Lebesgue integrable functions.