Show that $ \lim_{a \to \infty} \int_0^\infty f(x-a) \, dx = \int_{-\infty}^\infty f(x) \, dx $ for integrable functions $f \in L^1(\mathbb{R})$

Question

I noticed in a proof that he needed to state that one integral approached another:

$$ \lim_{a \to \infty} \int_0^\infty f(x-a) \, dx = \int_{-\infty}^\infty f(x) \, dx $$

Does this always hold for integrable functions (Riemann or Lebesgue) functions $f \in L^1(\mathbb{R})$ ? This seems rather obvious.

$$ \int_{-\infty}^\infty f(x) \, dx - \int_0^\infty f(x-a) \, dx =\int_{-\infty}^\infty f(x) \, dx - \int_{-a}^\infty f(x) \, dx =\int_{-\infty}^{-a} f(x) \, dx \stackrel{?}{\to} 0 $$

Here I would have used the linearity of the integral. Doesn't the integral on the right-hand-side tend to zero?

Answer 1

Yes, this is correct. The convergence of $\int_{-\infty}^{-a} f(x) \, dx$ to $0$ follows from the dominated convergence theorem, for instance (we are integrating the functions $f\cdot 1_{(-\infty,-a)}$ which are dominated by the integrable function $|f|$).