Using translational invariance of Lebesgue integrals to check integrability? I know that if a function f is integrable, then $\int_{\mathbb R} f(x) dx= \int_{\mathbb R} f(x-h) dx$ for any $h>0$. This of course does not hold in general if f is not integrable. However, can I use this to actually check integrability of a function? Suppose I check that $\int_{\mathbb R} |f(x-h)| dx$ is infinite, then that would imply $\int_{\mathbb R} |f(x)| dx$ is infinite as well, correct? Since if it were finite, then the integral of the translated f would give the same answer as the original integral, and would have to come out finite but it doesn't here. 
Thanks.
 A: Indeed, for either $f \in \mathrm L^{\!+}$ (the space of non-negative measurable functions) or $f \in \mathrm L^{\!1}$, we have
$$ \int_{\mathbb R} f(x) \mathrm dx = \int_{\mathbb R} f(x-h) \mathrm dx
$$
Edit: In the original post, $R$ had an ambiguous context, so the answer below can be ignored.
If $R$ represents a subset of the Euclidean space, observe that
$$
\int_R |f(x-h)| \mathrm dx = \int \mathbf 1_R(x) |f(x-h)| \mathrm dx = \int \mathbf 1_R(x+h) |f(x)| \mathrm dx
$$
while
$$
\int_R |f(x)| \mathrm dx = \int \mathbf 1_R(x) |f(x)| \mathrm dx
$$
These two integrals are in general not related.

Somewhat explicit counterexample:
Let
$$ f(x) = \begin{cases} 0 &\text{ if } x = 0 \\|x^{-1}| &\text{ else} \end{cases}
$$
Then $\int_{-1}^1 |f(x)| \mathrm dx = +\infty$ but $\int_{-1}^1 |f(x+3)| \mathrm dx < +\infty$.
A: Yes, that is correct. This reasoning is called contraposition. "P implies Q" is equivalent to "not Q implies not P".  If you are in Italy then you are in Europe. If you are not in Europe then you are not in Italy.
In this case, $f$ integrable implies for all $h$, $f(x-h)$ is integrable, which is equivalent to: If there exists $h$ such that $f(x-h)$ is not integrable, then $f$ is not integrable. 
