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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.

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    Sorry, R means real numbers. I just figured out how to type it as $\mathbb R $ in latex. – 2017-02-09

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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 $\int_{-1}^1 |f(x+3)| \mathrm dx < +\infty$.

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    Oh but I thought if f is integrable, then the two integrals are the same by translation invariance of lebesgue integrals? – 2017-02-09
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    @Socchi Translation invariance only works if you are integrating over the entire $\mathbb R^n$. – 2017-02-09
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    The integrals are over all of $\mathbb R$. – 2017-02-09
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    Oh I see now, Henry's example would not work since it would give infinity in both cases, assuming we integrating over all of $\mathbb R$, correct? Thanks. – 2017-02-09
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    Just to be sure, in the top edit that you have made, when you say f in $L^+$ then translational invariance holds, you mean it could hold but yield infinity= infinity correct? – 2017-02-09
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    @Socchi That is right. – 2017-02-09
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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.

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    Thanks, however Henry's counterexample confused me, because if we shift f(x) by 3 in the opposite direction, then we would indeed get integral of the shift in the other direction is infinite, but the original integral would be finite ? – 2017-02-09
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    @Socchi: Henry is not integrating over all of $\mathbb R$. Instead he is taking $f$ to be defined on $\mathbb R$, but restricting the domain of integration to a different set. You can take a bad function and shift the bad part away from a restricted domain, but then it isn't related to what you are asking. – 2017-02-09
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    Oh I see, so f has to be non zero on all of $\mathbb R$ for this to work? But the translation invariance theorem does not require that in the first place, just that you integrate over all of $\mathbb R$, and f defined on $\mathbb R$ but does not say f not 0 on a region of $\mathbb R$?? Where is the error in my reasoning? – 2017-02-09
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    @Socchi It depends on what you mean by $R$. – 2017-02-09
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    I edited it to mean the set of real numbers. Sorry for the confusion. – 2017-02-09