TITLE: Is it possible for a function to be in $L^p$ for only one $p$?
QUESTION [133 upvotes]: I'm wondering if it's possible for a function to be an $L^p$ space for only one value of $p \in [1,\infty)$ (on either a bounded domain or an unbounded domain).
One can use interpolation to show that if a function is in two $L^p$ spaces, (e.g. $p_1$ and $p_2$,with $p_1 \leq p_2$ then it is in all $p_1\leq p \leq p_2$). 
Moreover, if we're on a bounded domain, we also have the relatively standard result that if $f \in L^{p_1}$ for some $p_1 \in [1,\infty)$, then it is in $L^p$ for every $p\leq p_1$ (which can be shown using Hölder's inequality).
Thus, I think that the question can be reduced to unbounded domains if we consider the question for any $p>1$. 
Intuitively, a function on an unbounded domain is inside an $L^p$ space if it decrease quickly enough toward infinity. This makes it seem like we might be able to multiply the function by a slightly larger exponent. At the same time, doing this might cause the function to blow up near zero. That's not precise/rigorous at all though.   
So I'm wondering if it is possible to either construct an example or prove that this can't be true.

REPLY [3 votes]: Consider a real monomial: $x^\alpha$
Now poles give a restriction of the form $\alpha>\alpha_0$
while decay rates give a restriction of the form $\alpha<\alpha_0$.
This can be used to glue a desired example, e.g. if
$$f(x):=x^{\alpha_0}\chi_{[-1,1]}+x^{\alpha_\infty}\chi_{(-\infty,-1)\cup(1,\infty)}$$
then the integrability is restricted to
$$-\frac{1}{\alpha_0}<p<-\frac{1}{\alpha_\infty}$$
Unfortunately this technique only restricts up to a neighborhood like $p=7\pm\varepsilon$.