Lebesgue integrable p and q 
Be $E=[0,\infty]$ and $1\leq p\leq \infty$, Exhibit a function $f$ that $f\in \mathcal{L}^p$ and $f\notin \mathcal{L}^q$  when $q \neq p$

I try with simple functions but i dont know if are lebesgue integrable or dont
 A: Since your question is a bit ambiguous, I will give hints to get a stronger result that Arturo's. Let $p$ be in $[1, \infty]$. With Arturo's method, for any $q \neq p$, you can find a function $f$ such that $f \in \mathbb{L}^p$ but $f \notin \mathbb{L}^q$. Actually, you can do better, and find a function $f$ such that, for all $q \neq p$, one has $f \notin \mathbb{L}^q$.
Before starting, let us look at two examples.


*

*If we work on $[0,1]$, then $\mathbb{L}^p ([0,1]) \subset \mathbb{L}^q ([0,1])$ whenever $p \geq q$. This is a consequence of Jensen's or Hölder's inequality. For $p < q$, a function which belongs to $\mathbb{L}^p ([0,1])$ but not to $\mathbb{L}^q ([0,1])$ is, for instance, a function which has a pole of suitable order. A function such as $x^{-\frac{1}{2}\left(\frac{1}{p}+\frac{1}{q}\right)}$ does the trick. By taking function which are barely integrable, you can do better. For instance, the function $(x \ln (x/2)^2)^{-\frac{1}{p}}$ belongs to $\mathbb{L}^p ([0,1])$, but to no $\mathbb{L}^q ([0,1])$ for $q > p$.

*If we work on $\mathbb{N}$ with the counting measure, then $\ell^p \subset \ell^q$ whenever $p \leq q$. For $p > q$, a function which belongs to $\ell^p$ but not to $\ell^q$ is a function which has a suitable decay at infinity. A function which behaves like $n^{-\frac{1}{2}\left(\frac{1}{p}+\frac{1}{q}\right)}$ works (but recall that, this time, $p > q$, and not the other way around).  By taking function which are barely integrable, you can again do better. For instance, the function $(x \ln (2x)^2)^{-\frac{1}{p}}$ belongs to $\ell^p$, but to no $\ell^q$ for $q < p$.
The spaces $\mathbb{L}^p ([0,+\infty))$ are kind of a mix between the $\mathbb{L}^p ([0,1])$ spaces and the $\ell^p$ spaces. That's why you have no inclusions between the $\mathbb{L}^p$ spaces. If $q > p$, a function with a suitable pole at $0$ will belong to $\mathbb{L}^p ([0,+\infty))$ but not to $\mathbb{L}^q ([0,+\infty))$; it is as if we were working with $\mathbb{L}^p ([0,1])$. If $q < p$, a function with a suitable decay at infinity will belong to $\mathbb{L}^p ([0,+\infty))$ but not to $\mathbb{L}^q ([0,+\infty))$; it is as if we were working with $\ell^p$. So, if we take, for instance...
$$f(x) = \frac{1}{(x \ln (x/2)^2)^{\frac{1}{p}}}, \ x \leq 1;$$
$$f(x) = \frac{1}{(x \ln (2x)^2)^{\frac{1}{p}}}, \ x \geq 1,$$
I'll let you see what happens...
NB: the factors $2$ in the logarithm are here only to ensure that the functions are not dumbly not-integrable at $1$. What interests us is their behavior at $0$ and at infinity, no some kind of accident that might happen in-between.
A: Hint. $f(x) = \frac{1}{x^{\alpha}}$ is integrable on $[1,\infty]$ if and only if $\alpha\gt 1$, and on $(0,1)$ if and only if $\alpha\lt 1$. If $p\gt q$, Maybe you can find a function whose $p$th power has a large enough exponent but whose $q$th power does not; andif $q\lt p$, maybe you can find a function whose $p$th power is large enough but whose $q$th power is not?
