Finding a function $f\in L^p \setminus L^q$ for $0Given a measure space such that for all $\epsilon>0$, there exists a measurable set $E$ satisfying $\epsilon> \mu(E)>0$, Show that $L^p \nsubseteq L^q$.
I've been going about this by constructing a function:
$f:=\underset{n=1}{\overset{\infty}{\sum}} \Big( \frac{1}{a_n\cdot n^\alpha} \Big)^{\frac{1}{p}}\cdot \chi_{A_n}$
Where $\{ A_n \}_{n=1}^\infty$ are disjoint measurable sets with measures $a_n$ accordingly, such that $0<  a_{n+1} \leq \frac{a_n}{2}$. And I wanted to show that for a certain  $\alpha$, this will give that $f\in L^p$ but not in $L^q$. Somehow when I try to compute for which $\alpha$ does this work for, it doesn't make sense. 
I Would be thankful if someone were to able to tell if this is a dead end or suggest how to fix this attempt.
 A: Since 
$$
\left\lvert f\right\rvert^r=\sum_{n=1}^{\infty}\left(\frac{1}{a_n\cdot n^\alpha} \right)^{\frac{r}{p}}\cdot \chi_{A_n},
$$
it follows that $f$ belongs to $L^r$ if and only if 
$$
\sum_{n=1}^{\infty}\left(\frac{1}{a_n\cdot n^\alpha} \right)^{\frac{r}{p}} a_n<+\infty,
$$
or equivalently,
$$
\sum_{n=1}^{\infty}a_n^{1-\frac rp}n^{-\alpha r/p}<+\infty.
$$
We want $f\in L^p$ but not in $L^q$ hence we want 
$$
\sum_{n=1}^{\infty} n^{-\alpha }<+\infty\mbox{ and }\sum_{n=1}^{\infty}a_n^{1-\frac qp}n^{-\alpha q/p}=+\infty.
$$
If we choose $a_n$ such that $a_n\leqslant n^{-\beta}$, then take $\alpha\gt 1$ and $\beta$ such that 
$$
\sum_{n=1}^{\infty} n^{\beta\left(\frac qp-1\right)-\alpha q/p}=+\infty.
$$
A: I was able to solve it eventually using the ratio test, and the fact that $0<  a_{n+1} \leq \frac{a_n}{2}$ for all $n\in \mathbb{N}$. As suggested above for $\alpha>1$, $f\in L^p$. As for $f\notin L^q$, one can say that:
$\int\vert f\vert^qd\mu = \sum_{n=1}^{\infty}a_n^{ \frac{p-q}{p} }n^{-\alpha q/p}$
It is sufficient to show that  $a_n^{ \frac{p-q}{p} }n^{-\alpha q/p}$ does not tend to $0$ as $n\rightarrow \infty$. However:
$\dfrac{a_{n+1}^{ \frac{p-q}{p} }(n+1)^{-\alpha q/p}}{a_n^{ \frac{p-q}{p} }n^{-\alpha q/p}}=  \Big( \frac{a_{n+1}}{a_n} \Big)^{ \frac{p-q}{p} } \cdot \Big( \frac{n+1}{n} \Big)^{-\alpha q/p} \leq 2^{\frac{q-p}{p}} \cdot \Big( \frac{n+1}{n} \Big)^{-\alpha q/p} \rightarrow 2^{\frac{q-p}{p}} \cdot 1\overset{q>p}{>}2 $
And by the ratio test, $a_n^{ \frac{p-q}{p} }n^{-\alpha q/p} \rightarrow \infty$, which implies that the series diverges. Hence $f\notin L^q$.
