Application of Holder's inequality. Let $(X,\mathfrak{X},\mu)$ be a finite measure space and let $f \in L^p(\mu)$. Use Holder's inequality to show that: 
\begin{equation}
f \in L^r(\mu)
\end{equation} 
for $1 \leq r \leq p < \infty$. 
Applying Holder's inequality show that:
\begin{equation}
||f||_r \leq ||f||_p \mu(X)^s
\end{equation}
where $s = (1/r) - (1/p)$. Thus if $\mu(X) = 1$ then $||f||_r \leq ||f||_p$.
A finite measure space satisfies $\mu(X) < \infty$. 
Furthermore, Holder's inequality states that:
\begin{equation*}
\int |fg| \,d\mu \leq \left( \int |f|^p \,d\mu \right)^{1/p} \left(\int |g|^q \,d\mu \right)^{1/q}
\end{equation*}
with $f,g$ measurable and $ 1 < p < \infty$ and $q = \frac{p}{p-1}$
Since $f \in L^p(\mu)$ we know that:
\begin{equation}
\left( \int |f|^p \,d\mu \right)^{1/p} < \infty
\end{equation}
However, I do not see how to apply it in this case, as there are restrictions on $q$, which do not appear in the result I am trying to prove. Any help is much appreciated. I am stuck.
 A: A little bit of motivation: we are going to use Holder's inequality on $|f|^r$ multiplied by $g \equiv 1$. The trick here is that we want to get rid of the $r$ and replace it with a $p$, so we will manufacture a power that precisely produces this. In particular, notice that $|f|^p = (|f|^r)^{p/r}$. Let $s$ be such that $1/s + r/p = 1$. Then, using Holder,
$$ \int |f|^r \, d\mu \le \left(\int (|f|^r)^{p/r} \, d\mu\right)^{r/p}\left( \int \, d\mu \right)^{1/s} = \|f\|_{p}^r \mu(X)^{1/s} < \infty.$$
I suppose, ultimately, the sneaky thing in this problem is that the $p,q$ chosen for Holder's inequality are not the same as the given $p$ in the setup of the problem.
A: In general given $f\in L^p(\mu)$ you can't say anything about inclusions like $f\in L_r(\mu)$ with $r\in[1,p)$. 
Moreover, for any $p\in[1,+\infty)$ there exist function $f\in L_p(\mathbb{R}_+)$, such that $f\notin L_r(\mathbb{R}_+)$ for all $r\in[1,+\infty)\setminus\{p\}$. For details see the question Is it possible for a function to be in $L^p$ for only one $p$?
