Question on an exercise on $L^p$ spaces Let $(X,\mathcal{A},\mu)$ be a measure space such that $\mu(X)=1$ and $f\in L^\infty(\mu)$. 
Show $||f||_p\leq ||f||_q\qquad p\leq q$ 
and 
$\lim_{p\to\infty}||f||_p=||f||_\infty.$
Hint: for $\epsilon>0$ consider the set $A_\epsilon=\{x\in X:|f(x)|\geq ||f_\infty||-\epsilon\}.$
I really am lost regarding these $L^p$ spaces. No idea what to do with the hint either, but it's probably there to show $\lim_{p\to\infty}||f||_p=||f||_\infty.$ However, my gut and this $\mu(X)=1$ tells me I should somehow use Holder inequality. But how? 
If $f\in L^\infty(\mu)$ then $|f|^p\in L^{???}(\mu)?$
Is it true that if $f\in L^\infty(\mu)$ then $f\in L^p(\mu)$?
 A: If $1 < r < \infty$ then Holder's inequality gives you $$\int_X |g| \, d\mu \le \left( \int_X |g|^r \, d\mu \right)^{1/r} \left( \int_X \, d\mu \right)^{1/r'} = \mu(X)^{1/r'} \left( \int_X |g|^r \, d\mu \right)^{1/r}.$$
In particular if $\mu(X) = 1$ and $p < q$ you can take $g = |f|^p$ and $r = \frac qp$ to find $$\int_X |f|^p \, d\mu \le \left( \int_X |f|^q \, d\mu \right)^{p/q}.$$
Now take the $p$-root on each side.
If $f \in L^\infty(\mu)$ then $|f|\le \|f\|_\infty$ $\mu$-almost everywhere. Consequently $$\int_X |f|^p \, d\mu \le \|f\|_\infty^p \int_X \, d\mu = \mu(X) \|f\|_\infty^p$$ 
so that $\|f\|_p \le \mu(X)^{1/p} \|f\|_\infty$.
Finally assume $f \in l^\infty(\mu)$. Note that by definition $\mu(\{|f| > \|f\|_\infty - \epsilon\}) > 0$ for every $\epsilon > 0$.  Since $$\|f\|_p^p \ge \int_{\{|f| > \|f\|_\infty - \epsilon\}} |f|^p \, d\mu \ge (\|f\|_\infty - \epsilon)^p \mu(\{|f| > \|f\|_\infty - \epsilon\})$$
you get $$\|f\|_p \ge (\|f\|_\infty - \epsilon)\mu(\{|f| > \|f\|_\infty - \epsilon\})^{1/p}.$$
If $t > 0$ then $t^{1/p} \to 1$ as $p \to \infty$. Thus $$\limsup_{p \to \infty} \|f\|_p \ge \|f\|_\infty - \epsilon.$$ Now take $\epsilon \to 0^+$.
