Showing that $\|f\|_p\to\|f\|_{\infty}$ I know this question has been asked a lot in this site, I've been checking those questions myself, however according to the theory we use in class there are things that I can't use and/or I don't know how to use (for example, we don't work with probability spaces, and other things that are very measure theory). So this is another version of this same old problem.

If $|\Omega|<\infty$, $f\in L^{\infty}(\Omega)$, then $$\|f\|_{\infty}=\lim_{p\to{\infty}}|\Omega|^{-1/p}\|f\|_p$$

I see, that this problem really is like this one, since is easy to see that $\lim_{p\to\infty}|\Omega|^{-1/p}=1$.
Now, we are given two hints:

Prove that: (a) $|\Omega|^{-1/p}\|f\|_p\le\|f\|_{\infty},\forall p$. 
  (b) If $\varepsilon>0$ then $|\Omega|^{-1/p}\|f\|_p\ge\|f\|_{\infty}-\varepsilon$ for $p$ sufficiently big.

For the first one, it comes easily from a proposition that says, if $|\Omega|<\infty$ and $p<\infty$ then $\|f\|_p\le|\Omega|^{1/p}\|f\|_{\infty}$, if $|\Omega|\not=0$ then we have that $|\Omega|^{-1/p}\|f\|_p\le\|f\|_{\infty}$.
Now onto proving the second one. I'm having a lot of trouble, I used an idea from here, but instead of getting the measure of $\Omega$, I get the measure of a subset, and I'm not using that $p$ has to be big or not, that part of the hint confuses me really, I can't see how do you use it for the problem.
 A: The adjoint teacher solved the problem, I'm writting down just in case somebody got stuck like myself even though the solution is pretty simple (I know that now). I hope I don't mess it up.
Let's prove the hints:
(a) Is already solved in the question by that proposition.
(b) Let $\varepsilon>0$. Let's take the following set: $A=\{x\in\Omega: |f(x)|\ge\|f\|_\infty -\varepsilon\}$, wich satisfies $|A|>0$ because of the definition of $\|\cdot\|_\infty$, then it follows that: 
$$|\Omega|^{-1/p}\|f\|_p=\left(\frac{1}{|\Omega|}\|f\|_p\right)^{-1/p}\le\left(\frac{(\|f\|_\infty-\varepsilon)^p}{|\Omega|}\int_A 1_\Omega\right)^{1/p}=(\|f\|_\infty-\varepsilon)\left(\frac{|A|}{|\Omega|}\right)^{1/p}:=(\star)$$
but since $\left(\frac{|A|}{|\Omega|}\right)^{1/p}\to 1$ when $p\to\infty$, then $(\star)\to(\|f\|_\infty-\varepsilon)$.
From here the rest is history, and is pretty easy to finish.
Now we have another way to see $$\lim_{p\to\infty}\|f\|_p=\|f\|_\infty$$
with relation to the measure of the set $$\lim_{p\to\infty}|\Omega|^{-1/p}\|f\|_p=\|f\|_\infty$$ (well, maybe is not that impressive, but it's another way to see it).
