Show that $\lim_{p \to \infty} \| f \|_p = \| f \|_\infty$ In a space with measure $1$, $||f||_p$ is an increasing function with respect of $p$. To show that $\lim_{p \rightarrow \infty} ||f||_p=||f||_{\infty}$ we have to show that  $||f||_{\infty}$ is the supremum, right?? 
To show that, we assume that $||f||_{\infty}-\epsilon$ is the supremum.
From the essential supremum we have that $m(\{|f|>||f||_{\infty}-\epsilon\})=0$.
So, we have to show that $m(\{|f|>||f||_{\infty}-\epsilon\})>0$. 
Let  $A=\{|f|>||f||_{\infty}-\epsilon\}$. 
We have that $\int_A |f|^p \leq \int |f|^p \leq ||f||_{\infty}^p$. 
$\int_A |f|^p >\int_A (||f||_{\infty}-\epsilon)^p=(||f||_{\infty}-\epsilon)^p m(A)$ 
So, $m(A)^{1/p} (||f||_{\infty}-\epsilon)<||f||_p \leq ||f||_{\infty}$
How could we continue to show that $m(A)>0$?? 
EDIT: 
Is it as followed??
We have that $0<||f||_{\infty}-\epsilon<||f||_{\infty}$ for some $\epsilon>0$.
$||f||_{\infty}$ is the essential supremum. So, from the definition we have that $m\left ( \{|f(x)>||f||_{\infty}-\epsilon\} \right )>0$. 
Let  $A=\{|f|>||f||_{\infty}-\epsilon\}$. 
We have that $\int_A |f|^p \leq \int |f|^p \leq ||f||_{\infty}^p$. 
$\int_A |f|^p >\int_A (||f||_{\infty}-\epsilon)^p=(||f||_{\infty}-\epsilon)^p m(A)$ 
So, $m(A)^{1/p} (||f||_{\infty}-\epsilon)<||f||_p \leq ||f||_{\infty}$
Taking the limit $p \rightarrow +\infty$ we have the following:
$$\lim_{p \rightarrow +\infty}m(A)^{1/p} (||f||_{\infty}-\epsilon)<\lim_{p \rightarrow +\infty} ||f||_p \overset{ m(A)>0 \Rightarrow \lim_{p \rightarrow +\infty}m(A)^{1/p}=1}{\Longrightarrow} \\ ||f||_{\infty}-\epsilon<\lim_{p \rightarrow +\infty} ||f||_p$$ 
Is this correct?? How do we conclude that $\lim_{p \rightarrow +\infty} ||f||_p=||f||_{\infty}$ ?? 
 A: First of all you have to show that $||f||_\infty$ is a superior bound. But this is immediate by the definition of $\| \cdot\|_\infty$. 
Now you want to show that, for all $\epsilon$ we have $$\lim_{p \rightarrow \infty} \|f\|_p > ||f||_\infty - \epsilon$$
As already observed and using your notation, $m(A) = m(|f| > ||f||_\infty -\epsilon)>0$. Now putting the things together you got
$$\|f\|_p > (||f||_\infty- \epsilon)m(A)^{1/p}$$ for all values of $p$. Now just take the limit, which exists since the sequence is increasing and bounded, to obtain what you desire.
A: The final step follows from the definition of the essential supremum.

Proposition
  If $m(\{|f| > C\}) = 0$ then necessarily $C \ge \|f\|_\infty$.

Proof
Let $N := \{|f|>C\}$. By assumption $m(N) = 0$. This means
$$\|f\|_\infty = \inf_{m(N') = 0} \sup_{x\notin N'} |f(x)| \le \sup_{x\notin N} |f(x)| = \sup_{\{|f| > C\}} |f(x)| \le C$$
as was to be shown.
Now use this proposition together with your initial steps to conclude a contradiction for $C = \|f\|_\infty - \epsilon < \|f\|_\infty$.
