real analysis - 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 $\lim_{p \rightarrow +\infty} ||f||_p=||f||_{\infty}$ ??