Suppose $m(E)<\infty$ and $f\in\mathcal{L}^{\infty}(E)$. The goal of this problem is to show \begin{align*} \lim_{p\to \infty}\|f\|_p=\|f\|_{\infty}. \end{align*} First, prove that \begin{align*} \lim_{p\to\infty}\|f\|_p\leq \|f\|_{\infty}. \end{align*}Proof: Assume $m(E)<\infty$ and $f\in\mathcal{L}^{\infty}(E)$. By definition, \begin{align*} \|f\|_p=\left(\int_E|f|^pdm\right)^{1/p}\leq \left(\int_E\|f\|_{\infty}^p\right)^{1/p}=\|f\|_{\infty}\cdot m(E)^{1/p}<\infty. \end{align*} Letting $p\to \infty$, we have that $m(E)^{1/p}\to 1$. Therefore, \begin{align*} \lim_{p\to\infty}\|f\|_p\leq \|f\|_{\infty}. \end{align*}
I'm not sure if I am missing anything here or need to justify being able to bring the limit inside and applying to $m(E)$.
Next, I need to prove that \begin{align*} \lim_{p\to\infty}\|f\|_p\geq \|f\|_{\infty}-\epsilon \end{align*} for any $\epsilon>0$. Hint: Look at the set \begin{align*} F=\{x\in E:|f|>\|f\|_{\infty}-\epsilon\}. \end{align*}
I'm stuck on this part and would appreciate any help, thanks.