The problem this:
Assume that $f$ is a complex measurable function on $X$, $\mu$ is a positive measure on $X$, $\|f\|_\infty > 0$ and $\|f\|_r<\infty$ for some $r<\infty$. Prove that $$\|f\|_p\to\|f\|_\infty ~~~\mbox{as}~~~p\to\infty.$$
This is the Exercise 4(e) from Real and Complex Analysis (Rudion), Chapter 3 ($L^p$-Spaces). In the internet, I've found some proofs. One is this:
I couldn't understand why one can conclude $\lim_{p\to\infty}\|f\|_p=\infty?$ How can I fix this problem in the proof?
EDIT: Okay, what Frank Lu did justify $\lim_{p\to\infty}\|f\|_p=\infty$. Now I have other question. Since $\|f\|_p\leq \|f\|^{\frac{r}{p}}_r$, one can conclude that $\limsup \|f\|_p\leq 1.$ How can I conclude the equality?
Here's the idea: Let $t>0$ be any real number, set $$E_t=\{x\in X: |f(x)|\geq t\}.$$ If you assume that $\|f\|_\infty=+\infty$, then $\mu(E_t)>0$. Then it follows from Chebyshev's inequality that $$t^p\mu(E_t)\leq\|f\|_p^p\quad\Leftrightarrow\quad t\mu(E_t)^{\frac{1}{p}}\leq\|f\|_p.$$ Now if $\mu(E_t)=\infty$ then we can see that $\|f\|_p=\infty$ for every $p$ thus $\lim_{p\to\infty}\|f\|_p=+\infty$. If $\mu(E_t)<\infty$, then one can see that $$t\leq\liminf_{p\to\infty}\|f\|_p.$$ Observe that $E_s\subseteq E_t$ whenever $s\geq t$ so in this case $0<\mu(E_s)\leq\mu(E_t)<\infty$, thus by the same argument we can prove that $$s\leq\liminf_{p\to\infty}\|f\|_p,\quad\forall s\geq t.$$ Therefore $\liminf_{p\to\infty}\|f\|_p=\infty$. Note that here we take $\liminf$ because we don't know whether the limit exist or not.
For the second question, what we want to prove is $\liminf_{p\to\infty}\|f\|_p\geq\|f\|_\infty$. For any $0<t<1$ let $E_t$ be the same as above. Then we have $$\|f\|_p\geq t\mu(E_t)^{\frac{1}{t}}.$$ If $\mu(E_t)<\infty$ then we see that $\liminf_{p\to\infty}\|f\|_p\geq t$, then let $t\to 1^-$. If $\mu(E_t)=\infty$ then we still have $\liminf_{p\to\infty}\|f\|_p\geq t$. In both cases, we have $\liminf_{p\to\infty}\|f\|_p\geq 1.$