Given $u:\mathbb{R}^N \rightarrow \mathbb{R}$ is continuous and has compact support, we define the set $$K_u: = \{x\in \mathbb{R}^N : u(x) = \|u\|_\infty\}.$$
Looking at the following limit $$\lim_{p\rightarrow \infty} \int_{\mathbb{R}^N} \bigg(\frac{|u|} {\|u\|_p}\bigg)^p d\mu = \lim_{p\rightarrow \infty} \frac{\int_{\mathbb{R}^N} |u|^p d\mu}{\|u\|_p^p} = 1.$$
When $K$ defined as above, and if the Lebesgue measure $\mu(K) >0$, we break up the above integral into $K$ and $K^c$ $$1 = \lim_{p\rightarrow \infty} \int_{\mathbb{R}^N} \bigg(\frac{|u|} {\|u\|_p}\bigg)^p d\mu = \lim_{p\rightarrow \infty} \int_{K} \bigg(\frac{|u|} {\|u\|_p}\bigg)^p d\mu+ \lim_{p\rightarrow \infty}\int_{K^c} \bigg(\frac{|u|} {\|u\|_p}\bigg)^p d\mu.$$ The first integral over $K$ $$\lim_{p\rightarrow \infty} \int_{K} \bigg(\frac{|u|} {\|u\|_p}\bigg)^p d\mu = \frac{1}{\mu(K)} \int_{K} 1 d\mu = 1.$$ This is because on $K$, the integrand is the following \begin{align*} &\left(\frac{\| u\|_\infty}{\|u\|_p}\right)^{p}\\ =& \frac{\|u\|_\infty^p}{\int_{K} |u|^p dx + \int_{K^c} |u|^p dx }\\ =& \frac{\|u\|_\infty^p}{\| u\|_\infty^p m(K) + \|u\|_\infty^p \int_{K^c} \left(\frac{| u|}{\| u\|_\infty}\right)^p dx}\\ =& \frac{1}{\mu(K) + \int_{K^c} \left(\frac{| u|}{\| u\|_\infty}\right)^p dx} \end{align*} which converges to $\frac{1}{\mu(K)}$ as $p$ goes to infinity.
Therefore the limit of the second integral over $K^c$ has to be zero $$ \lim_{p\rightarrow \infty}\int_{K^c} \bigg(\frac{|u|} {\|u\|_p}\bigg)^p d\mu = 0. \quad\quad \text{ if } \mu(K)>0 \quad (\star)$$
However when $\mu(K) = 0$, integral over $K^c$ is the same as integral over $\mathbb{R}^N$ we have
$$\lim_{p\rightarrow \infty}\int_{K^c} \bigg(\frac{|u|} {\|u\|_p}\bigg)^p d\mu = 1 \quad\quad \text{ if } \mu(K)=0 \quad (\star\star)$$ which is different from $(\star)$ above.
Can you give me some insight about cause of the two limits being different. Thank you!
The best intuition I can get for your result comes from one dimension. Think of $u$ as being $1$ on an interval $(a,b)$ and then diminishing down to $0$ outside a bigger interval $(c,d)\supset (a,b)$. As you exponentiate $u$ to higher and higher powers, it becomes dominated by what happens on $(a,b)$ and goes to zero outside $(a,b)$. Also, no matter how large you make $p$, $||u||_p$ is bounded below by $(b-a)^{1/p}\to 1$. In contrast, if $a=b$, so the function reaches a peak at $a=b$ and falls to $0$ on both sides, the set of measure zero $[a,b]$ won't influence anything, and what is more $||u||_p^p\to0$ as $p\to\infty$, quite different behavior.
By the way, in your second displayed equation, not only is the limit $1$, but the integrals are $1$ for every $p$.