Question about integral inequality

Question

Please why we have this $$\int_{\Omega} |u|^{r(x)}\xi(x) dx\leq \int_{\Omega}|u|^r dx+|\Omega| $$ where $r=\max{r(x)}$ and $\xi(x)\leq 1,\forall x$

Answer 1

Split $\Omega$ up: $$\Omega = \{ x \in \Omega : \lvert u(x) \rvert \ge 1 \} \cup \{ x \in \Omega : \lvert u(x) \rvert < 1 \} =: \Omega_1 \cup \Omega_2.$$ These sets are disjoint so \begin{align*} \int_\Omega \lvert u \rvert^{r(x)}\xi(x) dx &= \int_{\Omega_1} \lvert u \rvert^{r(x)}\xi(x) dx + \int_{\Omega_2} \lvert u \rvert^{r(x)}\xi(x) dx \end{align*} In both of the above, we use $\xi(x) \le 1$. On $\Omega_1$, since $u \ge 1$, we have $\lvert u\rvert^{r(x)} \le \xi \lvert u \rvert^{r}$ and on $\Omega_2$, we have $\lvert u \rvert < 1$. Thus \begin{align*} \int_\Omega \lvert u \rvert^{r(x)}\xi(x) dx &= \int_{\Omega_1} \lvert u \rvert^{r(x)}\xi(x) dx + \int_{\Omega_2} \lvert u \rvert^{r(x)}\xi(x) dx \\ &\le \int_{\Omega_1} \lvert u \rvert ^r dx + \int_{\Omega_2} 1 \, dx \\ &\le \int_\Omega \lvert u \rvert ^r dx + \lvert \Omega \rvert. \end{align*}