 Solution

Work on $\mathbb{R}$ with Lebesgue measure. Without a uniform $L^p$ bound, useIt has $\Vert f_n{\Vert }_q=1$ and $\Vert f_n{\Vert }_r=n^{q/r-1}\le 1$, with $1/r=0$ for $r=\infty$. Each function still belongs to $L^p\cap L^r$, but its $L^p$ norm is unbounded. For every fixed $\epsilon >0$, its level set above $\epsilon$ is eventually empty, defeating any positive uniform $M$.

Without a uniform $L^r$ bound, use insteadNow $\Vert g_n{\Vert }_q=1$ and $\Vert g_n{\Vert }_p=n^{1-q/p}\le 1$, but for any fixed $\epsilon >0$ the high-amplitude level set has measure $n^{-q}\to 0$. These give the two distinct failures of the nonvanishing level-set bound between three Lp norms when either control is removed.