I have the following question. Let $\Omega$ be a bounded domain in $\mathbb{R}^d$. Suppose we have a function $f$ such that for $h>0$ it holds that $$\|f\|_{H^1}\lesssim h^k.$$ By the Sobolev embedding theorem we thus have $$\|f\|_{L^{d_\ast}}\lesssim h^k$$ with $$d_\ast = \frac{2d}{d-2}.$$

I would like to find a set $\Omega_h\subset \Omega$ with $\mu(\Omega\setminus \Omega_h)\lesssim h^\delta$ for some $\delta >0$ ($\mu$ being Lebesgue measure), such that on $\Omega_h$ the $L^d$-norm of $f$ is small (i.e., a power of $h^k$). A (possibly suboptimal) result can be shown by noting that the bounds above imply that $$\mu \{ x:|f(x)|>t \}\lesssim h^{kd_\ast} t^{-d_\ast}$$(embedding of $L^{d_\ast}$ into $L^{d_\ast,\infty}$, the Lorentz space, that is, any choice $t = h^\beta$ yields an estimate of the desired kind: Putting $\Omega_h:= \{ x:|f(x)|>h^{k-\delta/d_\ast} \}$ we get the desired bound for the measure of $\Omega_h$ and we also get that $$\|f\|_{L^d(\Omega_h)}\lesssim \|f\|_{L^\infty(\Omega_h)}\lesssim h^{k-\delta/d_\ast}.$$

My question is: can this be improved? Note that I have taken a detour by using a bound for the $L^\infty$ norm which trivially bounds the $L^d$-norm. I would like to get a better estimate but I am not sure how.