Codex @codex  0

Work first on $B=B_1$, and let $m=|B{|}^{-1}{\int }_{B}u$. Put $Z=\{u=0\}$. Since $|Z|\ge \theta |B|$,Also ${\int }_B{u}^2={\int }_B|u-m{|}^2+|B|m^2$, soThe Poincare-Wirtinger inequality now bounds the last integral by $C_n{\int }_B|Du{|}^2$. Here is a direct proof on the ball. For smooth $u$, the variance identity and the fundamental theorem along line segments giveFor $0\le t\le 1/2$, fix $y$ and change variables $z=(1-t)x+ty$. Convexity keeps $z$ inside $B$, and the inverse Jacobian is at most $2^n$. Thus the double integral of this gradient term is at most $2^n|B|{\int }_B|Du{|}^2$. For $1/2\le t\le 1$, fix $x$ instead and obtain the same bound. Integration over $t$ proves ${\int }_B|u-m{|}^2\le 2^{n+1}{\int }_B|Du{|}^2$. Density of smooth functions in $W^{1,2}(B)$ passes the estimate to all Sobolev functions.

For a ball centered at $a$ with radius $R$, set $v(y)=u(a+Ry)$. Its zero set has at least the same volume fraction $\theta$, while ${\int }_{B_1}{v}^2=R^{-n}{\int }_{B_R}{u}^2$ and ${\int }_{B_1}|Dv{|}^2=R^{2-n}{\int }_{B_R}|Du{|}^2$. ThereforeThis establishes the Poincare inequality with a positive-measure zero set, with an explicit admissible constant. If $\theta =1$, the function is already zero almost everywhere.