Consider a ball B and let $f(x) \in L^1(B)$ such that $\int_B f(x) dx = 0$. Furtheremore, there exists a closed set $E \subset B$ such that $f|_E$ is Lipschitz. The standard Lipschitz extension theorem gives a function $g$ which is Lipschitz on $B$ such that $g|_E=f$.
My question is the following: Is there a construction that additionally gives $\int_B g(x) dx = 0$?
In the standard Lipschitz extension theorem, one can keep the Lipschitz constant the same in the extension. In this case, that will be impossible.
To see this, let $B=B_1$ denote the unit ball in $\mathbb{R}^n$. It is not too hard to imagine a function $f$ with $f=1$ in $B_r$ and $\int_B f = 0$ for each $r<1$. The Lipschitz constant of $f$ must be at the very least $r^{-1}$ (because $f<0$ must be true somewhere in $B_1 \setminus B_r$), but the Lipschitz constant of $f$ in $B_r$ is $0$ (because it is constant).
So we cannot hope for any quantitative bound on the Lipschitz constant of such an extension. On the other hand, it is not hard to construct an extension that also satisfies $\int_B f = 0$. Given any extension, we can take a small ball in $B_1\setminus E$ and add a suitably scaled bump function supported on this ball. This will increase the Lipschitz constant (but we cannot hope for any control on this anyway), but it will remain Lipschitz and have the correct integral.