Vtyjkyuk

Suppose I have a Lipschitz function $f$ defined on $[a,b]$. Suppose there is a measure zero set on which $f$ is equal to $0$. Can the following happen?

For any $epsilon > 0$, if $f(x) = 0$ for some $x$, $f$ takes both positive and negative values in $(x, x+ epsilon)$ and also in $(x-epsilon ,x)$?

What if $f$ were also differentiable?

Obviously, the above would not happen if $f$ never becomes negative, i.e. bounces off $0$. A counterexample regarding one sided derivatives of Lipschitz functions posted here shows that at one point it can happen. But, I am interested in showing the following:

If, for some $x$ such that $f(x) = 0$, and for any $epsilon > 0$, $f$ takes both positive and negative values in $(x, x+ epsilon)$ and also in $(x-epsilon ,x)$ then there are at least two points where $f$ changes sign, i.e. in some closed interval to the left it's negative and to the right it's positive,

Suppose that $f(c)=0.$ Then $f$ on $(c,b)$ takes positive and negative values. Let $d$ be such that $f(d)>0.$ Now let define $$e=supxin (c,d).$$ It is clear that $e<d$ and $f(e)=0.$ Since $f(e)=0$ we have that on $(e,d)$ the function takes positive and negative values. Then there exist $zin (e,d)$ such that $f(z)=0.$ This contradicts the definition of $e.$ So, we conclude that such a function can't exist.