Can a monotone decreasing function intersect $y=-x$ on a set of more than measure $0$? Can a monotone decreasing continuous function intersect $y=-x$ on a set of more than measure $0$ unless it equals to $y=-x$ on an interval?
I am unsure whether something like Cantor's function can be massaged into satisfying this property. This comes from a research question in an applied area. Maybe I will need to prove more properties about the original function.

Take a fat Cantor set $C \subset [0,1]$ with $0,1\in C$. Set 
$$f(x) = \begin{cases}-x/2\quad &\text{for $x < 0$}, \\ 
-1-(x-1)/2 \quad & \text{for $x > 1$}, \\
-x\quad & \text{for $x\in C$.} \end{cases}$$ 
On each open interval of $[0,1]\setminus C$, fill the gap by a scaled and translated copy of $x\mapsto -x^2$. Then $f(x) = -x \iff x\in C$, and $f$ is (strictly) monotonically decreasing. -- Daniel Fischer

I will add another argument, not self-contained.  For every closed subset of $\mathbb{R}$ there exists a $C^\infty$-smooth function which vanishes only on that set. Let $\phi$ be such a function for the aforementioned fat Cantor set $C$.  Since $C$ is bounded, we can make $\phi'$ bounded on $\mathbb{R}$. Then for sufficiently small $\epsilon>0$, the function $f(x)=\epsilon\phi(x)-x$ is the desired example. 
