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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.

Ken
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Lost1
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    Take a fat Cantor set $C \subset [0,1]$ with $0,1\in C$. Set $f(x) = -x/2$ for $x < 0$, $f(x) = -1-(x-1)/2$ for $x > 1$, and $f(x) = -x$ for $x\in C$. 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 Jun 09 '15 at 11:58
  • @DanielFischer urgh... Thanks – Lost1 Jun 09 '15 at 12:10
  • @DanielFischer but doesnt the cantor set have measure 0? – Lost1 Jun 09 '15 at 12:19
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    I said to take a fat Cantor set, that means of positive measure. – Daniel Fischer Jun 09 '15 at 12:20
  • Ah, i see, i did not know what one of those was. Thanks again – Lost1 Jun 09 '15 at 13:22

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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.

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