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Consider the set $[0,1]$ with Lebesgue measure. We say that a measurable function $f$ is non-decreasing in a full-measure set if there is a set E with $\mu(E)=1$ such that for every $y,z\in E$ $$z\leq y \implies f(z)\leq f(y)$$

If $f$ is not non-decreasing in a full measure set, can I say that there exists an $x\in[0,1]$, an $A\subseteq [0,x]$ and $B\subseteq[x,1]$, $A$ and $B$ with positive Lebesgue measure such that $f(z)>f(y)$ for every $z\in A$ and $y \in B$?

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Yes, this is correct. Consider the upper level sets $U_t=\{x\in [0,1] :f(x)>t\}$. There are two possibilities:

  1. For every $t\in\mathbb R$, the set $U_t$ agrees with some interval $[x,1]$ up to a null set $N_t$.
  2. There is $t\in\mathbb R$ such that the set $U_t$ does not agree with any interval of the form $[x,1]$ up to a null set.

If Case 1 holds, then take $N=\bigcup_{t\in\mathbb Q}N_t$ and observe that $f$ is nondecreasing on the set $[0,1]\setminus N$, which has full measure.

Suppose Case 2 holds. Let $\alpha=\mathrm{ess\,inf}\,U_t$ where $t$ is as in Case 2. There is $\epsilon>0$ such that the set $B:=[\alpha+\epsilon,1]\setminus U_t$ has positive measure. Let $A=[\alpha,\alpha+\epsilon]\cap U_t$ and you are done.

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