Proving Constant CTP of a Function on [0,1]

[p33rz]

Suppose that $$f:[0,1]\longrightarrow{\mathbb{R}}$$ is a function such that $$f(.,y)$$ is constant for almost all $$y$$, and $$f(.,y)$$ is constant for almost every $$x$$. Prove that $$f$$ is constant ctp (with respect to u, where u is the Lebesgue measure).

Hint: Assume the contrary. Then it sets you and you have positive measure. Use Fubini to prove that each of these sets contains at least one vertical and one horizontal interval. Conclude.

Note: A function is constant ctp, if not constant in a set of measure zero.

Help please!

 

[Valkarie]

Proof: Assume, for the sake of contradiction, that f is not constant ctp. Then there exists a set A with positive measure such that f is not constant on A. Let A' be the projection of A onto the y-axis, and A'' the projection of A onto the x-axis. By Fubini's theorem, both A' and A'' have positive measure. Since A' and A'' are both intervals, there exist x_0,x_1 in A'' such that x_0<x_1 and y_0,y_1 in A' such that y_0<y_1. Since f(.,y) is constant for almost all y, we must have f(x_0,y_0)=f(x_0,y_1). Similarly, since f(x,.) is constant for almost all x, we must have f(x_0,y_0)=f(x_1,y_0). Hence, f(x_0,y_0)=f(x_0,y_1)=f(x_1,y_0), which is a contradiction since f is not constant on A. This completes the proof.