I have three questions regarding this example in Durret:
"Our last example shows that measurability is important or maybe that some of the axioms of set theory are not as innocent as they seem.
Example 1.7.8. By the axiom of choice and the continuum hypothesis one can define an order relation <' on (0,1) so that {x : x <' y} is countable for each y. Let X = Y = (0, 1), let A = B = the Borel sets and µ1 = µ2 = Lebesgue measure. Let f(x, y) = 1 if x <' y, = 0 otherwise. Since {x : x <'y} and $\{y : x <'y\}^{c}$ are countable,
$\int_X f(x, y) µ1(dx)=0 \ \ \ \ for\ all \ y$
$\int_Y f(x, y) µ2(dy)=1 \ \ \ \ for\ all \ x$"
- How can we define an order relation like $<'$ on $(0,1)$ that $\{x:x<'y\}$ is countable for each y?(any example)
- why isn't $\{x:x<'y\}$ a Borel set?
- When f is not a measurable function, Is looking at this example as a counterexample meaningful? the integral on $\mu1\times\mu2$ measure is not even defined.
Here also is the image of the example:
https://i.stack.imgur.com/714Cm.png
I really appreciate your help.(Sorry if this looks easy I'm new to Real analysis!)