Borel/ Lebesgue on $\chi_D$ for $D = \{(x,x):x \in E \}$

Question

Let $E \subset [0,1]$.

Define $D = \{(x,x):x \in E \}$.

Let $f= \chi_D$.

a) Are $f_x, f^y$ Borel for all $x,y \in [0,1]$?

b) Is it true that $f$ is Borel iff $E$ is Borel?

c) If we have both $\mu$ and $\nu$ being the Lebesgue measure, is it true that $f$ is $\mu \times \nu $- measurable iff $E$ is Lebesgue measurable?

My thinking:

a)Yes, since $f_x, f^y$ can only be $0$ or $1$.

b) Yes, since $f$ is Borel iff $D$ is Borel.

c) Yes, since $f$ is Lebesgue measurable iff D is.

Could someone let me know if I have the above correct? Thank you!

Thanks!

Answer 1

Your answers are not corrcet.

For fixed $x \in E$ we have $f_x(y)=1$ if $y=x$ and $0$ otherwise.

For fixed $x \notin E$ we have $f_x(y)=0$ for all $y$.

This makes $f_x$ Borel, hence also Lebesgue measurable always. [No condition on $E$].

Similarly, $f^{y}$ is Borel, hence also Lebesgue measurable always.

Now can write down the answers to all parts of the question.