Constant a.e on both $\sigma$-finite measure spaces implies constant a.e w.r to the product space

Question

Let $(X,\Sigma_X,\mu)$, $(Y,\Sigma_Y,\nu)$ be $\sigma$-finite measure spaces. Let $(X \times Y,\Sigma_X \otimes \Sigma_Y,\mu \times \nu)$ be their product space, and let $f:X \times Y \to \Bbb R$ be a $\mu \times \nu$ measurable function in that product space.

Assume that:

1. For almost every $x \in X$, the function $y \mapsto f(x,y)$ is constant.
2. For almost every $y \in Y$, the function $x \mapsto f(x,y)$ is constant.

Prove that there exists a constant $c \in \Bbb R$ such that $f=c$, almost everywhere with respect to $\mu \times \nu$.

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I have written down two solutions, but I'm not sure that they are air tight. I would be grateful for any comment or advice to either confirm the validity of my proofs or to point out why they are wrong.

By assumption:

$\exists N_x \subset X, c_x \in \Bbb R$ s.t $N_x$ is $\mu$-negligible and $\forall x \notin N_x, y \in Y$ $f(x,y)=c_x$.

$\exists N_y \subset Y, c_y \in \Bbb R$ s.t $N_y$ is $\nu$-negligible and $\forall x \in X, y \notin N_y$ $f(x,y)=c_y$.

Now, $N_x, N_y$ are $\mu,\nu$ measurable respectively, therefore $N_x \times N_y$ is $\mu \times \nu$ measurable.

Here are my solution attempts:

Method (1):

We have that $f(x,y)=\begin{cases} c_x, & \text{$(x,y) \in N_x^\complement \times Y$} \\ c_y, & \text{$(x,y) \in X \times N_y^\complement$} \\ unknown, & \text{$(x,y) \notin N_x^\complement \times N_y^\complement$} \\ \end{cases}$

Furthermore, $N_x^\complement \times N_y^\complement \subset X \times N_y^\complement$ and $N_x^\complement \times N_y^\complement \subset N_x^\complement \times Y$. Hence, $\forall (x,y) \in N_x^\complement \times N_y^\complement$ $c_x=f(x,y)=c_y$, and that is what we wanted to prove.

Method (2):

Define $g(x,y):=|f(x,y)-c_x|$. $g$ is $\mu \times \nu$ measurable, and non-negative, so we can use Tunelli's theorem to get:

$\int_{X \times Y}g(x,y) = \int_{X} \int_{Y}g(x,y) = \int_{N_x} \int_{Y}g(x,y) + \int_{N_x^\complement} \int_{Y}|f(x,y)-c_x| \leq \mu(N_x) \max_{Y} g(x,y) + \int_{N_x^\complement} \int_{Y}|c_x-c_x| \leq 0$.

Therefore $g(x,y)=0$ a.e with respect to $\mu \times \nu$, which means $f(x,y)=c_x$ a.e as requested.

Answer 1

How about this?

Let $F \subset X$ be the null set on which $y \mapsto f(x,y)$ is not constant. Let $G \subset Y$ be the null set on which $x \mapsto f(x,y)$ is not constant.

Notice that $F \times Y$ and $G \times X$ are null, and so $F \times Y \cup G \times X$ is null too.

Restricting $f$ to $E = X \times Y \backslash (F \times Y \cup G \times X)$, $f$ is genuinely constant in both variables, hence is constant on this set. Said another way, for $(x,y), (x',y') \in E$, we have $f(x,y) = f(x,y')$ and $f(x,y') = f(x',y')$, hence $f(x,y) = f(x',y')$.