Let $f$ be a locally integrable function on $\mathbb{R}^2$ such that for any real numbers $a,b$ outside a set of measure zero, we have that $f(a,x)=f(b,x)$ for almost every $x$ and $f(y, a)=f(y,b)$ for almost every $y$. I would like to show that $f$ is constant almost everywhere on $\mathbb{R}^2$.
A hint I received was to compare the integrals of $f$ on two squares that are translates of each other and use Lebesgue's differentiation theorem. I if Let $Q=[0,a]\times[0,b]$ and its translate $Q_{\varepsilon, \delta}=[\varepsilon,a+\varepsilon]\times[\delta,b+\delta]$, then I find that
$$\int_Q f=\int_{Q_{\varepsilon, \delta}}f. $$
However, I'm unsure of how to appeal to Lebesgue's differentiation theorem since it usually applies to balls.
Best Answer
As said @FShrike, Lebesgue differentiation theorem works on larger classes of sets than balls. If you want to use Lebesgue differentiation theorem only with balls, notice that the bounded open squares are a basis of open subsets of $\mathbb{R}^2$, thus they generate the Borel subsets of $\mathbb{R}^2$. Let $c = \int_{]0,1[^2} f \in \mathbb{R}$. If we show that for all square $Q$, $\int_Q f = \int_Q c = c|Q|$, then for all Borel subset $A$ of $\mathbb{R}^2$, $\int_A f = \int_A c = c|A|$ so $f = c$ almost everywhere (by Lebesgue differentiation theorem).
Let $Q = ]a,a + 1[ \times ]b,b + 1[$ be a unit square. Then, \begin{align*} \int_Q f & = \int_a^{a + 1}\int_b^{b + 1} f(y,x) \, dx\, dy \textrm{ by Fubini-Tonelli,}\\ & = \int_a^{a + 1}\int_0^1 f(y,x) \, dx\, dy\\ & \textrm{ because for almost every $y$, $f(y,\cdot)$ is constant,}\\ & = \int_0^1 \int_0^1 f(y,x) \, dy \, dx\\ & \textrm{ because for almost every $x$, $f(\cdot,x)$ is constant,}\\ & = c. \end{align*} Then, if $Q = ]a,a + n[ \times ]b,b + m[$ where $n,m$ are positive integers, we have $Q = \bigsqcup_{i = 0}^{n - 1}\bigsqcup_{j = 0}^{m - 1} ]a + i,a + i + 1[ \times ]b + j,b + j + 1[$ almost everywhere (in the sens that their symmetric difference is negligible). We deduce that $\int_Q f = nmc = \int_Q c$.
If $Q = \left]a,a + \frac{p}{q}\right[ \times \left]b,b + \frac{n}{m}\right[$ for positive integers $p,q,n,m$, we use the previous inequality with $y,x \mapsto f\left(\frac{y}{q},\frac{x}{m}\right)$ that verifies the same properties as $f$, to deduce that $\int_Q f = \frac{pn}{qm}c = \int_Q c$. Then, by density of $\mathbb{Q}$ in $\mathbb{R}$, we deduce this equality for any bounded open square $Q$, and conclude that $f$ is constant and equals $c$.