Apply Fubini's theorem to measurable equation to switch variables I would like to understand the following reasoning: Let $m$ be a measurable function on $\mathbb R$ and let E be an equation which is described by $m$ where we insert on boths sides of the equations a combination of real values $a$ and $b$ into $m$, that is, $E$ is of the form $$E: m(expr_1(a,b)) = m(expr_2(a,b))$$
where $expr_1, expr_2$ are simple combinations of $a$ and $b$ (finite sums, products,..). Let's assume that $E$ holds for almost all $b$ for each $a$. How does Fubini's theorem tell us that $E$ holds for some $b=b_0$ for almost every $a$?
 A: Let
$$
M := \{(x,y) \in \Bbb{R}^2 \,\mid\, m(expr_1 (a,b)) \neq m(expr_2 (a,b))\}.
$$
Since $m$ is measurable and since $expr_1 (a,b), expr_2 (a,b)$ are of the form you describe, $M$ is a measurable subset of $\Bbb{R}^2$.
Let $\chi_M$ denote the indicator function/characteristic  function of $M$.
Now Fubini's theorem implies
$$
0=\int_{\Bbb{R}} 0 \, da\overset{(\ast)}{=}\int_{\Bbb{R}}\int_{\Bbb{R}} \chi_M (a,b) \, db \, da = \lambda_2 (M) = \int_{\Bbb{R}}\int_{\Bbb{R}} \chi_M (a,b) \, da \, db.
$$
Here, at $(\ast)$, I used your assumption that for every $a$, $E$ holds for almost all $b$. This means $\chi_M (a,b) = 0$ for almost all $b$ for each $a$ and hence $\int \chi_M (a,b) \, db = 0$ for all $a$.
But the function $F: b \mapsto \int_{\Bbb{R}} \chi_M (a,b) \, da$ which is integrated on the right-hand side is a nonnegative measurable function. Since its integral vanishes, we get $F(b) = 0$ for almost all $b$. In particular, there is $b_0$ with $F(b_0) = 0$.
With the same argument, for each such $b$, we get $0=F(b) =\int_{\Bbb{R}} \chi_M (a,b) \, da$ and thus $\chi_M (a,b) = 0$ for almost all (depending on $b$) $a$.
But this means that $E$ is satisfied for almost all $a$ for each such $b$, in particular for $b_0$.
