$E \subset [0, 1]^{2}$ s.t. $m(E_{x}) = 1$ a.e. $x \in [0, 1]$. Show $m(E_{y}) = 1$ a.e. $y \in [0, 1]$.

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Let $E \subset [0, 1]^{2}$ be measurable and let $E_{x} := \{y \in [0, 1] : (x, y) \in E\}$, $E_{y} := \{x \in [0, 1] : (x, y) \in E\}$. If $m(E_{x}) = 1$ a.e. $x \in [0, 1]$ then show that $m(E_{y}) = 1$ a.e. $y \in [0, 1]$.

Is it true that

$$(m \otimes m)(E) = \int_{[0, 1]} m(E_{x}) dx, \quad ?$$

If so, then can I also say

$$(m \otimes m)(E) = \int_{[0, 1]} m(E_{y}) dy,$$

which must be $1$ by hypothesis? Then I want to say this implies that $m(E_{y}) = 1$ a.e. $y \in [0, 1]$. I'm not sure about the integral representations above, or even the last implication.