I am working on a problem$^{(1)}$ like this:
Suppose $A$ is a Lebesgue measurable subset of [0, 1]$^2$ with 2-dimensional Lebesgue measure $m_2 (A) = 1$. Show that for almost every $x \in [0, 1]$ (with respect to 1-dimensional Lebesgue measure) the set $S_x(A)$ has 1-dimensional Lebesgue measure one.
Here are how I look at this problem intuitively:
- The Lebesgue measure $m_1, m_2$ and $m_3$ relate intuitively to length, area and volume. Therefore $m_2(A) = 1$ means that it has area of 1.
- $S_x(A)$ is defined as $\{y \in [0,1] \mid (x, y) \in A\}$, therefore $m_1(S_x) = 1$ means that one side of rectangle $A$ has length of 1.
- Here I think I need to prove $m_1 \left ( x \in [0, 1] \mid m_1 (S_x (A)) \neq 1 \right ) = 0.$
Here are what I did so far to the best of my ability:
(1) Suppose by way of contradiction that $m_1 \left ( x \in [0, 1] \mid m_1 (S_x (A)) \neq 1 \right ) \neq 0.$ This implies that $\exists x \in [0, 1]$ such that $0 \le m_1(S_x (A)) < 1.$
(2) However, it is given in the hypothesis that $m_2(A) = 1.$ In order for this to be true, then $m_1(T_y(A))$ has to be greater than 1, which contradicts the fact that $T_y(A) \subseteq [0, 1].$
(3) Hence it has to be $m_1 \left ( x \in [0, 1] \mid m_1 (S_x (A)) \neq 1 \right ) = 0.$
Please let me know if I did this problem correctly. Thank you for your time and effort.
(1) Richard F. Bass' Real Analysis, 2nd. edition, chapter 11: Product Measure, Exercise 11.4, page 88.