x-section of two-dimensional Lebesgue measurable subset

Question

Let $A \subseteq [0,1]^2$ be measurable with $m_2(A) = 1$ ($m_2$ is a two-dimensional Lebesgue measure). Define the $x$-section of $A$ : $$s_x(A) = \{y \in \mathbb{R}| (x,y) \in A\}.$$ Show that for almost every $x$,$$m(s_x(A)) = 1.$$

This is my idea :

Observe $$1 = m_2(A) = \int_{[0,1]}\int_{[0,1]} 1_A(x,y) dydx = \int\int_{[0,1]^2}1_{s_x(A)}(y) dydx = \int_{[0,1]}m(s_x(A)) dx.$$ Since for each $x$, $s_x(A) \subseteq [0,1],$ $$\int_{[0,1]}m(s_x(A))dx\leq \int_{[0,1]} m([0,1]) dx = m([0,1])=1.$$ So $$\int_{[0,1]}m(s_x(A))dx \leq 1, \ \mbox{but}\ \int_{[0,1]}m(s_x(A))dx = 1.$$

I seem that I should have $m(s_x(A)) = 1$ for almost every $x$, but it is just "seem". How should I proceed rigorously from the conclusion I get above ?

Answer 1

So you have (from the inequalities)

$$\int_{[0,1]} 1 dx = 1 = \int_{[0,1]} m(s_x (A)) dx $$

and $m(s_x (A)) \le 1$. So the function $f(x) = 1- m(s_x(A))$ is non-negative and

$$\int_{[0,1]} f(x) dx = 0.$$

So $f = 0$ almost everywhere (Try to argue this by contradiction).