Assume $A \subset \mathbb{R^{n+m}}$ has Lebesgue measure zero then prove for almost every $x \in \mathbb{R^n}$ the set
$$A_x=\{y\in \mathbb{R^m} ; (x,y) \in \mathbb{R^{n+m}} \}$$
is in Lebesgue $\sigma$-Algebra of $\mathbb{R^m}$ and has $m$-dimensional Lebesgue measure zero.
First I thought the Lebesgue $\sigma$-Algebra of $\mathbb{R^{m+n}}$ is the product of Lebesgue $\sigma$-Algebra of $\mathbb{R^{n}}$ and Lebesgue $\sigma$-Algebra of $\mathbb{R^{m}}$ and I proved the statement using this fact which I realised is wrong after reading this: Is the n-dim lebesgue measure the product of the lebesgue measure?
Now I am not sure how to prove this using definition.
Lebesgue measure on $\mathbb R^{n+m}$ is not the product of the Lebesgue measures on $\mathbb R^{n}$ and $\mathbb R^{m}$ but it is the completion of this product. You have to use the definition of completion of measures to prove this result. There exist Borel sets $B$ and $D$ and a set $C$ such that $A = B \cup C$, $C \subset D$ and both $B$ and $D$ have measure $0$. Now $A_x = B_x \cup C_x$ and $C_x \subset D_x$ from which it follows that $A_x$ has measure $0$ since $B_x$ and $D_x$ have measure $0$..