I was following the definitions from Royden, where the measurability is given as
We say $A \subseteq \mathbb R$ is measurable if $\forall X \subseteq \mathbb R$ $$m^*(X) = m^*(X \cap A) + m^*(X \setminus A)$$
Similarly,
$A\subseteq \mathbb R^n$ is said to be measurable in $\mathbb R^n$ if for all $X\subseteq \mathbb R^n$, $$m^\ast(X)=m^\ast(A\cap X)+m^\ast(X\backslash A).$$
An open rectangle in $\mathbb R^n$ is a set of the form $R=(a_1,b_1)\times\dots\times (a_n,b_n)$. The volumn of the open rectangle is given as $\ell(R)=(b_1-a_1)(b_2-a_2)\cdots(b_n-a_n)$.
We then define Lebesgue outer measure on $\mathbb R^n$ by $m^\ast: P(\mathbb R^n)\rightarrow [0,\infty)\cup\lbrace \infty\rbrace$, $$m^\ast(A)=\inf\left\lbrace \sum_{i=1}^\infty \ell(R_i): A\subseteq \bigcup_{i=1}^\infty R_i\right\rbrace,$$
I am wondering how can we show something in $\mathbb R^n$ is measurable if each of its component is measurable in $\mathbb R$.
That is:
Let $A=A_1\times A_2\times\cdots \times A_n\subseteq \mathbb R^n$, where each $A_i\subseteq \mathbb R$ is measurable. Prove that $A$ is measurable.
Hint: Consider sets $B_i=\mathbb{R}\times \ldots \times A_i \times \mathbb{R}\times\ldots\times\mathbb{R}$ so that if you project $B_i$ onto its $i^{th}$ component you get $A_i$ and if you project onto any other component you get $\mathbb{R}$. Then observe that $A = \bigcap_i B_i$ and that the countable intersection of measurable sets is measurable.