Measurable with respect to the product measure Let $X=Y=[0,1]$, and let $\mu = \nu$ be Lebesgue measure. Show that each open set in $X\times Y$ is measurable, and hence each Borel set in $X\times Y$ is measurable. Is every continuous real-valued function on $[0,1]\times [0,1]$ measurable with respect to the product measure?
 A: Let $O$ be an open set in $X\times Y$. If $x\in O$ then there is an open ball $O_x$ containing $x$.  You can construct an open rectangle, $R_x$, with rational endpoints to contain $x$ but contained in $O_x$. Now 
\begin{eqnarray*}
O=\bigcup_{x\in O} R_x.
\end{eqnarray*}
You assert that it is countable because there are at most $\mathbb{Q}^4$ such rectangles.  So there are at most countably many rectangles with rational endpoints.
Thus,
\begin{eqnarray*}
O=\bigcup_{x\in O} R_x=\bigcup_{n=1}^\infty R_n.
\end{eqnarray*}
Every Borel set is just the union of open sets, thus measurable as well.
That $f$ should be continuous means that the inverse image of open sets are open.
But that means the inverse image are measurable.  Thus, the inverse image of open sets are measurable which is equivalent to a function being measurable.
A: Let A be an open subset of $X\times Y=[0,1]\times [0,1]$. Then you can write it as a countble union : $$A=\bigcup_{i=1 }^\infty  I_i \times J_i$$
in which $I_i,J_i$ are open intervals of [0,1]. 
Of course, $I_i\times J_i$ are all measurable subsets of $[0.1]^2$, and therefore their union is measurable (since the measurable subsets of $[0,1]^2$ form a σ-Algebra).
Each Borel set in $X\times Y$ is produced by countable unions and intersections of sets of the form of A (above), therefore it is measurable ( again, use the fact that the measurable subsets of $[0,1]^2$ form a σ-Algebra).
Now, for the second question, consider a continuous function $f:[0,1]\times[0,1]\longrightarrow \mathbb R$.
Then, we know that it is measurable if and only if for each borel set $B\subset \mathbb R$, it holds that $f^{-1}(B)$ is measurable.
Take a borel set $B\subset\mathbb R$. We know that the inverse image of a borel set through a continuous function is a borel set , therefore $f^{-1}(B)$ is a Borel set, thus it is measurable, QED.
