Showing a function is measurable in a product measure space

I'm working on problem 10.G in Bartle's Elements of Integration and Lebesgue Measure:

Let $f$ and $g$ be real-valued functions on $X$ and $Y$, respectively; suppose that $f$ is $\mathcal{X}$-measurable and $g$ is $\mathcal{Y}$-measurable. If $h$ is defined on $X\times Y$ by $h(x,y) = f(x) g(y)$, show that $h$ is $(\mathcal{X} \times \mathcal{Y})$-measurable.

My problem is that I can't really think of a way to show that any function is $(\mathcal{X} \times \mathcal{Y})$-measurable. The only thing I can think to do is try to show that

$$h^{-1}(\alpha,\infty] \in \mathcal{X} \times \mathcal{Y},$$

but I can't make any progress in this direction. Any advice would be appreciated.

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Do you know how to show that the product of two measurable functions (in the same sigma algebra) are measurable? – Christopher A. Wong Dec 2 '12 at 23:31

What you can do is consider functions $\tilde f, \tilde g:X\times Y\to\mathbb R$, where $$\tilde f(x,y)=f(x),\ \ \tilde g(x,y)=g(y).$$ These are easily seen to be measurable: $$\tilde f^{-1}(\alpha,\infty)=f^{-1}(\alpha,\infty)\times Y, \ \ \tilde g^{-1}(\alpha,\infty)=X\times g^{-1}(\alpha,\infty).$$ As $h=\tilde f\,\tilde g$, it is measurable.
Oh, and then since products of measurable functions on the same space are measurable, $h$ is measurable. Thanks, good tip :) – Antonio Vargas Dec 2 '12 at 23:32