Understanding measurability on a product space

Question

I am having trouble understanding the concept of borel algebra on $\mathbb R^n$ and how it applies to measurable functions. I learned in class that the sigma algebra on $\mathbb R^n$ is generated by $O_1\times O_2\times\ldots O_n$, where each $O_i$ is a collection of open sets in $\mathbb R$. When applied to a measurable function $g:\mathbb R^n\rightarrow \mathbb R$, this amounts to saying that for any $B\in\mathcal B(\mathbb R)$, $g^{-1}(B)=C_1\times C_2\times\ldots C_n$, for some $C_1,\ldots C_n\in\mathcal B(\mathbb R)$. However, this doesn't seem to be correct to me! To me, it seems that a possible value for, say, the $n$th co-ordinate, must depend on the values of the other co-ordinates and so the set cannot be a product.

For example, consider a function $g:\mathbb R^2\rightarrow\mathbb R, \,g(x) = x_1+x_2$. We know that this is measurable. But then, how do you represent, say, $g^{-1}((0,\infty))$ as a product of anything? Clearly, $(-2,3),(3,-2)\in g^{-1}((0,\infty))$, and so if it is a product of two sets, it must contain $(-2,-2)$, which it doesn't!

I realize that I must be making some very basic conceptual error here, and I apologize for that. Can someone please help me understand this? Thanks in advance.

Answer 1

[T]his amounts to saying that for any $B\in\mathcal{B}(\mathbb{R})$, $g^{-1}(B)=C_1\times C_2\times\cdots\times C_n$, for some $C_1,\ldots,C_n\in \mathcal{B}(\mathbb{R})$.

It does not. To say that $\mathcal{B}(\mathbb{R}^n)$ is generated by such Cartesian products says only that $\mathcal{B}(\mathbb{R}^n)$ is the smallest $\sigma$-algebra that contains all such Cartesian products. The variety of sets found in $\mathcal{B}(\mathbb{R}^n)$ is bewildering.

Chapter 5 of Lebesgue Integration on Euclidean Space, by Frank Jones, contains the folllowing advice.

Do not think that every Borel set can be constructed from the open sets by countably many of the standard set operations. Instead, proofs involving Borel sets in this book will rely on the basic definition of the Borel sets as the smallest $\sigma$-algebra containing the open sets.

For example, when proving that

if

• $E\in\mathcal{B}(\mathbb{R}^n)$,
• $f:E\to\mathbb{R}^m$ is continuous, and
• $A\in\mathcal{B}(\mathbb{R}^m)$,

then

• $f^{-1}(A)\in\mathcal{B}(\mathbb{R}^n)$,

the same book states the following.

Remember! We don't know how to construct Borel sets, so we are unable to provide a proof which begins by analyzing the structure of $A$ and then progresses to a statement about the structure of $f^{-1}(A)$. Instead, we must give a cleverer (and easier!) proof based on the definition of the Borel sets. In fact, our proof will ignore the set $A$ entirely, and will instead prove the desired conclusion for all Borel sets in $\mathbb{R}^m$.