Locally finite Borel measure on $\mathbb{R}^2$ that is not a product measure. In the construction of Lebesgue-Stieltjes measures on $\mathbb{R}$, I have learned that a Borel measure that is finite on bounded intervals corresponds to a right-continuous increasing real-valued function up to constants. Then we can construct a product measure in $\mathbb{R}^2$ from two such locally finite Borel measures.
My question is whether every locally finite Borel measure on $\mathbb{R}^{d}$ can be expressed as a product measure from those on $\mathbb{R}$ in this way.
The question arose from a separate construction of Lebesgue-Stieltjes measures on $\mathbb{R}^2$ by generating it from function $F:\mathbb{R}^2\rightarrow \mathbb{R}$ satisfying the following properties:

*

*F is increasing in the sense that $a_0\leq a_1$, $b_0\leq b_1$ then $F((a_1,b_1))-F((a_0,b_0))\geq 0$.

*$F((a_1,b_1))-F((a_0,b_1))-F((a_1,b_0))+F((a_0,b_0))\geq 0$ for $a_0\leq a_1$, $b_0\leq b_1$.

*$F$ is right-continuous for each coordinate.

Then $\mu_F((a_0,a_1] \times (b_0,b_1])=F((a_1,b_1))-F((a_0,b_1))-F((a_1,b_0))+F((a_0,b_0))$ can be extended to a Borel measure using a similar proof in $\mathbb{R}$, and similar correspondence to locally finite Borel measures hold as well.
It is easy to see that a product measure from $f$ and $g$ satisfies these conditions by defining $F((a,b))=f(a)g(b)$.
But I was not able to show the converse, that is, this construction is just a product measure construction.
I tried to come up with a locally finite Borel measure on $\mathbb{R}^2$ that cannot be constructed as a product measure, but I was not able to do so.
The only thing I could see was that if $F((x,y))=f(x)+g(y)$ where each function depends on each variable only, then the resulting measure generated by $F$ is just zero measure.
I have also found that there is a theory of disintegration of measures, but I am not familiar with the topic.
Any help would be appreciated.
 A: Let $\mu$ be the image measure of the Lebesgue measure, $\lambda$, on $\Bbb R$ by the map $d: \Bbb R \rightarrow \Bbb R ^2$ such that $d(x) = (x,x)$. Then we have $\mu((a_0,a_1] \times (b_0,b_1])=\lambda((a_0,a_1] \cap (b_0,b_1])$.
Let us prove that there are no functions $f$ and $g$ such that $\mu((a_0,a_1] \times (b_0,b_1])= (f(a_1)-f(a_0))(g(b_1)-g(b_0))$. So $\mu$ is not a product measure of Borel measures on $\Bbb R$.
Proof:
Suppose such functions $f$ and $g$ exist. It is clear that $f$ can not be constant. So there are $a_1$ and $a_0$ such that $f(a_1)-f(a_0) \neq 0$. So given any $b_0$ and $b_1$ such that $(a_0,a_1] \cap (b_0,b_1] = \emptyset$, we have
$$ (f(a_1)-f(a_0))(g(b_1)-g(b_0)) = \mu((a_0,a_1] \times (b_0,b_1]) = \lambda(\emptyset)=0$$
Since  $f(a_1)-f(a_0) \neq 0$, we must have $g(b_1)-g(b_0)=0$.
But then for any $b_0$ and $b_1$ such that $(a_0,a_1] \cap (b_0,b_1] = \emptyset$, we have
$$\lambda((b_0,b_1])= \mu((b_0,b_1] \times (b_0,b_1])= (f(b_1)-f(b_0))(g(b_1)-g(b_0))= 0 $$
Contradiction.
So, there are no functions $f$ and $g$ such that $\mu((a_0,a_1] \times (b_0,b_1])= (f(a_1)-f(a_0))(g(b_1)-g(b_0))$. So $\mu$ is not a product measure of Borel measures on $\Bbb R$.
