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Let $X$ be a locally compact Hausdorff space. Let $\mathcal B$ be the $\sigma$-algebra generated by the family of open subsets of $X$. A measure $\mu$ on $\mathcal B$ is called a (positive) Radon measure if it satisfies the following conditions.

1) $\mu(K) \lt \infty$ for every compact set $K$.

2) $\mu(U) = sup \{\mu(K) : K \subset U$, where $K$ is compact$\}$ for every open $U$.

3) $\mu(E) = inf \{\mu(U): E \subset U$, where $U$ is open $\}$ for every $E\in \mathcal B$.

Let $X$ and $Y$ be locally compact Hausdorff spaces, and let $\mu$ and $\nu$ be Radon measures on $X$ and $Y$ respectively. I would like to prove the following assertions without using the Riesz Representation Theorem(see for example Rudin's Real and Complex Analysis).

1) There exists a unique Radon measure $\lambda$ on $X \times Y$ such that $\lambda (A\times B$) = $\mu(A)\times\nu(B)$ where $\mu(A) \lt \infty$ and $\nu(B) \lt \infty$.

2) Fubini's theorem holds on $X\times Y$.

The motivation is as follows. If $X$ and $Y$ have countable bases, the assertions can be proved by the usual method of product measures and Fubini's theorem.

Here's what I tried to solve the problem. Let $\mathcal B'(X)$ be the $\sigma$-ring generated by the family of compact subsets of $X$. Then the product measure $\lambda = \mu\times\nu$ can be defined on $\mathcal B'(X)\times \mathcal B'(Y)$(see Halmos's Measure Theory). Let $\Gamma$ be the family of finite unions of sets of the form $K\times L$ where $K$(resp. $L$) is a compact subset of $X$(resp. $Y$). Define $\lambda^*(U) = sup \{\lambda (H): H\subset U, H\in \Gamma\}$, where $U$ is an open subset of $X\times Y$. And define $\lambda^*(E) = inf\{\lambda^*(U): E\subset U\}$ where $E$ is a Borel subset of $X\times Y$. Then $\lambda^*$ should be the Radon measure we are looking for.

Makoto Kato
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  • Why do you want to do that without using Riesz representation theorem? As you probably know, one problem with the usual product measure in this context is that the product sigma algebra is strictly smaller than the sigma algebra generated by the product topology. Further, you want a Radon measure and the usual way to construct these is to use the Riesz representation theorem. – PhoemueX May 17 '15 at 05:52
  • @PhoemueX "Further, you want a Radon measure and the usual way to construct these is to use the Riesz representation theorem." Not necessarily so. For example, we usually construct the Lebesgue measure on $\mathbb R^n$ without using the Riesz representation theorem. – Makoto Kato May 20 '15 at 01:36
  • Yes, but in that case you are in a very concrete situation, because you know exactly how $\Bbb{R}^n$ looks like and you can calculate with objects like "boxes", i.e. $[a_1, b_1] \times \cdots \times [a_n, b_n]$. In the general setting of arbitrary LCH spaces $X,Y$, this is no longer possible. – PhoemueX May 20 '15 at 08:20
  • @PhoemueX "In the general setting of arbitrary LCH spaces X,Y, this is no longer possible." So? I don't see why the Riesz theorem is necessary to construct a Radon measure in the general setting. – Makoto Kato May 21 '15 at 00:21
  • Of course it is not. The proof of the Riesz representation theorem for example does not use the Riesz representation theorem. The "this is no longer possible" meant that you can not calculate concretely (with "boxes" or something like that) in general LCH spaces. There is probably another construction, but if the Riesz representation theorem is so convenient, why not use it? – PhoemueX May 21 '15 at 06:49
  • @PhoemueX "but if the Riesz representation theorem is so convenient, why not use it?" I don't see why it is so convenient to solve the current problem. Would you explain? – Makoto Kato May 22 '15 at 00:38
  • I think the basic idea is that bounded linear functionals are easier to work with than measures. Who wants to go through the detail of explicitly constructing a Haar measure on a product of locally compact Hausdorff groups, when instead one can just identify the appropriate positive linear functional and work with that? – D_S Jan 08 '16 at 05:16

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