# Is Riesz measure an extension of product measure?

Suppose $X$ and $Y$ are compact Hausdorff spaces and $(X, \mathcal A, \mu)$ and $(Y, \mathcal B, \nu)$ are finite regular Borel measure spaces. (By regular I mean that every measurable set can be approximated from above by open measurable sets and from below by compact measurable sets.) Let $\mu \times \nu$ be product measure on $(X \times Y, \mathcal A \times \mathcal B)$. For $f \in C(X \times Y)$ define $\psi(f) = \int f \; d(\mu \times \nu)$. By the Riesz–Markov–Kakutani representation theorem there is a unique regular Borel measure $\lambda$ on $X \times Y$ such that $\psi(f) = \int f \; d\lambda$. I believe I can show that $\lambda$ must be an extension of $\mu \times \nu$, but I am looking for a simpler or more elegant proof.

• I guess $f$ should be in $C(X\times Y)$. – Michael Greinecker May 10 '15 at 14:10
• Perhaps you could give a sketch of your proof, so that prospective answerers know whether theirs is actually simpler or more elegant. – Nate Eldredge May 13 '15 at 15:31
• – PhoemueX Aug 10 '15 at 21:06

Here is a proof. I do not know if it is simpler or more elegant than whatever you have in mind.

Let us begin by establishing the following

Lemma. Let $g : Y \to [0,\infty)$ be a bounded nonnegative Borel function. For Borel sets $A \in \mathcal{A}$, set $\mu_g(A) = \int 1_A(x) g(y) \,d\lambda$. Then $\mu_g$ is a finite regular Borel measure on $X$.

Proof. Showing that $\mu_g$ is a measure is routine (use the monotone convergence theorem). To see it is regular, since $X$ is compact, it suffices to show it is inner regular. Let $A \in \mathcal{A}$ be Borel and fix $\epsilon > 0$. Since $\lambda$ is regular, we can find a compact $E \subset A \times Y$ with $\lambda((A \times Y) \setminus E) < \epsilon/\|g\|_\infty$. Let $F \subset X$ be the projection of $E$ onto the first coordinate, so that $F$ is also compact. Then $E \subset F \times Y \subset A \times Y$, so we also have $\mu((A \times Y) \setminus (F \times Y)) < \epsilon/\|g\|_\infty$.

Now \begin{align*}\mu_g(A \setminus F) &= \int (1_A(x) - 1_F(x)) g(y)\,d\lambda \\ &\le \|g\|_\infty \int (1_A(x) - 1_F(x))\,d\lambda \\ &= \|g\|_\infty \lambda((A \times Y) \setminus (F \times Y)) \\ &< \epsilon \end{align*} and thus $\mu_g$ is regular. $\square$

Now fix $g \in C(Y)$ with $g \ge 0$. I claim $\mu_g(A) = \mu(A) \int g\,d\nu$. Since both sides are regular Borel measures (thanks to our lemma), then by the uniqueness in the Riesz representation theorem, it suffices to show that $\int f\,d\mu_g = \int f\,d\mu \int g\,d\nu$ for all $f \in C(X)$. So let $f \in C(X)$. By approximation with simple functions, we have \begin{align*}\int f\,d\mu_g &= \int f(x) g(y)\,d\lambda \\ &= \int f(x) g(y) \, d(\mu \times \nu) && \text{by definition of \lambda} \\ &= \int f(x)\,d\mu \int g(y)\,d\nu && \text{by Fubini's theorem} \end{align*} as desired.

Now fix $A \in \mathcal{A}$. For Borel sets $B \subset \mathcal{B}$, set $\nu_A(B) = \lambda(A \times B) = \int 1_A(x) 1_B(y)\,d\lambda$. I claim $\nu_A(B) = \mu(A) \nu(B)$. Again, both sides are regular Borel measures, so it suffices to show $\int g\,d\nu_A = \mu(A) \int g\,d\nu$ for all $g \in C(Y)$. By taking positive and negative parts, we can assume $g \ge 0$. Then by approximating $g$ by simple functions, we have $$\int g\,d\nu_A = \int 1_A(x) g(y)\,d\lambda = \mu_g(A) = \mu(A) \int g\,d\nu$$ as desired.

Thus we have shown, for all $A \in \mathcal{A}$, $B \in \mathcal{B}$, that $\lambda(A \times B) = \mu(A) \nu(B) = (\mu \times \nu)(A \times B)$. It now follows, by a monotone class or $\pi$-$\lambda$ argument, that $\lambda(C) = (\mu \times \nu)(C)$ for all $C \in \mathcal{A} \times \mathcal{B}$, which is the desired statement. For instance, set $\mathcal{P} = \{ A \times B : A \in \mathcal{A}, B \in \mathcal{B}\}$ and $\mathcal{L} = \{C \in \mathcal{A} \times \mathcal{B} : \lambda(C) = (\mu \times \nu)(C)\}$. It is easy to check that $\mathcal{P}$ is a $\pi$-system and $\mathcal{L}$ is a $\lambda$-system. We have just shown $\mathcal{P} \subset \mathcal{L}$, so the $\pi$-$\lambda$ lemma implies $\sigma(\mathcal{P}) \subset \mathcal{L}$. But $\sigma(\mathcal{P}) = \mathcal{A} \times \mathcal{B}$.

• At first I thought this result should be quite easy to prove. Nate's approach is more straightforward than my own. I would welcome other answers too. – Richard Hevener May 13 '15 at 21:14