Let $X$ be a locally compact Hausdorff space. We denote by $\mathcal B(X)$ the $\sigma$-algebra of Borel sets of $X$. A positive Radon measure $\mu$ on $X$ is a measure defined on $\mathcal B(X)$ with the following properties.

(1) $\mu(K) \lt \infty$ for all compact sets $K$.

(2) For any $E\in \mathcal B(X)$, $\mu(E) = \text{inf}\ \mu(U)$ for all open $U$ such that $E \subset U$.

(3) For any open $U$, $\mu(U) = \sup\ \mu(K)$ for all compact $K$ such that $K \subset U$.

It can be proved(see, for example, Rudin's Real and Complex Analysis) that a positive Radon measure has the following property.

(3') For any $\sigma$-finite set $E\in \mathcal B(X)$, $\mu(E) = \sup\ \mu(K)$ for all compact $K$ such that $K \subset E$.

Now let $\mu, \nu$ be positive Radon measures on locally compact Hausdorff spaces $X$ and $Y$. We denote by $\mathcal B(X)\otimes \mathcal B(Y)$ the smallest $\sigma$-algebra on $X\times Y$ containg all the sets of the form $A\times B, A\in \mathcal B(X), B \in \mathcal B(Y)$. Suppose $\mu$ and $\nu$ are $\sigma$-finite. Let $\mu\otimes \nu$ be the product measure on $\mathcal B(X)\otimes \mathcal B(Y)$. Let $f:X\times Y \rightarrow \mathbb C$ be a continuous function with compact support. It can be proved using Stone-Weierstrass theorem that $f$ is measurable with respect to $\mathcal B(X)\otimes \mathcal B(Y)$. Hence it is integrable with respect to $\mu\otimes \nu$. Hence $f \rightarrow \int f d(\mu\otimes\nu)$ is a positive linear functional defined on the vector space of continuous functions of compact support on $X\times Y$. By Riesz representation theorem there is a unique positive Radon measure $\lambda$ on $X\times Y$ such that $\int f d(\mu\otimes\nu) = \int f d\lambda$ for all continuous functions of compact support on $X\times Y$. Lang claims without explanation in his book Real and Functional Analysis that $\lambda$ and $\mu\otimes\nu$ coincide on $\mathcal B(X)\otimes \mathcal B(Y)$(Th. 6.3, Ch. IX, Sect. 7, p.274). I don't understand why this is so. Would anyone elaborate on this?

EDIT(29 July 2015) For the convenience of the reader, I reproduce Theorem 6.3 and its proof in Lang's book.

Theorem 6.3. Let $X, Y$ be locally compact Hausdorff spaces and let $\mu, \nu$ be positive $\sigma$-regular Borel measures on $X$ and $Y$, respectively. Assume that $\mu$ and $\nu$ are $\sigma$-finite with respect to these measures. Then all functions in $C_c(X\times Y)$ are in $\mathcal L^1(\mu\otimes\nu)$, and there exists a unique $\sigma$-regular Borel measure on $X\times Y$ which restricts to $\mu\otimes\nu$ on $\mathcal B(X)\otimes\mathcal B(Y)$.

Proof. Lemma 6.1 shows that functions in $C_c(X\times Y)$ are $(\mu\otimes\nu)$-measurable, and combined with Fubini's theorem shows that these functions are in $\mathcal L^1(\mu\otimes\nu)$. The map $f \rightarrow \int_{X\times Y} f d(\mu\otimes\nu)$ is obviously a positive functional on $C_c(X\times Y)$, and we can therefore apply Theorem 2.3 to get a $\sigma$-regular Borel measure having the desired properties. The Corollary 2.8 gives the uniquness, thus proving our theorem.

Remark. Lemma 6.1 just proves by using Stone-Weierstrass theorem that functions in $C_c(X\times Y)$ are $(\mu\otimes\nu)$-measurable. Theorem 2.3 is a part of Riesz representation theorem. A $\sigma$-regular Borel measure is just a positive Radon measure in our terminology.


Lemma: Let $K\subset X$ and $L\subset Y$ be compact and let $U\subset X\times Y$ be open with $K\times L\subset U$. Then there are open sets $V\subset X$ and $W\subset Y$ with $$ K\times L\subset V\times W\subset U. $$ Proof: For each $\left(x,y\right)\in K\times L$ there are -- by definition of the product topology -- open sets $V_{x,y}\subset X$ and $W_{x,y}\subset Y$ with $\left(x,y\right)\in V_{x,y}\times W_{x,y}\subset U$.

By compactness of $K$, there is thus for each $y\in L$ some $n_{y}\in\mathbb{N}$ and suitable $x_{1}^{\left(y\right)},\dots,x_{n_{y}}^{\left(y\right)}\in X$ with $$ K\subset\bigcup_{i=1}^{n_{y}}V_{x_{i}^{\left(y\right)},y}=:V_{y}. $$ Note that $$ W_{y}:=\bigcap_{i=1}^{n_{y}}W_{x_{i}^{\left(y\right)},y}\subset Y $$ is open with $y\in W_{y}$. By compactness of $L$, there are thus $y_{1},\dots,y_{m}\in L$ with $L\subset\bigcup_{j=1}^{m}W_{y_{j}}=:W$. Set $$ V:=\bigcap_{j=1}^{m}V_{y_{j}} $$ and note that $V\subset X$ and $W\subset Y$ are open with $K\times L\subset V\times W$, so that it remains to show $V\times W\subset U$.

To this end, let $\left(x,y\right)\in V\times W$ be arbitrary. By definition of $W$, there is $j\in\left\{ 1,\dots,m\right\} $ with $y\in W_{y_{j}}$. By definition of $V$, we have $x\in V_{y_{j}}=\bigcup_{i=1}^{n_{y}}V_{x_{i}^{\left(y_{j}\right)},y_{j}}$ and hence $x\in V_{x_{i}^{\left(y_{j}\right)},y_{j}}$ for some $i\in\left\{ 1,\dots,n_{y}\right\} $. But by definition of $W_{y_{j}}$, we have $y\in W_{x_{i}^{\left(y_{j}\right)},y_{j}}$ and hence $\left(x,y\right)\in V_{x_{i}^{\left(y_{j}\right)},y_{j}}\times W_{x_{i}^{\left(y_{j}\right)},y_{j}}\subset U$. $\square$

Now, we can use the usual proof to show $\lambda\left(K\times L\right)=\mu\left(K\right)\nu\left(L\right)$ for all compact $K,L$. Let $\varepsilon\in\left(0,1\right)$ be arbitrary. By regularity, there are open sets $U\subset X\times Y$ and $V_{1}\subset X$ as well as $W_{1}\subset Y$ with $K\times L\subset U$, $K\subset V_{1}$ and $L\subset W_{1}$ and with \begin{align*} \lambda\left(K\times L\right) & \leq\lambda\left(U\right)<\lambda\left(K\times L\right)+\varepsilon,\\ \mu\left(K\right) & \leq\mu\left(V_{1}\right)<\mu\left(K\right)+\frac{\varepsilon}{1+\nu\left(L\right)},\\ \nu\left(L\right) & \leq\nu\left(W_{1}\right)<\nu\left(L\right)+\frac{\varepsilon}{1+\mu\left(K\right)}. \end{align*} By the Lemma, there are open sets $V_{2}\subset X$ and $W_{2}\subset Y$ with $K\times L\subset V_{2}\times W_{2}\subset U$. Set $V_{3}:=V_{1}\cap V_{2}$ and $W_{3}:=W_{1}\cap W_{2}$ and note $K\times L\subset V_{3}\times W_{3}\subset U$. By Uryshon's Lemma, there are $\varphi\in C_{c}\left(X\right)$ and $\psi\in C_{c}\left(Y\right)$ with $0\leq\varphi\leq1$, $0\leq\psi\leq1$ and $\varphi\equiv1$ on $K$ as well as ${\rm supp}\,\varphi\subset V_{3}$. Analogously, $\psi\equiv1$ on $L$ and ${\rm supp}\,\psi\subset W_{3}$. Define $\varphi\otimes\psi\left(x,y\right):=\varphi\left(x\right)\psi\left(y\right)$. Then $$ \mu\left(K\right)\nu\left(L\right)\leq\int\varphi\,{\rm d}\mu\cdot\int\psi\,{\rm d}\nu\leq\mu\left(V_{3}\right)\nu\left(W_{3}\right)\leq\mu\left(V_{1}\right)\nu\left(W_{1}\right)<\mu\left(K\right)\nu\left(L\right)+3\varepsilon. $$ Furthermore, $$ \lambda\left(K\times L\right)\leq\int\varphi\otimes\psi\,{\rm d}\lambda\leq\lambda\left(V_{3}\times W_{3}\right)\leq\lambda\left(U\right)<\lambda\left(K\times L\right)+\varepsilon. $$ But $\int\varphi\otimes\psi\,{\rm d}\lambda=\int\varphi\,{\rm d}\mu\cdot\int\psi\,{\rm d}\nu$, which finally implies $$ \left|\mu\left(K\right)\nu\left(L\right)-\lambda\left(K\times L\right)\right|\leq3\varepsilon+\left|\int\varphi\,{\rm d}\mu\cdot\int\psi\,{\rm d}\nu-\int\varphi\otimes\psi\,{\rm d}\lambda\right|+\varepsilon=4\varepsilon. $$ Since $\varepsilon>0$ was arbitrary, we obtain the claim for compact sets $K,L$.

Now, let $U\subset X$ and $V\subset Y$ be open and of finite measure. Let $\alpha<\lambda\left(U\times V\right)$ be arbitrary. By inner regularity of $\lambda$, there is a compact set $T\subset U\times V$ with $\alpha<\lambda\left(T\right)\leq\lambda\left(U\times V\right)$. Let $\pi_{1}:X\times Y\to X$ be the usual projection and define $\pi_{2}$ analogously. Note that $K_{1}:=\pi_{1}\left(T\right)\subset U$ and $L_{1}:=\pi_{2}\left(T\right)\subset V$ are compact with $T\subset K_{1}\times L_{1}\subset U\times V$. Hence, $$ \alpha<\lambda\left(T\right)\leq\lambda\left(K_{1}\times L_{1}\right)\leq\lambda\left(U\times V\right). $$

By inner regularity of $\mu,\nu$, there are also compact sets $K_{2}\subset U$ and $L_{2}\subset V$ with

\begin{align*} \mu\left(K_{2}\right) & \leq\mu\left(U\right)<\mu\left(K_{2}\right)+\varepsilon,\\ \nu\left(L_{2}\right) & \leq\nu\left(V\right)<\nu\left(L_{2}\right)+\varepsilon. \end{align*} Now, let $K_{3}:=K_{1}\cup K_{2}$ and $L_{3}:=L_{1}\cup L_{2}$. We then have $K_{3}\times L_{3}\subset U\times V$ and $$ \alpha<\lambda\left(K_{1}\times L_{1}\right)\leq\lambda\left(K_{3}\times L_{3}\right)\leq\lambda\left(U\times V\right) $$ as well as $$ \alpha<\lambda\left(K_{3}\times L_{3}\right)=\mu\left(K_{3}\right)\nu\left(L_{3}\right)\leq\mu\left(U\right)\nu\left(V\right). $$ Since $\alpha<\lambda\left(U\times V\right)$ was arbitrary, we see $\lambda\left(U\times V\right)\leq\mu\left(U\right)\nu\left(V\right)<\infty$.

Furthermore, we get $$ \left(\mu\left(U\right)-\varepsilon\right)\left(\nu\left(V\right)-\varepsilon\right)\leq\mu\left(K_{2}\right)\nu\left(L_{2}\right)\leq\mu\left(K_{3}\right)\nu\left(L_{3}\right)=\lambda\left(K_{3}\times L_{3}\right)\leq\lambda\left(U\times V\right). $$ By sending $\varepsilon\downarrow0$, we see $\mu\left(U\right)\nu\left(V\right)\leq\lambda\left(U\times V\right)$ and hence $$ \lambda\left(U\times V\right)=\mu\left(U\right)\nu\left(V\right) $$ for all open sets $U\subset X$ and $V\subset Y$ of finite measure.

Now, by $\sigma$-finiteness and outer regularity of $\mu,\nu$, we can find increasing families $\left(U_{n}\right)_{n\in\mathbb{N}},\left(V_{n}\right)_{n\in\mathbb{N}}$ of open sets with $X=\bigcup_{n}U_{n}$ and $Y=\bigcup_{n}V_{n}$ as well as $\mu\left(U_{n}\right)<\infty$ and $\nu\left(V_{n}\right)<\infty$. For $n\in\mathbb{N}$ and any open set $V\subset Y$ of finite measure, $$ \mathcal{G}_{n,V}:=\left\{ M\subset X\,\mid\, M\text{ Borel and }\lambda\left(\left[M\cap U_{n}\right]\times V\right)=\mu\left(M\cap U_{n}\right)\nu\left(V\right)\right\} . $$ By what we just showed, $\mathcal{G}_{n,V}$ contains the $\pi$-system of all open sets. Furthermore, it is not hard to see that $\mathcal{G}_{n,V}$ is closed under disjoint countable unions. Finally, for $M\in\mathcal{G}_{n,V}$, we have \begin{align*} \lambda\left(\left[M^{c}\cap U_{n}\right]\times V\right)+\mu\left(M\cap U_{n}\right)\nu\left(V\right) & =\lambda\left(\left[M^{c}\cap U_{n}\right]\times V\right)+\lambda\left(\left[M\cap U_{n}\right]\times V\right)\\ & =\lambda\left(U_{n}\times V\right)\\ & =\mu\left(U_{n}\right)\nu\left(V\right)\\ & =\mu\left(M\cap U_{n}\right)\nu\left(V\right)+\mu\left(M^{c}\cap U_{n}\right)\nu\left(V\right), \end{align*} so that subtraction (everything is finite) shows $\lambda\left(\left[M^{c}\cap U_{n}\right]\times V\right)=\mu\left(M^{c}\cap U_{n}\right)\nu\left(V\right)$ and hence $M^{c}\in\mathcal{G}_{n,V}$. Thus, $\mathcal{G}_{n,V}$ is a $\lambda$-system.

By Dynkin's $\pi$-$\lambda$-Theorem, we conclude $\mathcal{G}_{n,V}=\mathcal{B}\left(X\right)$, the $\sigma$-Algebra of all Borel sets. By continuity of $\lambda,\mu$ we thus get $$ \lambda\left(M\times V\right)=\lim_{n}\lambda\left(\left[M\cap U_{n}\right]\times V\right)=\lim_{n}\mu\left(M\cap U_{n}\right)\nu\left(V\right)=\mu\left(M\right)\nu\left(V\right) $$ for all Borel sets $M\subset X$ and all open sets $V\subset Y$ of finite measure.

Now, define analogously $$ \mathcal{H}_{m,M}:=\left\{ N\subset Y\,\mid\, N\text{ Borel and }\lambda\left(M\times\left[N\cap V_{m}\right]\right)=\mu\left(M\right)\cap\nu\left(N\cap V_{m}\right)\right\} $$ for each Borel set $M\subset X$ of finite measure. Just as above, we see that $\mathcal{H}_{m,M}$ contains all open sets and then all Borel sets. Using the same limit argument as above, we get $$ \mu\left(M\times N\right)=\mu\left(M\right)\nu\left(N\right) $$ for all Borel sets $N\subset Y$ and all Borel sets $M\subset X$ of finite measure. A final limit argument (using $\sigma$-finiteness) removes the assumption that $M$ has to have finite measure. $\square$

  • $\begingroup$ Thanks. By the way I would like to know whether Fubini's theorem holds for a Borel integrable function on $X\times Y$ or not. Do you have any idea? $\endgroup$ – JHW Aug 13 '15 at 3:03
  • $\begingroup$ @JHW: Yes, it does hold if the integrand $f(x,y)$ vanishes outside of a $\sigma$-finite set (I think, maybe it has to be $\sigma$-compact). One book in which this is shown is Folland's "Real Analysis". The argument is not quite trivial and proceeds through several lemmata. One of the steps is to show that $\int f \, d\mu = \sup \{\int g \, d\mu \, \mid \, g \in C_c, 0\leq g \leq f\}$ for a lower semicontinuous function $f \geq 0$ (in particular for $f = \chi_U$. $\endgroup$ – PhoemueX Aug 13 '15 at 12:37
  • $\begingroup$ Fubini's theorem can be used as long as the function in question $f$ is measurable and has a $\sigma$--finite carrier (i.e. $E_f:=\{(x,y)\in X\times Y:f(x,y)\neq0\}$ can be covered by countable integrable sets). Fubini--Tonelli's theorem then says: Suppose $f$ is measurable and has $\sigma$--finite carrier. $f$ is integrable iff any of the iterated integrals (and hence all) $\int_X\int_Y|f(x,y)|\mu(dy)\nu(dx)<\infty$ or $\int_Y\int_X|f(x,y)|\nu(dx)\mu(dy)<\infty$. In either case, $\int_{X\times Y}f d(\mu\otimes\nu) =\int_X\int_Y|f(x,y)|\mu(dy)\nu(dx)=\int_Y\int_X|f(x,y)|\nu(dx)\mu(dy)$. $\endgroup$ – Oliver Diaz Sep 30 '19 at 22:49

To elaborate on @MartinArgerami's argument, you want to prove that if $\mu_1$ and $\mu_2$ are two Radon measures on $X$ which agree on continuous functions, then they agree on all Borel sets. To do this, it suffices (by inner regularity) to prove that $\mu_1(K) = \mu_2(K)$ for all compacts $K \subset X$. For this you need the following lemma:

Lemma: Let $K \subset X$ be a compact set, then $\exists \varphi : X\to \mathbb{R}$ continuous bounded such that $$ \varphi \equiv 1 \text{ on } K \text{ and } \varphi < 1 \text{ on } X\setminus K $$ Proof: For each $n \in \mathbb{N}$, let $$ G_n = \{x \in X : d(x,K) < 1/n\} $$ there is a function $\varphi_n : X\to [0,1]$ such that $$ \varphi_0 \equiv 1 \text{ on } K \text{ and } \varphi_n \equiv 0 \text{ on } X\setminus G_n $$ Now check that $$ \varphi = \sum \frac{1}{2^n} \varphi_n $$ satisfies the required properties.

Now to complete the argument: If $$ \int_X f d\mu_1 = \int_X fd\mu_2 \quad\forall f\in C(X) $$ then for $K \subset X$ compact, choose a function $\varphi$ as above. Now note that $$ \varphi^n \to \chi_K $$ pointwise. So now you can apply Dominated convergence theorem to conclude that $\mu_1(K) = \mu_2(K)$

  • 1
    $\begingroup$ Thanks. Maybe $\lambda$ and $\mu\otimes\nu$ agree on compact subsets of $X\times Y$, but I don't see why they agree on $\mathcal B(X)\otimes \mathcal B(X)$. Regards, $\endgroup$ – JHW Jul 23 '15 at 20:02
  • $\begingroup$ I'm wondering why you have been silent to my last comment. $\endgroup$ – JHW Jul 26 '15 at 5:33
  • $\begingroup$ I assumed that @MartinArgerami answered you fine in the edit he made to this post. $\endgroup$ – Prahlad Vaidyanathan Jul 26 '15 at 12:36
  • $\begingroup$ He didn't answer me fine in the edit. Please read my comments to him about it. $\endgroup$ – JHW Jul 26 '15 at 19:50
  • $\begingroup$ This is not correct: 1) You seem to assume that there is some kind of metric on $X$, which is not true in general. $X$ might not even be first countable. 2) We do not know that the measure $\mu \otimes \nu$ is Radon. In general, it is not even defined on the whole Borel $\sigma$-algebra on $X\times Y$. $\endgroup$ – PhoemueX Aug 10 '15 at 15:53

I don't know how Lang's book is organized. But it is a general fact that in a locally compact Hausdorff space, if two measures agree as functionals on continuous functions, then they are equal. It is basically about using Urysohn to approximate characteristic functions by continuous functions.

Edit: after you show that $\lambda$ and $\mu\otimes\nu$ agree on compact sets, you know in particular that they agree on sets of the form $K_1\times K_2$, with $K_1\subset X$, $K_2\subset Y$, both compact.

From here, if $U\subset X$ is open, you have that $\mu(U)=\sup\{\mu(K):\ K\subset V,\ \text{compact }\}$. So \begin{align} \mu\otimes\nu(V\times K_2)&=\mu(V)\nu(K_2)=\sup\{\mu(K_1)\nu(K_2):\ K_1\subset V\}\\ &=\sup\{\mu\otimes\nu(K_1\otimes K_2):\ K_1\subset V\}\\ &=\sup\{\lambda(K_1\times K_2):\ K_1\subset V\}\\ &=\lambda(V\times K_2) \end{align} (the last equality because $\lambda$ is a Radon measure). By repeating the argument, you obtain that $\mu\otimes\nu$ and $\lambda$ agree on products of open sets. From here you fix an open set $W\in\mathcal B(Y)$ and consider the measures $E\longmapsto \mu\otimes\nu(E\times W)$ and $E\longmapsto \lambda(E\times W)$. These are two measures on $X$ that agree on open sets; as these generate $\mathcal B(X)$, we obtain that $\mu\otimes \nu(E\times W)=\lambda(E\times W)$ for all $E\in\mathcal B(X)$. With a similar argument, we get that $\mu\otimes\nu(E\times F)=\lambda(E\times F)$ for all $E\in\mathcal B(X)$, $F\in\mathcal B(Y)$. Then $\mu\otimes\nu$ and $\lambda$ agree on the sets that generate the $\sigma$-algebra $\mathcal B(X)\otimes \mathcal B(Y)$.

  • $\begingroup$ Was just about to post something to this effect. You use inner regularity/Urysohn to conclude that they agree on all open sets of the form $U\times V$, for $U\subset X$ and $V\subset Y$ open. It follows that they agree on $\mathcal{B}(X)\otimes \mathcal{B}(Y)$. $\endgroup$ – Moya Jul 23 '15 at 3:25
  • $\begingroup$ @Moya Would you please elaborate on why they agree on $\mathcal B(X)\otimes\mathcal B(Y)$? $\endgroup$ – JHW Jul 23 '15 at 19:45
  • $\begingroup$ "if two measures agree as functionals on continuous functions, then they are equal." Maybe they agree on compact subsets of $X\times Y$, but I don't see why they agree on $\mathcal B(X)\otimes \mathcal B(X)$. Regards, $\endgroup$ – JHW Jul 23 '15 at 19:57
  • $\begingroup$ I don't think it is hard. Please see the edit. $\endgroup$ – Martin Argerami Jul 23 '15 at 20:39
  • $\begingroup$ Let $X$ be a locally compact Hausdorff space. You seem to be saying that if two measures on $\mathcal B(X)$ agree on all open sets, they agree on $\mathcal B(X)$. Would you please explain why this is so? $\endgroup$ – JHW Jul 23 '15 at 21:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.