Two measures over different spaces with the same integral

Question

Suppose that $\mathcal{A}$, $\mathcal{B}$ are two compact Hausdorff spaces, $\mu_1$ is a measure in the Borel $\sigma$-algebra $\sigma(\mathcal{A})$ and $\mu_2$ is a measure in the Borel $\sigma$-algebra $\sigma(\mathcal{B})$. Let $p:\mathcal{A}\rightarrow\mathcal{B}$ continuous and surjective function. I need to prove that if

$$\int_{\mathcal{B}}f d\mu_2=\int_{\mathcal{A}}(f\circ p) d\mu_1$$

for every continuous function $f:\mathcal{B}\rightarrow\mathbb{C}$, then $\mu_2=\mu_1\circ p^{-1}$, where $p^{-1}$ is the inverse image, not the inverse function (for example $p^{-1}(U)$ is the inverse image of $U\subseteq\mathcal{B}$). I have search for similar problems, but none has the same conditions than this one. Moreover, I have tried to prove this using the monotone convergence theorem, approximating the functions at both sides by simple functions. My problem is that the integrals are over different spaces.

How can I prove this? Any suggestions would be welcome.

Answer 1

Since your integrals are supposed to make sense for all $\mathbb C$-valued continuous functions, I will assume you are dealing with finite measures $\mu_1$ and $\mu_2$.

Now, your statement is false without some regularity assumptions on the measures. To see this, let $\mathcal A=\mathcal B=\omega_1+1=[0,\omega_1]$, where $\omega_1$ is the first uncountable ordinal, and let $p$ be the identity. Give $[0,\omega_1]$ the order topology, which is compact and Hausdorff. Let $\mu_1$ be the Dieudonné measure described here (extended to $[0,\omega_1]$ by setting $\mu_1(\{\omega_1\})=0$):

Atoms of Dieudonné measure on $[0, \omega_1)$

That is, $\mu_1(A)=1$ if and only if $A\cap[0, \omega_1)$ contains a closed and unbounded (relative to $[0,\omega_1)$) set, and otherwise $\mu_1(A)=0$. Let $\mu_2$ be a dirac measure at $\omega_1$. It is easy to check that every continuous function $f\colon [0,\omega_1]\to \mathbb C$ must be constantly equal to $f(\omega_1)$ for all but countably many points in $[0,\omega_1]$, and so $\int f\,d\mu_1=f(\omega_1)=\int f\,d\mu_2$, but the measures are not the same.

Assuming the measures $\mu_1$ and $\mu_2$ are Borel regular (that is, inner and outer regular), then the statement is true. This really boils down to two key facts:

1. The pushforward measure $p_*\mu_1$ given by $p_*\mu_1(B)=\mu_1(p^{-1}(B))$ is also Borel regular.

2. A Borel regular measure on a compact Hausdorff space is completely determined by integration on continuous functions.

#1 is dealt with in this answer:

https://math.stackexchange.com/a/2475119/1210477

#2 is a pretty standard fact from measure theory, but I will give the argument here to make clear it applies in this generality.

Let $\mu$ and $\nu$ be finite Borel regular measures on a compact Hausdorff space $X$, where $\int f d\mu=\int f d\nu$ for every continuous $f\colon X\to \mathbb C$.

First we will establish $\mu(U)=\nu(U)$ whenever $U\subseteq \mathcal B$ open. To this end, fix such $U$.

By inner regularity, there are increasing sequences of compact sets $A_i\subseteq U$, $B_i\subseteq U$ with $\mu(A_i)\to \mu(U)$, $\nu(B_i)\to \nu(U)$. Then $C_i=A_i\cup B_i$ are also an increasing family of compact sets with $\mu(C_i)\to \mu(U)$ and $\nu(C_i)\to \nu(U)$.

Now by Urysohn's lemma there are continuous functions $f_i\colon X\to [0,1]$ that are identically equal to $1$ on $C_i$ and $0$ outside of $U$.

Then $\chi_U\in L^1(\mu)$, and since each $f_i$ is bounded by $\chi_U$, and $f_i\to\chi_U$ ($\mu$-a.e.), the dominated convergence theorem tells us that $$\int f_i \,d\mu\to \int \chi_U\,d\mu = \mu(U)\text{,}$$ and similarly for $\nu$, whereby $\mu(U)=\nu(U)$.

Finally, by outer regularity we immediately have $\mu(S)=\nu(S)$ for arbitrary Borel sets $S\subseteq X$.