Are convex functions enough to determine a measure?

Suppose we are talking about $\mathbb{R}^n$. We know that if $\mu$, $\nu$ are two finite Borel measures such that $$\int_{\mathbb{R}^n}f(x) \, d\mu(x)=\int_{\mathbb{R}^n}f(x) \, d\nu(x),$$ for all continuous functions $f$, then $\mu$ and $\nu$ are really the same measure.

Now, suppose the equation only holds for all convex functions. Is it still true that $\mu$ and $\nu$ are the same measure?

Edit: As Emanuele Paolini has pointed out, there is a counterexample to the original question. So, what if we further restrict $\mu$ and $\nu$ to have compact support?

Solution 1:

Take on $\mathbb R$ the measure $d\mu = \frac{dx}{1+x^2}$. Then for every non constant convex function $f(x)$ if the integral is well defined one has $$ \int f(x) d\mu(x) = +\infty $$ since $f(x)>mx$ for either $x\to +\infty$ or $x\to-\infty$. For constant functions the integral only depends on the total mass of the measure $\mu$.

Hence you cannot distinguish $\mu$ from a translation of itself.

On the other hand if, additionally, you suppose the measure has compact support I think that convex function are enough to distinguish it.

Solution 2:

The one dimensional case

The result is true if we assume that some integrals are finite. Below I will assume that $\int_0^\infty x \,\mu(dx)=\int_0^\infty x \,\nu(dx)<\infty.$ Emanuele Paolini's argument shows that some such assumption is necessary.

Fix $z\in\mathbb{R}$ and $m>0$. The functions $f(x)=(m(x-z)+1)_+$ and $g(x)=m(x-z)_+$ are both convex. Set $h(x)=f(x)-g(x)$ which is zero to the left of $z-1/m$, one to the right of $z$, and linear between.

By hypothesis, we have \begin{eqnarray*} \int h(x)\,\mu(dx)&=&\int f(x)\,\mu(dx) -\int g(x)\,\mu(dx)\\&=&\int f(x)\,\nu(dx)-\int g(x)\,\nu(dx)\\&=&\int h(x)\,\nu(dx),\end{eqnarray*} and letting $m\to\infty$ gives $\mu([z,\infty))=\nu([z,\infty))$.

Since this is true for any $z$, we find that $\mu(B)=\nu(B)$ for all Borel sets.

The multidimensional case

Suppose that we have two finite measures on $\mathbb{R}^n$ such that $\int \phi(x)\,\mu(dx)=\int \phi(x)\,\nu(dx)$ for every non-negative convex function $\phi$, with $\int \|x\|\,\mu(dx)=\int \|x\| \,\nu(dx)<\infty.$

Then for any $t\in\mathbb{R}^n$ and my non-negative convex functions $f,g$ on $\mathbb{R}$ we have $$\int f(w)\,\mu_t(dw)=\int f(\langle x,t\rangle)\,\mu(dx)=\int f(\langle x,t\rangle)\,\nu(dx) =\int f(w)\,\nu_t(dw)<\infty,$$ and similarly for $g$. Here $\mu_t$ and $\nu_t$ are the image measures of $\mu$ and $\nu$ respectively under the map $x\mapsto \langle x,t\rangle.$ The argument for $\mathbb{R}$ gives $\mu_t=\nu_t$.

This implies that $\int e^{i \langle x,t\rangle}\,\mu(dx)=\int e^{iw}\,\mu_t(dw)= \int e^{iw}\,\nu_t(dw)= \int e^{i \langle x,t\rangle}\,\nu(dx)$ and so, by Fourier uniqueness $\mu=\nu$.