[proofplan] The moment equalities first extend by linearity from monomials to all complex polynomial test functions on $[-R,R]$. We then use the [Weierstrass Approximation Theorem](/theorems/480) to approximate an arbitrary [continuous function](/page/Continuous%20Function) uniformly by polynomials. Since $\mu$ and $\nu$ are probability measures, uniform approximation controls the corresponding integrals, so the two measures have identical integrals against every continuous function on the compact interval. The [Riesz Representation Theorem](/theorems/218) for finite Borel measures on compact Hausdorff spaces then identifies the restricted measures on $[-R,R]$, and the support hypothesis extends the equality to all Borel subsets of $\mathbb{R}$. [/proofplan] [step:Extend equality from monomials to polynomial test functions] Let $K := [-R,R] \subset \mathbb{R}$. Let $p: K \to \mathbb{C}$ be a complex polynomial, so there are an integer $m \geq 0$ and coefficients $a_0,\dots,a_m \in \mathbb{C}$ such that \begin{align*} p(t) = \sum_{n=0}^{m} a_n t^n \end{align*} for every $t \in K$. Since $p$ is bounded on the compact set $K$ and $\mu,\nu$ are finite measures, $p$ is integrable with respect to both $\mu$ and $\nu$ on $K$. By linearity of the [Lebesgue integral](/page/Lebesgue%20Integral) for complex-valued integrable functions and by the assumed moment equalities, \begin{align*} \int_K p(t)\,d\mu(t) = \sum_{n=0}^{m} a_n \int_K t^n\,d\mu(t). \end{align*} Using the hypothesis for each $0 \leq n \leq m$ gives \begin{align*} \sum_{n=0}^{m} a_n \int_K t^n\,d\mu(t) = \sum_{n=0}^{m} a_n \int_K t^n\,d\nu(t). \end{align*} Applying linearity again, \begin{align*} \sum_{n=0}^{m} a_n \int_K t^n\,d\nu(t) = \int_K p(t)\,d\nu(t). \end{align*} Therefore \begin{align*} \int_K p(t)\,d\mu(t) = \int_K p(t)\,d\nu(t) \end{align*} for every complex polynomial $p: K \to \mathbb{C}$. [/step] [step:Approximate continuous functions uniformly by polynomials and pass to the integral] Let $f: K \to \mathbb{C}$ be continuous. By the Weierstrass Approximation Theorem applied separately to the real-valued continuous functions $\operatorname{Re} f: K \to \mathbb{R}$ and $\operatorname{Im} f: K \to \mathbb{R}$, for every $\varepsilon > 0$ there exists a complex polynomial $p_\varepsilon: K \to \mathbb{C}$ such that \begin{align*} \sup_{t \in K} |f(t) - p_\varepsilon(t)| < \varepsilon. \end{align*} Since $\mu(K)=\nu(K)=1$, the triangle inequality for integrals gives \begin{align*} \left|\int_K f(t)\,d\mu(t) - \int_K f(t)\,d\nu(t)\right| \leq \int_K |f(t)-p_\varepsilon(t)|\,d\mu(t) + \int_K |f(t)-p_\varepsilon(t)|\,d\nu(t). \end{align*} Using the uniform bound and the fact that both measures are probability measures, \begin{align*} \int_K |f(t)-p_\varepsilon(t)|\,d\mu(t) + \int_K |f(t)-p_\varepsilon(t)|\,d\nu(t) < 2\varepsilon. \end{align*} Because $\varepsilon > 0$ was arbitrary, \begin{align*} \int_K f(t)\,d\mu(t) = \int_K f(t)\,d\nu(t). \end{align*} [guided] Let $f: K \to \mathbb{C}$ be continuous. The moment hypothesis gives equality of integrals only for polynomials, so the goal is to replace $f$ by a polynomial without changing the integral by much. The correct topology here is the uniform norm, because if two functions are uniformly close and the measure has total mass $1$, then their integrals are close by at most the same uniform error. We apply the Weierstrass Approximation Theorem. Its hypothesis is that the target function is continuous on a compact interval. The set $K=[-R,R]$ is a compact interval in $\mathbb{R}$, and the functions $\operatorname{Re} f: K \to \mathbb{R}$ and $\operatorname{Im} f: K \to \mathbb{R}$ are continuous real-valued functions. Hence, for every $\varepsilon > 0$, there are real polynomials $q_\varepsilon: K \to \mathbb{R}$ and $r_\varepsilon: K \to \mathbb{R}$ such that \begin{align*} \sup_{t \in K} |\operatorname{Re} f(t) - q_\varepsilon(t)| < \frac{\varepsilon}{2} \end{align*} and \begin{align*} \sup_{t \in K} |\operatorname{Im} f(t) - r_\varepsilon(t)| < \frac{\varepsilon}{2}. \end{align*} Define the complex polynomial $p_\varepsilon: K \to \mathbb{C}$ by \begin{align*} p_\varepsilon(t) = q_\varepsilon(t) + i r_\varepsilon(t) \end{align*} for $t \in K$. Then \begin{align*} \sup_{t \in K} |f(t)-p_\varepsilon(t)| < \varepsilon. \end{align*} We now compare the two integrals of $f$ by inserting the polynomial $p_\varepsilon$. The polynomial part cancels because polynomial integrals were already proved equal: \begin{align*} \int_K p_\varepsilon(t)\,d\mu(t) = \int_K p_\varepsilon(t)\,d\nu(t). \end{align*} Therefore the triangle inequality gives \begin{align*} \left|\int_K f(t)\,d\mu(t) - \int_K f(t)\,d\nu(t)\right| \leq \int_K |f(t)-p_\varepsilon(t)|\,d\mu(t) + \int_K |f(t)-p_\varepsilon(t)|\,d\nu(t). \end{align*} The uniform estimate bounds the integrands by $\varepsilon$ on all of $K$. Since $\mu$ and $\nu$ are probability measures supported on $K$, their masses on $K$ are both $1$. Hence \begin{align*} \int_K |f(t)-p_\varepsilon(t)|\,d\mu(t) + \int_K |f(t)-p_\varepsilon(t)|\,d\nu(t) < \varepsilon \mu(K) + \varepsilon \nu(K) = 2\varepsilon. \end{align*} Thus \begin{align*} \left|\int_K f(t)\,d\mu(t) - \int_K f(t)\,d\nu(t)\right| < 2\varepsilon. \end{align*} Since this holds for every $\varepsilon > 0$, the nonnegative number on the left must be $0$. Therefore \begin{align*} \int_K f(t)\,d\mu(t) = \int_K f(t)\,d\nu(t). \end{align*} [/guided] [/step] [step:Identify the restricted measures on the compact interval] Let $\mu_K$ and $\nu_K$ denote the restrictions of $\mu$ and $\nu$ to the Borel $\sigma$-algebra $\mathcal{B}(K)$. The preceding step proves that \begin{align*} \int_K f(t)\,d\mu_K(t) = \int_K f(t)\,d\nu_K(t) \end{align*} for every continuous function $f: K \to \mathbb{C}$. Since $K$ is compact Hausdorff and $\mu_K,\nu_K$ are finite Borel measures, the [Riesz Representation Theorem](/theorems/221) for finite Borel measures on compact Hausdorff spaces implies \begin{align*} \mu_K = \nu_K. \end{align*} [/step] [step:Use the support hypothesis to conclude equality on $\mathbb{R}$] Let $A \in \mathcal{B}(\mathbb{R})$ be an arbitrary Borel set. Since $\operatorname{supp}\mu \subset K$ and $\operatorname{supp}\nu \subset K$, both measures assign zero mass to $\mathbb{R}\setminus K$. Hence \begin{align*} \mu(A) = \mu(A \cap K) \end{align*} and \begin{align*} \nu(A) = \nu(A \cap K). \end{align*} The set $A \cap K$ belongs to $\mathcal{B}(K)$, and $\mu_K=\nu_K$ on $\mathcal{B}(K)$. Therefore \begin{align*} \mu(A) = \mu_K(A \cap K) = \nu_K(A \cap K) = \nu(A). \end{align*} Since $A$ was arbitrary, $\mu=\nu$ as Borel measures on $\mathbb{R}$. [/step]