[proofplan] We use the [Weierstrass Approximation Theorem](/theorems/480) to show that $\int_a^b g \, d\mu = 0$ for every $g \in C([a,b])$. Since the hypothesis gives $\int_a^b p \, d\mu = 0$ for every polynomial $p$ (by linearity), and every continuous function on $[a,b]$ is a uniform limit of polynomials, the bounded convergence theorem — applied to the finite signed measure $\mu$ — transfers the vanishing from polynomials to all continuous functions. The Riesz representation theorem for $C([a,b])^*$ then forces $\mu = 0$. [/proofplan]

[step:Extend the vanishing hypothesis from monomials to all polynomials] By assumption, $\int_a^b x^k \, d\mu(x) = 0$ for every $k \geq 0$. Since integration against $\mu$ is linear, for any polynomial $p(x) = c_0 + c_1 x + \cdots + c_m x^m$ with $c_0, \ldots, c_m \in \mathbb{R}$, \begin{align*} \int_a^b p(x) \, d\mu(x) = \sum_{j=0}^{m} c_j \int_a^b x^j \, d\mu(x) = \sum_{j=0}^{m} c_j \cdot 0 = 0. \end{align*} [/step]

[step:Approximate any continuous function uniformly by polynomials]Let $g \in C([a,b])$. By the [Weierstrass Approximation Theorem](/theorems/480), there exists a sequence of polynomials $(p_m)_{m=1}^\infty$ such that \begin{align*} \|p_m - g\|_\infty := \max_{x \in [a,b]} |p_m(x) - g(x)| \to 0 \quad \text{as } m \to \infty. \end{align*} In particular, $p_m(x) \to g(x)$ for every $x \in [a,b]$, and $|p_m(x) - g(x)| \leq \|p_m - g\|_\infty$ for all $x$ and $m$.[/step]

[guided]The [Weierstrass Approximation Theorem](/theorems/480) states that for any continuous function on a compact interval $[a,b]$ and any $\varepsilon > 0$, there exists a polynomial within $\varepsilon$ of the function in the supremum norm. Applying this with $\varepsilon = 1/m$ for $m = 1, 2, 3, \ldots$ produces a sequence $(p_m)$ with $\|p_m - g\|_\infty \leq 1/m \to 0$. This is the step where the Weierstrass theorem enters: without it, we would have no way to pass from polynomials (where the integral vanishes) to arbitrary continuous functions.[/guided]

[step:Pass the vanishing to all continuous functions via dominated convergence]Since $\mu$ is a signed Borel measure on the compact interval $[a,b]$, its total variation $|\mu|$ is a finite positive measure with $|\mu|([a,b]) < \infty$. For each $m$, the previous step gives $\int_a^b p_m \, d\mu = 0$. The sequence $p_m \to g$ pointwise, and for all sufficiently large $m$ (say $m \geq M$), $\|p_m - g\|_\infty \leq 1$, so $|p_m(x)| \leq \|g\|_\infty + 1$ for all $x \in [a,b]$. The constant function $\|g\|_\infty + 1$ is integrable with respect to $|\mu|$ (since $|\mu|([a,b]) < \infty$). By the dominated convergence theorem applied with respect to the finite measure $|\mu|$, \begin{align*} \int_a^b g \, d\mu = \lim_{m \to \infty} \int_a^b p_m \, d\mu = \lim_{m \to \infty} 0 = 0. \end{align*}[/step]

[guided]We need to justify passing the limit inside the integral. Since $\mu$ is a signed measure, integrals against $\mu$ can be expressed via the Jordan decomposition $\mu = \mu^+ - \mu^-$, so $\int g \, d\mu = \int g \, d\mu^+ - \int g \, d\mu^-$. The dominated convergence theorem applies to each integral separately: both $\mu^+$ and $\mu^-$ are finite positive measures on $[a,b]$ (since $|\mu| = \mu^+ + \mu^-$ is finite), and the dominating function $\|g\|_\infty + 1$ is integrable with respect to either. The dominated convergence theorem then gives $\int p_m \, d\mu^{\pm} \to \int g \, d\mu^{\pm}$, and by linearity $\int p_m \, d\mu \to \int g \, d\mu$. Alternatively, one can use the estimate $\left|\int_a^b (p_m - g) \, d\mu\right| \leq \|p_m - g\|_\infty \cdot |\mu|([a,b]) \to 0$ directly, which avoids the dominated convergence theorem entirely. Either approach yields $\int_a^b g \, d\mu = 0$ for every $g \in C([a,b])$.[/guided]

[step:Conclude $\mu = 0$ from the vanishing of all continuous integrals]We have shown that $\int_a^b g \, d\mu = 0$ for every $g \in C([a,b])$. A signed Borel measure on a compact metric space that annihilates every continuous function must be the zero measure. To verify this, let $F \subset [a,b]$ be any closed set. For each $m \geq 1$, define \begin{align*} g_m: [a,b] &\to [0,1] \\ x &\mapsto \max\{1 - m \cdot \operatorname{dist}(x, F), \, 0\}, \end{align*} where $\operatorname{dist}(x, F) = \inf_{y \in F} |x - y|$. Each $g_m$ is continuous (as the composition of the Lipschitz function $\operatorname{dist}(\cdot, F)$ with piecewise-linear functions), $g_m = 1$ on $F$, and $g_m(x) \to \mathbb{1}_F(x)$ for every $x \in [a,b]$ as $m \to \infty$ (since $\operatorname{dist}(x, F) > 0$ when $x \notin F$, and $\operatorname{dist}(x, F) = 0$ when $x \in F$). Since $|g_m| \leq 1$ and $|\mu|([a,b]) < \infty$, the dominated convergence theorem gives \begin{align*} \mu(F) = \int_a^b \mathbb{1}_F \, d\mu = \lim_{m \to \infty} \int_a^b g_m \, d\mu = 0. \end{align*} Since $\mu$ vanishes on every closed subset of $[a,b]$ and the Borel $\sigma$-algebra on $[a,b]$ is generated by the closed sets, the uniqueness of measures (applied to $\mu^+$ and $\mu^-$ separately on this compact space) gives $\mu^+ = \mu^- = 0$, and therefore $\mu = 0$.[/step]

[guided]This final step uses a regularity argument. Having shown $\int g \, d\mu = 0$ for every $g \in C([a,b])$, we need to conclude that $\mu$ itself is zero — not merely that it annihilates continuous functions. The key observation is that the indicator function of any closed set $F$ can be approximated from above by continuous functions. The functions $g_m(x) = \max\{1 - m \cdot \operatorname{dist}(x, F), 0\}$ satisfy: - $g_m$ is continuous (since $\operatorname{dist}(\cdot, F)$ is $1$-Lipschitz and the map $t \mapsto \max\{1 - mt, 0\}$ is continuous); - $0 \leq g_m \leq 1$; - $g_m \equiv 1$ on $F$; - $g_m(x) \to 0$ for $x \notin F$ (since $\operatorname{dist}(x,F) > 0$ implies $1 - m \cdot \operatorname{dist}(x,F) < 0$ for $m > 1/\operatorname{dist}(x,F)$). So $g_m \to \mathbb{1}_F$ pointwise. The dominated convergence theorem (with dominating function $1$, integrable since $|\mu|([a,b]) < \infty$) gives $\mu(F) = \lim_m \int g_m \, d\mu = 0$. Since every Borel set in a compact metric space can be generated from closed sets, and $\mu$ vanishes on all closed sets, the measures $\mu^+$ and $\mu^-$ (from the Jordan decomposition) agree on the $\pi$-system of closed sets. Since $\mu^+([a,b]) = \mu^-([a,b])$ (both equal zero, as $\mu([a,b]) = 0$ and $\mu^+, \mu^-$ are concentrated on disjoint sets summing to $|\mu|$), the $\pi$-$\lambda$ theorem implies $\mu^+ = \mu^-$ on all Borel sets, hence $\mu = 0$.[/guided]