[proofplan] We show $\mathbb{E}[h(X)] = \mathbb{E}[h(Y)]$ for every bounded continuous $h$ via Gaussian smoothing. Replace $h$ by $h * g_t$, express the smoothed expectations via Fourier transforms (where $\phi_X = \phi_Y$ ensures equality), and let $t \to 0$ using the Dominated Convergence Theorem. [/proofplan] [step:Show smoothed expectations agree when $\phi_X = \phi_Y$] [claim:Smoothed Expectations Agree] If $\phi_X = \phi_Y$, then $\mathbb{E}[h(X + \sqrt{t}\,Z)] = \mathbb{E}[h(Y + \sqrt{t}\,Z)]$ for all $t > 0$ and bounded continuous $h$, where $Z \sim N(0, I_d)$ is independent of $X$ and $Y$. [/claim] [proof] The smoothed function $(h * g_t)(x) = \mathbb{E}[h(x + \sqrt{t}\,Z)]$ has Fourier transform $\hat{h}(u)\,e^{-t|u|^2/2}$ by the [Convolution Theorem](/theorems/527). Since $\widehat{h * g_t}$ is integrable, the [Fourier Inversion Formula](/theorems/528) gives \begin{align*} (h * g_t)(x) = \frac{1}{(2\pi)^d}\int_{\mathbb{R}^d} \hat{h}(u)\,e^{-t|u|^2/2}\,e^{-i\langle u, x\rangle}\,d\mathcal{L}^d(u). \end{align*} Integrating against $\mu_X$ and exchanging by [Fubini](/theorems/513) (the integrand is bounded by $\|h\|_\infty \cdot e^{-t|u|^2/2} \in L^1$): \begin{align*} \mathbb{E}[(h * g_t)(X)] = \frac{1}{(2\pi)^d}\int_{\mathbb{R}^d} \hat{h}(u)\,e^{-t|u|^2/2}\,\overline{\phi_X(u)}\,d\mathcal{L}^d(u). \end{align*} Since $\phi_X = \phi_Y$, the same computation with $Y$ gives the same value. [/proof] [/step] [step:Show $(h * g_t)(x) \to h(x)$ pointwise as $t \to 0$] [claim:Pointwise Convergence Of Smoothed Functions] For bounded continuous $h$ and every $x \in \mathbb{R}^d$, $(h * g_t)(x) \to h(x)$ as $t \to 0$. [/claim] [proof] For fixed $x$, $|(h * g_t)(x) - h(x)| \leq \varepsilon + 2\|h\|_\infty \int_{|y| \geq \delta} g_t(y)\,d\mathcal{L}^d(y)$ for $\delta$ chosen so that $|h(x-y) - h(x)| < \varepsilon$ when $|y| < \delta$. The Gaussian tail integral tends to $0$ as $t \to 0$. [/proof] [/step] [step:Conclude $\mu_X = \mu_Y$ by applying DCT] By the first step, $\mathbb{E}[(h * g_t)(X)] = \mathbb{E}[(h * g_t)(Y)]$ for all $t > 0$. As $t \to 0$, $(h * g_t)(X) \to h(X)$ a.s. and $|(h * g_t)(X)| \leq \|h\|_\infty$. By the Dominated Convergence Theorem, $\mathbb{E}[(h * g_t)(X)] \to \mathbb{E}[h(X)]$, and similarly for $Y$. Therefore $\mathbb{E}[h(X)] = \mathbb{E}[h(Y)]$ for every bounded continuous $h$, which is the definition of $\mu_X = \mu_Y$. [/step]