[proofplan] We prove convergence in distribution by establishing tightness and identifying the limit. Tightness is shown via a Fourier-analytic bound on the probability mass outside compact sets, using the pointwise convergence of characteristic functions and DCT. Prokhorov's theorem extracts weakly convergent subsequences, and the characteristic function equality identifies any subsequential limit as having characteristic function $\phi_X$. A subsequence-uniqueness argument then gives full convergence. [/proofplan] [step:Establish tightness of the sequence $(X_n)$ using the Fourier inversion bound] Using the identity \begin{align*} \int_{[-\lambda,\lambda]^d} \phi_X(u) \, d\mathcal{L}^d(u) = (2\lambda)^d \int_{\mathbb{R}^d} \prod_j \frac{\sin(\lambda x_j)}{\lambda x_j} \, \mu(dx) \end{align*} and the bound $|\sin(x)/x| \leq \sin 1$ for $|x| \geq 1$, one derives \begin{align*} \mathbb{P}(\|X_n\|_\infty \geq K) \leq C \left(\frac{K}{2}\right)^d \int_{[-K^{-1}, K^{-1}]^d} (1 - \phi_{X_n}(u)) \, d\mathcal{L}^d(u). \end{align*} Since $|1 - \phi_{X_n}(u)| \leq 2$, DCT gives convergence of the integrals to $\int_{[-K^{-1}, K^{-1}]^d} (1 - \phi_X(u)) \, d\mathcal{L}^d(u)$. The continuity of $\phi_X$ at $0$ with $\phi_X(0) = 1$ ensures this integral is small for large $K$, establishing tightness. [/step] [step:Extract subsequential limits via Prokhorov and identify them using characteristic functions] Prokhorov's theorem extracts a subsequence $\mathcal{L}(X_{n_k}) \Rightarrow \nu$ for some probability measure $\nu$. Since $x \mapsto e^{i\langle u, x\rangle}$ is bounded and continuous, weak convergence gives $\mathbb{E}[e^{i\langle u, X_{n_k}\rangle}] \to \int e^{i\langle u, x\rangle} \, \nu(dx)$. But by hypothesis, $\mathbb{E}[e^{i\langle u, X_{n_k}\rangle}] = \phi_{X_{n_k}}(u) \to \phi_X(u)$. Therefore $\nu$ has characteristic function $\phi_X$, so $\nu = \mu_X$. [/step] [step:Conclude full convergence by a subsequence-uniqueness argument] If $\mathcal{L}(X_n)$ did not converge weakly to $\mathcal{L}(X)$, there would exist a subsequence $(m_k)$ and a bounded continuous $f$ with $|\mathbb{E}[f(X_{m_k})] - \mathbb{E}[f(X)]| > \varepsilon$ for all $k$. Tightness of $(m_k)$ gives a further subsequence converging weakly, and its limit must also have characteristic function $\phi_X = \phi_X$, contradicting the $\varepsilon$-separation. Therefore $X_n \xrightarrow{d} X$. [/step]