[proofplan] We prove both distributional identities by characteristic functions. First we compute the characteristic function of the sample mean using the independence of $X_1,\dots,X_n$ and the characteristic function of a $\mathcal{N}_p(\mu,\Sigma)$ random vector. The resulting expression is exactly the characteristic function of $\mathcal{N}_p(\mu,\Sigma/n)$. Finally, applying the same characteristic-function computation to the affine transformation $\sqrt{n}(\bar{X}-\mu)$ gives the centred normal law with covariance matrix $\Sigma$. [/proofplan] [step:Declare the characteristic functions used in the proof] For every random vector $Y : (\Omega,\mathcal{F}) \to (\mathbb{R}^p,\mathcal{B}(\mathbb{R}^p))$, define its characteristic function \begin{align*} \varphi_Y : \mathbb{R}^p &\to \mathbb{C} \\ t &\mapsto \mathbb{E}\left[\exp\left(i\,t^\top Y\right)\right]. \end{align*} Since $X_i \sim \mathcal{N}_p(\mu,\Sigma)$ for each $i \in \{1,\dots,n\}$, the characteristic-function characterization of the multivariate normal distribution gives, for every $t \in \mathbb{R}^p$, \begin{align*} \varphi_{X_i}(t) &= \exp\left(i\,t^\top \mu-\frac{1}{2}t^\top \Sigma t\right). \end{align*} [/step] [step:Compute the characteristic function of the sample mean] Fix $t \in \mathbb{R}^p$. By the definition of $\bar{X}$ and linearity of the Euclidean inner product, \begin{align*} t^\top \bar{X} &= t^\top \left(\frac{1}{n}\sum_{i=1}^{n}X_i\right) = \sum_{i=1}^{n}\left(\frac{t}{n}\right)^\top X_i. \end{align*} Therefore \begin{align*} \varphi_{\bar{X}}(t) &= \mathbb{E}\left[\exp\left(i\,t^\top \bar{X}\right)\right] \\ &= \mathbb{E}\left[\exp\left(i\sum_{i=1}^{n}\left(\frac{t}{n}\right)^\top X_i\right)\right] \\ &= \mathbb{E}\left[\prod_{i=1}^{n}\exp\left(i\left(\frac{t}{n}\right)^\top X_i\right)\right]. \end{align*} The random variables \begin{align*} \omega \mapsto \exp\left(i\left(\frac{t}{n}\right)^\top X_i(\omega)\right) \end{align*} are independent because $X_1,\dots,X_n$ are independent and each displayed random variable is a Borel-measurable function of the corresponding $X_i$. Hence expectation factors over the product: \begin{align*} \varphi_{\bar{X}}(t) &= \prod_{i=1}^{n} \mathbb{E}\left[\exp\left(i\left(\frac{t}{n}\right)^\top X_i\right)\right] \\ &= \prod_{i=1}^{n} \varphi_{X_i}\left(\frac{t}{n}\right). \end{align*} Using the normal characteristic function from the previous step, \begin{align*} \varphi_{\bar{X}}(t) &= \prod_{i=1}^{n} \exp\left(i\left(\frac{t}{n}\right)^\top \mu -\frac{1}{2}\left(\frac{t}{n}\right)^\top \Sigma \left(\frac{t}{n}\right)\right) \\ &= \exp\left(n\left[ \frac{i}{n}t^\top \mu -\frac{1}{2n^2}t^\top \Sigma t \right]\right) \\ &= \exp\left(i\,t^\top \mu-\frac{1}{2}t^\top \left(\frac{1}{n}\Sigma\right)t\right). \end{align*} This is the characteristic function of $\mathcal{N}_p(\mu,\Sigma/n)$. Therefore \begin{align*} \bar{X} \sim \mathcal{N}_p\left(\mu,\frac{1}{n}\Sigma\right). \end{align*} [guided] Fix $t \in \mathbb{R}^p$. The characteristic function of $\bar{X}$ is determined by the scalar random variable $t^\top \bar{X}$. Since $\bar{X}$ is the average of the $X_i$, bilinearity of matrix multiplication gives \begin{align*} t^\top \bar{X} &= t^\top \left(\frac{1}{n}\sum_{i=1}^{n}X_i\right) = \sum_{i=1}^{n}\left(\frac{t}{n}\right)^\top X_i. \end{align*} Substituting this expression into the characteristic function gives \begin{align*} \varphi_{\bar{X}}(t) &= \mathbb{E}\left[\exp\left(i\,t^\top \bar{X}\right)\right] \\ &= \mathbb{E}\left[\exp\left(i\sum_{i=1}^{n}\left(\frac{t}{n}\right)^\top X_i\right)\right] \\ &= \mathbb{E}\left[\prod_{i=1}^{n}\exp\left(i\left(\frac{t}{n}\right)^\top X_i\right)\right]. \end{align*} The point of rewriting the exponential as a product is that independence can now be used. For each $i$, the map \begin{align*} \mathbb{R}^p &\to \mathbb{C} \\ x &\mapsto \exp\left(i\left(\frac{t}{n}\right)^\top x\right) \end{align*} is Borel-measurable, so the corresponding complex-valued random variables are independent because $X_1,\dots,X_n$ are independent. Their absolute values are all equal to $1$, so they are integrable. Thus the expectation of the product factors: \begin{align*} \varphi_{\bar{X}}(t) &= \prod_{i=1}^{n} \mathbb{E}\left[\exp\left(i\left(\frac{t}{n}\right)^\top X_i\right)\right] \\ &= \prod_{i=1}^{n} \varphi_{X_i}\left(\frac{t}{n}\right). \end{align*} Now use that each $X_i$ has distribution $\mathcal{N}_p(\mu,\Sigma)$. For every $i \in \{1,\dots,n\}$, \begin{align*} \varphi_{X_i}\left(\frac{t}{n}\right) &= \exp\left(i\left(\frac{t}{n}\right)^\top \mu -\frac{1}{2}\left(\frac{t}{n}\right)^\top \Sigma \left(\frac{t}{n}\right)\right). \end{align*} Multiplying the $n$ identical factors gives \begin{align*} \varphi_{\bar{X}}(t) &= \prod_{i=1}^{n} \exp\left(i\left(\frac{t}{n}\right)^\top \mu -\frac{1}{2}\left(\frac{t}{n}\right)^\top \Sigma \left(\frac{t}{n}\right)\right) \\ &= \exp\left(n\left[ \frac{i}{n}t^\top \mu -\frac{1}{2n^2}t^\top \Sigma t \right]\right) \\ &= \exp\left(i\,t^\top \mu-\frac{1}{2}t^\top \left(\frac{1}{n}\Sigma\right)t\right). \end{align*} This final expression is exactly the characteristic function of a $\mathcal{N}_p(\mu,\Sigma/n)$ random vector. Therefore \begin{align*} \bar{X} \sim \mathcal{N}_p\left(\mu,\frac{1}{n}\Sigma\right). \end{align*} [/guided] [/step] [step:Center and rescale the sample mean] Define the centred and rescaled random vector \begin{align*} Z : \Omega &\to \mathbb{R}^p \\ \omega &\mapsto \sqrt{n}\bigl(\bar{X}(\omega)-\mu\bigr). \end{align*} For every $t \in \mathbb{R}^p$, \begin{align*} \varphi_Z(t) &= \mathbb{E}\left[\exp\left(i\,t^\top \sqrt{n}(\bar{X}-\mu)\right)\right] \\ &= \exp\left(-i\sqrt{n}\,t^\top\mu\right) \mathbb{E}\left[\exp\left(i\,(\sqrt{n}t)^\top \bar{X}\right)\right] \\ &= \exp\left(-i\sqrt{n}\,t^\top\mu\right) \varphi_{\bar{X}}(\sqrt{n}t). \end{align*} Using the formula for $\varphi_{\bar{X}}$ proved above, \begin{align*} \varphi_Z(t) &= \exp\left(-i\sqrt{n}\,t^\top\mu\right) \exp\left(i(\sqrt{n}t)^\top\mu -\frac{1}{2}(\sqrt{n}t)^\top\left(\frac{1}{n}\Sigma\right)(\sqrt{n}t)\right) \\ &= \exp\left(-\frac{1}{2}t^\top\Sigma t\right). \end{align*} This is the characteristic function of $\mathcal{N}_p(0,\Sigma)$. Hence \begin{align*} \sqrt{n}(\bar{X}-\mu) \sim \mathcal{N}_p(0,\Sigma). \end{align*} Together with the distribution of $\bar{X}$, this proves both asserted identities. [/step]