# How can I get $\int_{\Bbb R^n}e^{itu_1x_1}dP^{X}=\int_{\Bbb R}e^{iu_1x_1}dP^{X_1}$?

Let $(\Omega,\mathcal{A},P)$ be a probability space and $X=(X_1,X_2,\cdots,X_n)$ a random vector on $(\Omega,\mathcal{A},P)$ such that $$X_k:(\Omega,\mathcal{A})\to({\Bbb R},{\mathcal B}_{\Bbb R}), k=1,2,\cdots, n$$ are random variables, but not necessarily independent. What I need to show is that $$\phi_X(u)=\phi_{X_1}(u_1)$$ where $u=(u_1,0,0,\cdots,0)\in{\Bbb R}^n$, and $\phi_X$ is the characteristic function of $X$: $$\phi_X(u)=\int_{\Bbb R^n}e^{it\langle u,x\rangle}dP^{X}$$ where $P^X$ is the probability distribution measure of $X$. By simple calculation, $$\phi_X(u)=\int_{\Bbb R^n}e^{iu_1x_1}dP^X.$$

My question is: How can I relate this integral with $$\phi_{X_1}(u_1)=\int_{\Bbb R}e^{iu_1x_1}dP^{X_1} ?$$

Formally, what I tried is that (in the case $n=2$) \begin{align} \phi_{X_1}(u_1)&=\int_{\Bbb R}e^{iu_1x_1}dP^{X_1}\\ &=\int_{\Bbb R}e^{iu_1x_1}dP^{X_1}\int_{\Bbb R}dP^{X_2}\\ &=\int_{\Bbb R}\left(\int_{\Bbb R}e^{iu_1x_1}dP^{X_1}\right)dP^{X_2}\\ &=\int_{\Bbb R^2}e^{iu_1x_1}P^{X_1}\otimes P^{X_2} \end{align} But $X_1,X_2$ are not necessarily independent and I'm not able to get $$P^{(X_1,X_2)}=P^{X_1}\otimes P^{X_2}.$$

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If you treat them as integrals on $\Omega$ instead of integrals on $\mathbb{R}^n$ and $\mathbb{R}$, then $$\phi_X(u)=\int_\Omega e^{i\langle u,X \rangle}\,\mathrm{d}P=\int_\Omega e^{iu_1X_1}\,\mathrm{d}P=\phi_{X_1}(u).$$