Why do distributional isomorphisms preserve joint distribution?

Question

Let $(\Omega,\mathcal{A},\mu)$ and $(\Omega',\mathcal{A}',\mu')$ be probability spaces and $$f_1,\ldots,f_n:\Omega\to\mathbb R,\; f_1',\cdots, f_n':\Omega'\to\mathbb{R}$$ be integrable random variables such that for every $\vec{\lambda} = (\lambda_1,,\cdots,\lambda_n)\in\mathbb{R}^n$, the random variables $$f_{\vec\lambda} = \sum_{i=1}^N\lambda_if_i\text{ and } f'_{\vec\lambda} = \sum_{i=1}^n\lambda_i f_i'$$ have the same distribution, i.e., for every measurable $A\subset\mathbb{R}$, $\mu([f_{\vec\lambda}\in A]) = \mu'([ f'_{\vec\lambda}\in A])$. In other words, the mapping $f_i\mapsto f'_i$ extends to a linear distributional isomorphism between the linear span of $f_1,\ldots,f_n$ in $L_1(\Omega,\mathcal{A},\mu)$ and the linear span of $f_1',\ldots, f_n'$ in $L_1(\Omega',\mathcal{A}',\mu')$.

Is it true that the joint distribution of $f_1,\ldots,f_n$ coincides with the joint distribution of $f_1',\ldots,f_n'$, i.e., for every measurable subsets $A_1,A_2,\ldots,A_n$ of $\mathbb{R}$, $$\mu\big(\cap_{i=1}^n[f_i\in A_i]\big) = \mu'\big(\cap_{i=1}^n[f_i'\in A_i]\big)?$$

If yes, is there a citable reference for this fact?

I can sketch a relatively elementary proof for finitely valued random variables that also works for countably valued ones, but the general proof eludes me.

Here is the sketch of the proof when the random variables are finitely valued:

For $\omega\in \Omega$, define the linear operator $T_\omega:\mathbb{R}^n\to\mathbb{R}$ given by $T_\omega(\vec\lambda) = f_\lambda(\omega)$. Because the set $\mathcal{T} = \{T_\omega:\omega\in\Omega\}$ is finite, the subset $K = \cup_{T\neq S\in\mathcal{T}}\mathrm{ker}(T-S)$ of $\mathbb{R}^n$ has Lebesgue measure zero, as a finite union of hyperplanes. Choose $\vec\lambda_0\in\mathbb{R}^n\setminus K$. Then, for $\omega_1,\omega_2\in\Omega$, $T_{\omega_1}(\vec\lambda_0) = T_{\omega_2}(\vec\lambda_0)$ if and only if $T_{\omega_1} = T_{\omega_2}$, i.e., $(f_1(\omega_1))_{i=1}^n = (f_1(\omega_2))_{i=1}^n$. It follows that for $\vec a = (a_1,\ldots,a_n)\in\mathbb{R}^n$, $$\cap_{i=1}^n[f_i=a_i] = [f_{\vec\lambda_0}=\langle\vec\lambda_0,\vec a \rangle].$$ Defining, for $\omega'\in\Omega'$, $T_{\omega'}\to\mathbb{R}^n\to\mathbb{R}$ in the same way, we obtain $\mathcal{T} = \{T_{\omega'}:\omega'\in\Omega'\}$, and thus, for $\vec a = (a_1,\ldots,a_n)\in\mathbb{R}^n$, $$\cap_{i=1}^n[f'_i=a_i] = [f'_{\vec\lambda_0}=\langle\vec\lambda_0,\vec a \rangle].$$ The conclusion easily follows.

Answer 1

I cannot resist to provide an easily citable reference to Rudin's beautiful paper $L_p$-Isometries and equimeasurability, where Rudin proves the much stronger result that if $p$ is not an even integer and $\|1 + \sum_i \lambda_i f_i\|_p = \|1 + \sum_i \lambda_i f'_i\|_p$ for every $p$, then the joint distributions agree. Take $p=1$ here.

But this is silly, because there is a much easier proof: your assumptions imply that $\mathbf{E}(e^{i \sum \lambda_k f_k}) = \mathbf{E}(e^{i \sum \lambda_k f'_k})$, that is the random variables $(f_1,\dots,f_n)$ and $(f'_1,\dots,f'_n)$ have the same Fourier transform (characteristic function). Any basic probability textbook should have a citable reference for the fact that the Fourier transform of an $\mathbf{R}^d$-valued random variable characterizes its distribution.