Function $g:\mathbb{R}\to \mathbb{R}^n$ such that $g(\sum_{i=1}^nx_i)=(x_1,\dotsc,x_n)$ a.e

Question

Is there a measurable function $g:\mathbb{R}\to \mathbb{R}^n$ such that $g(\sum_{i=1}^nx_i)=(x_1,\dotsc,x_n)$ a.e.?

Due to the papers [1], [2], and [3] I'm obtaining a result that I think it's false. I posted a question in stats.exchange in order to find where the mistake is, but I didn't receive any answer. Below I'll describe the result I found.

For all $\theta\in\mathbb{R}$ let $P_\theta:\mathfrak{B}_\mathbb{R}\to \mathbb{R}$ be the probability measure induced by the density $f_\theta (x):=(2\pi)^{-1/2}e^{-\frac{(x-\theta)^2}{2}}$ (normal distribution).

Let $n\in\mathbb{N}^\times$. We denote by $P_\theta^{\otimes n}$ the product measure $\underbrace{P_\theta\otimes \cdots \otimes P_\theta}_{n\text{ times}}$.

The result I obtained says that there's a measurable function $g:\mathbb{R}\to \mathbb{R}^n$ such that for all $\theta\in\mathbb{R}$ there's $N_\theta\in\mathfrak{B}_{\mathbb{R}^n}$ with $P_{\theta}^{\otimes n}(N_\theta)=0$ and $g(\sum_{i=1}^n x_i)=(x_1,\dotsc,x_n)$ for all $(x_1,\dotsc,x_n)\in\mathbb{R}^n\setminus N_\theta$.

This result sounds absurd but I don't know how to prove that this can’t happen.

My question: is this result indeed false?

I tried to find different elements $(x_1,\dotsc,x_n),(y_1,\dotsc,y_n)\in \mathbb{R}^n\setminus N_\theta $ with $\sum _{i=1}^n x_i=\sum_{i=1}^ny_i$, but I failed.

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Below is how I obtained the result.

Consider the statistical model $(\mathbb{R},\mathfrak{B}_\mathbb{R},\{P_\theta\}_{\theta\in\mathbb{R}})$.

It's easy to see that $T:\mathbb{R}\to \mathbb{R}$ given by $T(x):=x$ is a minimal sufficient statistic, therefore, using Theorem 1 of [3] (or Theorems 4 and 1 of [1] and [2], respectively), we can conclude that $S:\mathbb{R}^n\to\mathbb{R}^n$ given by $S(x_1,\dotsc,x_n):=(x_1,\dotsc,x_n)$ is a minimal sufficient statistics w.r.t. the model $(\mathbb{R}^n,\mathfrak{B}_{\mathbb{R}^n},\{P_\theta^{\otimes n}\}_{\theta\in\mathbb{R}})$.

It's easy to see that $R:\mathbb{R}^n\to \mathbb{R}$ given by $R(x_1,\dotsc,x_n):=\sum_{i=1}^nx_i$ is sufficient statistic w.r.t. $(\mathbb{R}^n,\mathfrak{B}_{\mathbb{R}^n},\{P_\theta^{\otimes n}\}_{\theta\in\mathbb{R}})$ which implies, by the minimality of $S$, that there's a measurable function $g:\mathbb{R}\to \mathbb{R}^n$ such that for all $\theta\in\mathbb{R}$ there's $N_\theta\in\mathfrak{B}_{\mathbb{R}^n}$ with $P_{\theta}^{\otimes n}(N_\theta)=0$ and $S(x)=g(R(x))$ for all $x\in\mathbb{R}^n\setminus N_\theta$.

Answer 1

This indeed cannot happen. For an element $v\in\mathbb R^n$, let $\sum v$ be the sum of its coefficients.

Let $A=\mathbb R^n\setminus N_\theta$. For any nonzero vector $v\in\mathbb R^n$ with $\sum v=0$, and any $a\in A$, we have $a+v\neq a=g(\sum a)=g(\sum(a+v))$, hence $a+v\not\in A$. Therefore $A$ has uncountably many pairwise disjoint translates, all of the same (Lebesgue) measure, which imples $A$ is necessarily null, so $N_\theta$ cannot be null.

I'm afraid I cannot comment on your argument proving the contrary and how it fails.