Let $\{X\_i\}\_{i \in \mathbb{R}-\{0\}}$ be a set of subsets of a separable *infinite-dimensional* Fréchet space $X$ and $I$ be uncountable. Moreover, suppose that 


* (Dense $G\_{\delta}$) $X\_i$ is a dense $G\_{\delta}$ subset of $X$ not containing $0$,
* (Almost Contains a Linear Subspace) For each $i$, there exists a dense linear subset $E\_i\subset X$ satisfying
$$
E\_i-\{0\}\subseteq X\_i
$$
* (Disjoint) $\bigcap\_{i \in I} X\_i=\emptyset$,
* (Not a Cover) $\cup\_{i \in I} X\_i \neq X-\{0\}$,


Can we conclude that:
$$
X - \bigcup\_{i \in \mathbb{R}-\{0\}} X\_i,
$$
is Haar-null, or at-least it is finite-dimensional? 


I have never seen this type of result and am pretty new to this type of thing but I ask here since it seems beyond the level of math-stack exchange. 


**Relevant Definitions:**
*Haar-null set:* A subset $A\subseteq X$ is Haar-null if there exists a Borel probability measure $\mu$ on $X$ and a Borel subset $A\subseteq B$ satisfying
$$
\mu\left(
B+x
\right)=0 \qquad (\forall x \in X).
$$




---


**Facts:**


* I do know that $X=X\_i -X\_i$ upon applying the Baire category theorem. (Also from the comments the Pettis Lemma). This means that every element in $X$ can be represented as a sum of elements from each $X\_i$.
* In the case *(not covered by my question)* where $I$ is a singleton, [this paper](https://pdfs.semanticscholar.org/1c6d/8dafb7ada6993c940c15118342e3752fd033.pdf) gives a counter-example.




---


**Intuitions:**


As intuition, it can be seen [here](https://arxiv.org/pdf/1606.06607.pdf), that if $X$ is locally compact, then a Borel set is Haar-null if and only if it is of Haar-measure $0$. 



Haar-null union of dense subsets
In the Frechet space $X:=\mathbb R^\omega$ consider the dense linear subspace $$L\_0:=\{(x\_n)\_{n\in\omega}\in\mathbb R^\omega:|\{n\in\omega:x\_n\ne0\}|<\omega\}.$$


Fix a countable base $\{V\_n\}\_{n\in\omega}$ of the topology of the space $L\_0$ and in each set $V\_n$ choose a point $x\_n$, which is not contained in the linear hull of the set $\{x\_i\}\_{i<n}$. Then $\{x\_n\}\_{n\in\omega}$ is a dense linearly independent set $\{x\_n\}\_{n\in\omega}$ in $X$. For every $n\in\mathbb N$ consider the linear hull $L\_n$ of the set $\{x\_m\}\_{m\ge n}$ and observe that $\{x\_m\}\_{m\ge n}$ and $L\_n$ are dense in $X$, and $\bigcap\_{n\in\omega}L\_n=\{0\}$.


Consequently, for every non-zero element $x\in X$ we can find a number $n\_x\in \omega$ such that $x\notin L\_{n\_x}$. 


It is easy to see that the closed convex set $F:=[1,\infty)^\omega$ in $X=\mathbb R^\omega$ is not Haar-null but is disjoint with the dense linear subspace $L\_0$ of $X$.


For any $x\in X\setminus\{0\}$ consider the open subset $W\_x:=X\setminus(F\cup \cup\{x,0\})$ and observe that $L\_{n\_x}\setminus\{0\}\subset W\_x\subset X\setminus\{x,0\}$, which implies $\bigcap\_{x\in X\setminus \{0\}}W\_x=\emptyset$.


Also $X\setminus \bigcup\_{x\in X\setminus\{0\}}W\_x\supset F$ is not Haar-null.


So, the family of dense open (and hence $G\_\delta$) sets $(W\_x)\_{x\in X\setminus\{x\}}$ has the properties required in the question.


