Geometry is not my area, and so, I am not sure the title accurately captures what I am interested in exactly... I hope the tags are appropriate.
For any vector $v$, denote it's $i$-th component by $v_{(i)}$. Fix $m, n \in \mathbb{N}$ with $m>n$.
Let $A$ be an $n \times m$ binary matrix, let $\mathcal{Y} \subseteq \mathbb{R}$, and let $b(y) \in \mathbb{R}_{\geq 0}^n$ for each $y \in \mathcal{Y}$. Let $\mu$ be some measure on $\mathcal{Y}$ and assume that each component of $b(y)$ is a measurable function wrt $\mu$.
Now, the set $$ \omega(y):=\left\{x \in \mathbb{R}_{\geq 0}^m: A x=b(y)\right\} $$ is a convex polytope for each $y \in \mathcal{Y}$. Assume that $A$ and $b(y)$ are such that $\omega(y) \neq \emptyset$ for each $y \in \mathcal{Y}$. Note that $b(y)$ and $x$ are required to have non-negative components.
Define $\varpi$ as the set of non-negative functions $g: \mathcal{Y} \rightarrow \mathbb{R}_{\geq 0}^m$ such that $g(y) \in \omega(y)$ for each $y \in \mathcal{Y}$.
If my understanding is correct, then the space $$\bigsqcup_{y \in \mathcal Y}\left( \left\{y\right\} \times \omega(y)\right)$$ can be seen as (almost?) a fiber bundle, with base field $\mathcal Y$, and the usual projection map (i.e. $(y, x(y)) \mapsto y$ for any $x(y) \in \mathbb{R}^m_{\geq 0}$ with $Ax(y) = b(y)$), but I am not sure what the "fiber" should be here since $\omega(y)$ depends on $y$.
If it can be seen as a fiber bundle, then we can see $\varpi$ as the (global) sections of this bundle.
Now, what I am looking for is a "tractable" characterization of the following set $$ \Pi:=\left\{\left(\int_{\mathcal{Y}} g_{(i)}(y) d \mu(y): i \in\{1, \ldots, m\}\right): g \in \varpi\right\}~, $$ which can be seen as the space generated by applying an integral transform to the sections of the "bundle" above. I am not sure what $\Pi$ is, and would appreciate any ideas that can help me better understand what $\Pi$ looks like.
I have the following conjecture that basically suggests that I can think of $\Pi$ in terms of integrals of "envelopes" that constrain $g \in \varpi$:
For each subset $\xi \subseteq\{1, \ldots, m\}$, define $\bar{g}_{\xi}$ as the function that satisfies, for each $y \in \mathcal{Y}$, $$ \bar{g}_{\xi}(y)=\underset{x \in \omega(y)}{\operatorname{argmax}} \sum_{i \in \xi} x_{(i)} $$ and $\underline{g}_{\xi}$ as the function that satisfies, for each $y \in \mathcal{Y}$, $$ \underline{g}_{\xi}(y)=\underset{x \in \omega(y)}{\operatorname{argmin}} \sum_{i \in \xi} x_{(i)} $$ Now, my conjecture is that $\hat \Pi = \Pi$, where $$ \hat \Pi := \left\{p \in \mathbb{R}_{\geq 0}^m: \sum_{i \in \xi} p_{(i)} \in\left[\int_{\mathcal{Y}} \underline g_{\xi}(y) d \mu(y), \int_{\mathcal{Y}} \bar{g}_{\xi}(y) d \mu(y)\right], \forall \xi \subseteq\{1, \ldots, m\}\right\}~. $$ It is easy to see that $\Pi \subseteq \hat \Pi$, so it comes down to whether the reverse inclusion holds.
I suspect the conjecture is false if $A$ is allowed to vary with $y$, but I think it may be true in my case given the convexity (and "similarity") of each $y$-section.
