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Julian Newman
Julian Newman
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Let $X$ be a Banach space, and for any compactly supported Borel probability measure $\mathbb{P}$ on $X$, define the mean $\mu_\mathbb{P}$ by $\mu_\mathbb{P}=\int_X x \, \mathbb{P}(dx)$.

I want to say that a Borel set $S \subset X$ is continuum-convex if every compactly supported Borel probability measure $\mathbb{P}$ on $X$ with $\mathbb{P}(S)=1$ has $\mu_\mathbb{P} \in S$.

Is there a recognised term for what I have called "continuum-convex"?

The point behind this definition:

The standard "set-theoretic" definition of a "convex subset of a vector space" is a subset that is closed under finite convex combinations, i.e. under weighted averages where the weighting is discrete. I want a stronger notion where the weighting is allowed to be completely general, i.e. not necessarily discrete (hence the term "continuum").

Note that

  • if $S$ is continuum-convex then $S$ is convex;
  • if $S$ is closed and convex then $S$ is continuum-convex.

However, neither of these implications is reversible. For example:

Suppose that $X$ is the completion of the vector space $\mathcal{M}(M)$ of finite signed measures on a compact metric space $M$ where $\mathcal{M}(M)$ is equipped with a norm $\|\cdot\|_{\mathcal{M}(M)}$ whose topology is the topology of weak convergence. (E.g. one can find a sequence of continuous functions $g_n \colon M \to [0,2^{-n}]$ such that $\|\mu\|_{\mathcal{M}(M)}:=\sum_n |\int_M g_n \, d\mu|$ is such a norm.) Then,

  • the set of all finitely supported probability measures on $M$ is convex, but not continuum-convex;
  • given a non-closed Borel set $A \subset M$ and a non-empty interval $I \subset (0,1]$, the set of all probability measures assigning to $A$ a value in $I$ is not closed in $X$, but nonetheless is continuum-convex.

Update. The question has essentially been answered by Gerald Edgar in a comment: there is a term "measure convex" whose exact definition varies between authors, but the idea behind it is exactly what I've presented above.

I realise that I should also have mentioned: every convex Borel subset of $\mathbb{R}^n$ is measure convex, so my question is really about the problem of infinite dimensions.

Let $X$ be a Banach space, and for any compactly supported Borel probability measure $\mathbb{P}$ on $X$, define the mean $\mu_\mathbb{P}$ by $\mu_\mathbb{P}=\int_X x \, \mathbb{P}(dx)$.

I want to say that a Borel set $S \subset X$ is continuum-convex if every compactly supported Borel probability measure $\mathbb{P}$ on $X$ with $\mathbb{P}(S)=1$ has $\mu_\mathbb{P} \in S$.

Is there a recognised term for what I have called "continuum-convex"?

The point behind this definition:

The standard "set-theoretic" definition of a "convex subset of a vector space" is a subset that is closed under finite convex combinations, i.e. under weighted averages where the weighting is discrete. I want a stronger notion where the weighting is allowed to be completely general, i.e. not necessarily discrete (hence the term "continuum").

Note that

  • if $S$ is continuum-convex then $S$ is convex;
  • if $S$ is closed and convex then $S$ is continuum-convex.

However, neither of these implications is reversible. For example:

Suppose that $X$ is the completion of the vector space $\mathcal{M}(M)$ of finite signed measures on a compact metric space $M$ where $\mathcal{M}(M)$ is equipped with a norm $\|\cdot\|_{\mathcal{M}(M)}$ whose topology is the topology of weak convergence. (E.g. one can find a sequence of continuous functions $g_n \colon M \to [0,2^{-n}]$ such that $\|\mu\|_{\mathcal{M}(M)}:=\sum_n |\int_M g_n \, d\mu|$ is such a norm.) Then,

  • the set of all finitely supported probability measures on $M$ is convex, but not continuum-convex;
  • given a non-closed Borel set $A \subset M$ and a non-empty interval $I \subset (0,1]$, the set of all probability measures assigning to $A$ a value in $I$ is not closed in $X$, but nonetheless is continuum-convex.

Let $X$ be a Banach space, and for any compactly supported Borel probability measure $\mathbb{P}$ on $X$, define the mean $\mu_\mathbb{P}$ by $\mu_\mathbb{P}=\int_X x \, \mathbb{P}(dx)$.

I want to say that a Borel set $S \subset X$ is continuum-convex if every compactly supported Borel probability measure $\mathbb{P}$ on $X$ with $\mathbb{P}(S)=1$ has $\mu_\mathbb{P} \in S$.

Is there a recognised term for what I have called "continuum-convex"?

The point behind this definition:

The standard "set-theoretic" definition of a "convex subset of a vector space" is a subset that is closed under finite convex combinations, i.e. under weighted averages where the weighting is discrete. I want a stronger notion where the weighting is allowed to be completely general, i.e. not necessarily discrete (hence the term "continuum").

Note that

  • if $S$ is continuum-convex then $S$ is convex;
  • if $S$ is closed and convex then $S$ is continuum-convex.

However, neither of these implications is reversible. For example:

Suppose that $X$ is the completion of the vector space $\mathcal{M}(M)$ of finite signed measures on a compact metric space $M$ where $\mathcal{M}(M)$ is equipped with a norm $\|\cdot\|_{\mathcal{M}(M)}$ whose topology is the topology of weak convergence. (E.g. one can find a sequence of continuous functions $g_n \colon M \to [0,2^{-n}]$ such that $\|\mu\|_{\mathcal{M}(M)}:=\sum_n |\int_M g_n \, d\mu|$ is such a norm.) Then,

  • the set of all finitely supported probability measures on $M$ is convex, but not continuum-convex;
  • given a non-closed Borel set $A \subset M$ and a non-empty interval $I \subset (0,1]$, the set of all probability measures assigning to $A$ a value in $I$ is not closed in $X$, but nonetheless is continuum-convex.

Update. The question has essentially been answered by Gerald Edgar in a comment: there is a term "measure convex" whose exact definition varies between authors, but the idea behind it is exactly what I've presented above.

I realise that I should also have mentioned: every convex Borel subset of $\mathbb{R}^n$ is measure convex, so my question is really about the problem of infinite dimensions.

Extended definition to allow $S$ to be non-relatively-compact; changed "measure-theoretic" to "continuum-"; added explanation of definition.
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Julian Newman
Julian Newman
  • 708
  • 8
  • 27

Is there a proper term for a "measure"continuum-theoretically convex"convex" set?

Let $X$ be a Banach space, and for any compactly supported Borel probability measure $\mathbb{P}$ on $X$, define the mean $\mu_\mathbb{P}$ by $\mu_\mathbb{P}=\int_X x \, \mathbb{P}(dx)$.

I want to say that a relatively compact Borel set $S \subset X$ is measurecontinuum-theoretically convexconvex if every compactly supported Borel probability measure $\mathbb{P}$ on $X$ with $\mathbb{P}(S)=1$ has $\mu_\mathbb{P} \in S$.

Is there a recognised term for what I have called "measure"continuum-theoretically convex"convex"?

The point behind this definition:

The standard "set-theoretic" definition of a "convex subset of a vector space" is a subset that is closed under finite convex combinations, i.e. under weighted averages where the weighting is discrete. I want a stronger notion where the weighting is allowed to be completely general, i.e. not necessarily discrete (hence the term "continuum").

Note that

  • if $S$ is measurecontinuum-theoretically convexconvex then $S$ is convex;
  • if $S$ is compactclosed and convex then $S$ is measurecontinuum-theoretically convexconvex.

To give some further examplesHowever, supposeneither of these implications is reversible. For example:

Suppose that $X$ is the completion of the vector space $\mathcal{M}(M)$ of finite signed measures on a compact metric space $M$ where $\mathcal{M}(M)$ is equipped with a norm $\|\cdot\|$$\|\cdot\|_{\mathcal{M}(M)}$ whose topology is the topology of weak convergence. (E.g. one can find a sequence of continuous functions $g_n \colon M \to [0,2^{-n}]$ such that $\|\mu\|:=\sum_n |\int_M g_n \, d\mu|$$\|\mu\|_{\mathcal{M}(M)}:=\sum_n |\int_M g_n \, d\mu|$ is such a norm.) Then,

  • the set of all finitely supported probability measures supported on a finite subset of $M$ is convex, but not measurecontinuum-theoretically convex;convex;
  • given a non-closed Borel set $A \subset M$ and a non-empty interval $I \subset (0,1]$, the set of all probability measures assigning to $A$ a value in $I$ is not closed in $X$, but nonetheless is measurecontinuum-theoretically convexconvex.

Is there a proper term for a "measure-theoretically convex" set?

Let $X$ be a Banach space, and for any compactly supported Borel probability measure $\mathbb{P}$ on $X$, define the mean $\mu_\mathbb{P}$ by $\mu_\mathbb{P}=\int_X x \, \mathbb{P}(dx)$.

I want to say that a relatively compact Borel set $S \subset X$ is measure-theoretically convex if every Borel probability measure $\mathbb{P}$ with $\mathbb{P}(S)=1$ has $\mu_\mathbb{P} \in S$.

Is there a recognised term for what I have called "measure-theoretically convex"?

Note that

  • if $S$ is measure-theoretically convex then $S$ is convex;
  • if $S$ is compact and convex then $S$ is measure-theoretically convex.

To give some further examples, suppose that $X$ is the completion of the vector space of finite signed measures on a compact metric space $M$ equipped with a norm $\|\cdot\|$ whose topology is the topology of weak convergence. (E.g. one can find a sequence of continuous functions $g_n \colon M \to [0,2^{-n}]$ such that $\|\mu\|:=\sum_n |\int_M g_n \, d\mu|$ is such a norm.) Then,

  • the set of all probability measures supported on a finite subset of $M$ is convex, but not measure-theoretically convex;
  • given a non-closed Borel set $A \subset M$ and a non-empty interval $I \subset (0,1]$, the set of all probability measures assigning to $A$ a value in $I$ is not closed in $X$, but nonetheless is measure-theoretically convex.

Is there a proper term for a "continuum-convex" set?

Let $X$ be a Banach space, and for any compactly supported Borel probability measure $\mathbb{P}$ on $X$, define the mean $\mu_\mathbb{P}$ by $\mu_\mathbb{P}=\int_X x \, \mathbb{P}(dx)$.

I want to say that a Borel set $S \subset X$ is continuum-convex if every compactly supported Borel probability measure $\mathbb{P}$ on $X$ with $\mathbb{P}(S)=1$ has $\mu_\mathbb{P} \in S$.

Is there a recognised term for what I have called "continuum-convex"?

The point behind this definition:

The standard "set-theoretic" definition of a "convex subset of a vector space" is a subset that is closed under finite convex combinations, i.e. under weighted averages where the weighting is discrete. I want a stronger notion where the weighting is allowed to be completely general, i.e. not necessarily discrete (hence the term "continuum").

Note that

  • if $S$ is continuum-convex then $S$ is convex;
  • if $S$ is closed and convex then $S$ is continuum-convex.

However, neither of these implications is reversible. For example:

Suppose that $X$ is the completion of the vector space $\mathcal{M}(M)$ of finite signed measures on a compact metric space $M$ where $\mathcal{M}(M)$ is equipped with a norm $\|\cdot\|_{\mathcal{M}(M)}$ whose topology is the topology of weak convergence. (E.g. one can find a sequence of continuous functions $g_n \colon M \to [0,2^{-n}]$ such that $\|\mu\|_{\mathcal{M}(M)}:=\sum_n |\int_M g_n \, d\mu|$ is such a norm.) Then,

  • the set of all finitely supported probability measures on $M$ is convex, but not continuum-convex;
  • given a non-closed Borel set $A \subset M$ and a non-empty interval $I \subset (0,1]$, the set of all probability measures assigning to $A$ a value in $I$ is not closed in $X$, but nonetheless is continuum-convex.
Source Link
Julian Newman
Julian Newman
  • 708
  • 8
  • 27

Is there a proper term for a "measure-theoretically convex" set?

Let $X$ be a Banach space, and for any compactly supported Borel probability measure $\mathbb{P}$ on $X$, define the mean $\mu_\mathbb{P}$ by $\mu_\mathbb{P}=\int_X x \, \mathbb{P}(dx)$.

I want to say that a relatively compact Borel set $S \subset X$ is measure-theoretically convex if every Borel probability measure $\mathbb{P}$ with $\mathbb{P}(S)=1$ has $\mu_\mathbb{P} \in S$.

Is there a recognised term for what I have called "measure-theoretically convex"?

Note that

  • if $S$ is measure-theoretically convex then $S$ is convex;
  • if $S$ is compact and convex then $S$ is measure-theoretically convex.

To give some further examples, suppose that $X$ is the completion of the vector space of finite signed measures on a compact metric space $M$ equipped with a norm $\|\cdot\|$ whose topology is the topology of weak convergence. (E.g. one can find a sequence of continuous functions $g_n \colon M \to [0,2^{-n}]$ such that $\|\mu\|:=\sum_n |\int_M g_n \, d\mu|$ is such a norm.) Then,

  • the set of all probability measures supported on a finite subset of $M$ is convex, but not measure-theoretically convex;
  • given a non-closed Borel set $A \subset M$ and a non-empty interval $I \subset (0,1]$, the set of all probability measures assigning to $A$ a value in $I$ is not closed in $X$, but nonetheless is measure-theoretically convex.