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My question is motivated by a general measure-theoretic problem that one frequently encounters in probability: the need to work with uncountably many mutually singular measures at once, and with functions that are initially only defined almost everywhere, not pointwise. Whenever such a situation arises, it seems like a common practice to resolve the problem in ad hoc ways, imposing requirements like continuity or quasi-continuity to pin down the "good enough" pointwise versions of the relevant functions. Broadly put, my question is whether there is a chance for a purely measure-theoretic (i.e. non-topological) way to construct a pointwise version of a function from its almost-everywhere "shadows" w.r.t. different measures.

Let $(X, \mathcal B(X))$ be a standard Borel space, and let $M$ be the Banach lattice of (signed, finite) Borel measures on it. We equip $M$ with the standard $\sigma$-algebra $\mathcal B(M)$ generated by the evaluation maps $\mu \mapsto \mu[A], A \in \mathcal B (X)$.

Let $I \subset M$ be a norm-closed ideal (equivalently: a subspace that, together with every measure $\mu \in I$ contains all measures that are $\mu$-absolutely continuous). Assume that, furthermore,

• $I$ is a $\mathcal{B}(M)$-measurable set (note that $\mathcal{B}(M)$ is much smaller than the Borel $\sigma$-algebra of the norm topology, so this doesn't follow from being norm-closed).
• $I$ is closed under taking barycenters. That is, if $\mu : [0,1] \to I$ is a measurable map and $\intop_0^1 \Vert \mu_t \Vert dt < \infty$ then $\intop_0^1 \mu_t dt \in I$.

Now assume that with every measure $\mu \in I$ we associate a $\mu$-equivalence class of a function: $f_\mu \in L^0(|\mu|)$. Call the family $(f_\mu, \mu \in I)$ consistent if:

• If $\nu \ll \mu$ then $f_\nu = f_\mu$ $\nu$-almost everywhere.
• For any measurably parametrized family $\mu : [0,1] \to I$ as above, the equivalence classes of functions $f_{\mu_t}$ are related to $f_\nu, \nu := \intop_0^1 \mu_t dt$ in the natural way: there is a Borel function $\tilde f : X \to \mathbb{R}$, such that $\tilde f = f_{\nu}$ $\nu$-almost everywhere and $\tilde f = f_{\mu_t}$ $\mu_t$-almost everywhere for Lebesgue-almost all $t$.

The latter condition allows to recover $f_{\mu_t}$ for Lebesgue-almost all $t$ from $f_{\intop \mu_t dt}$ and vice versa. Note that nothing really depends on the choice of the version $\tilde f$. The point is merely its existence, which imposes a strong compatibility requirement on $f_{\mu_t}$, even when $\mu_t$ are singular to each other.

Question: Obviously, if there is a universally measurable function $F : X \to \mathbb{R}$ that is a version of $f_\mu$ for all $\mu \in I$ simultaneously then the family $(f_\mu)$ is consistent. Is the converse true? If not, should I replace $I \in \mathcal{B}(M)$ by some better regularity condition in order for the answer to be positive?

Actually, I don't know the answer even for the ideal of all atomless measures.

My question is motivated by a general measure-theoretic problem that one frequently encounters in probability: the need to work with uncountably many mutually singular measures at once, and with functions that are initially only defined almost everywhere, not pointwise. Whenever such a situation arises, it seems like a common practice to resolve the problem in ad hoc ways, imposing requirements like continuity or quasi-continuity to pin down the "good enough" pointwise versions of the relevant functions. Broadly put, my question is whether there is a chance for a purely measure-theoretic (i.e. non-topological) way to construct a pointwise version of a function from its almost-everywhere "shadows" w.r.t. different measures.

Let $(X, \mathcal B(X))$ be a standard Borel space, and let $M$ be the Banach lattice of (signed, finite) Borel measures on it. We equip $M$ with the standard $\sigma$-algebra $\mathcal B(M)$ generated by the evaluation maps $\mu \mapsto \mu[A], A \in \mathcal B (X)$.

Let $I \subset M$ be a norm-closed ideal (equivalently: a subspace that, together with every measure $\mu \in I$ contains all measures that are $\mu$-absolutely continuous). Assume that, furthermore,

• $I$ is a $\mathcal{B}(M)$-measurable set (note that $\mathcal{B}(M)$ is much smaller than the Borel $\sigma$-algebra of the norm topology, so this doesn't follow from being norm-closed).
• $I$ is closed under taking barycenters. That is, if $\mu : [0,1] \to I$ is a measurable map and $\intop_0^1 \Vert \mu_t \Vert dt < \infty$ then $\intop_0^1 \mu_t dt \in I$.

Now assume that with every measure $\mu \in I$ we associate a $\mu$-equivalence class of a function: $f_\mu \in L^0(|\mu|)$. Call the family $(f_\mu, \mu \in I)$ consistent if:

• If $\nu \ll \mu$ then $f_\nu = f_\mu$ $\nu$-almost everywhere.
• For any measurably parametrized family $\mu : [0,1] \to I$ as above, the equivalence classes of functions $f_{\mu_t}$ are related to $f_\nu, \nu := \intop_0^1 \mu_t dt$ in the natural way: there is a Borel function $\tilde f : X \to \mathbb{R}$, such that $\tilde f = f_{\nu}$ $\nu$-almost everywhere and $\tilde f = f_{\mu_t}$ $\mu_t$-almost everywhere for Lebesgue-almost all $t$.

The latter condition allows to recover $f_{\mu_t}$ for Lebesgue-almost all $t$ from $f_{\intop |\mu_t| dt}$ and vice versa. Note that nothing really depends on the choice of the version $\tilde f$. The point is merely its existence, which imposes a strong compatibility requirement on $f_{\mu_t}$, even when $\mu_t$ are singular to each other.

Question: Obviously, if there is a universally measurable function $F : X \to \mathbb{R}$ that is a version of $f_\mu$ for all $\mu \in I$ simultaneously then the family $(f_\mu)$ is consistent. Is the converse true? If not, should I replace $I \in \mathcal{B}(M)$ by some better regularity condition in order for the answer to be positive?

Actually, I don't know the answer even for the ideal of all atomless measures.

My question is motivated by a general measure-theoretic problem that one frequently encounters in probability: the need to work with uncountably many mutually singular measures at once, and with functions that are initially only defined almost everywhere, not pointwise. Whenever such a situation arises, it seems like a common practice to resolve the problem in ad hoc ways, imposing requirements like continuity or quasi-continuity to pin down the "good enough" pointwise versions of the relevant functions. Broadly put, my question is whether there is a chance for a purely measure-theoretic (i.e. non-topological) way to construct a pointwise version of a function from its almost-everywhere "shadows" w.r.t. different measures.

Let $(X, \mathcal B(X))$ be a standard Borel space, and let $M$ be the Banach lattice of (signed, finite) Borel measures on it. We equip $M$ with the standard $\sigma$-algebra $\mathcal B(M)$ generated by the evaluation maps $\mu \mapsto \mu[A], A \in \mathcal B (X)$.

Let $I \subset M$ be a norm-closed ideal (equivalently: a subspace that, together with every measure $\mu \in I$ contains all measures that are $\mu$-absolutely continuous). Assume that, furthermore,

• $I$ is a $\mathcal{B}(M)$-measurable set (note that $\mathcal{B}(M)$ is much smaller than the Borel $\sigma$-algebra of the norm topology, so this doesn't follow from being norm-closed).
• $I$ is closed under taking barycenters. That is, if $\mu : [0,1] \to I$ is a measurable map and $\intop_0^1 \Vert \mu_t \Vert dt < \infty$ then $\intop_0^1 \mu_t dt \in I$.

Now assume that with every measure $\mu \in I$ we associate a $\mu$-equivalence class of a function: $f_\mu \in L^0(|\mu|)$. Call the family $(f_\mu, \mu \in I)$ consistent if:

• If $\nu \ll \mu$ then $f_\nu = f_\mu$ $\nu$-almost everywhere.
• For any measurably parametrized family $\mu : [0,1] \to I$ as above, the equivalence classes of functions $f_{\mu_t}$ are related to $f_\nu, \nu := \intop_0^1 \mu_t dt$ in the natural way: there is a Borel function $\tilde f : X \to \mathbb{R}$, such that $\tilde f = f_{\nu}$ $\nu$-almost everywhere and $\tilde f = f_{\mu_t}$ $\mu_t$-almost everywhere for Lebesgue-almost all $t$.

The latter condition allows to recover $f_{\mu_t}$ for Lebesgue-almost all $t$ from $f_{\intop \mu_t dt}$ and vice versa. Note that nothing really depends on the choice of the version $\tilde f$. The point is merely its existence, which imposes a strong compatibility requirement on $f_{\mu_t}$, even when $\mu_t$ are singular to each other.

Question: Obviously, if there is a universally measurable function $F : X \to \mathbb{R}$ that is a version of $f_\mu$ for all $\mu \in I$ simultaneously then the family $(f_\mu)$ is consistent. Is the converse true? If not, should I replace $I \in \mathcal{B}(M)$ by some better regularity condition in order for the answer to be positive?

My question is motivated by a general measure-theoretic problem that one frequently encounters in probability: the need to work with uncountably many mutually singular measures at once, and with functions that are initially only defined almost everywhere, not pointwise. Whenever such a situation arises, it seems like a common practice to resolve the problem in ad hoc ways, imposing requirements like continuity or quasi-continuity to pin down the "good enough" pointwise versions of the relevant functions. Broadly put, my question is whether there is a chance for a purely measure-theoretic (i.e. non-topological) way to construct a pointwise version of a function from its almost-everywhere "shadows" w.r.t. different measures.

Let $(X, \mathcal B(X))$ be a standard Borel space, and let $M$ be the Banach lattice of (signed, finite) Borel measures on it. We equip $M$ with the standard $\sigma$-algebra $\mathcal B(M)$ generated by the evaluation maps $\mu \mapsto \mu[A], A \in \mathcal B (X)$.

Let $I \subset M$ be a norm-closed ideal (equivalently: a subspace that, together with every measure $\mu \in I$ contains all measures that are $\mu$-absolutely continuous). Assume that, furthermore,

• $I$ is a $\mathcal{B}(M)$-measurable set (note that $\mathcal{B}(M)$ is much smaller than the Borel $\sigma$-algebra of the norm topology, so this doesn't follow from being norm-closed).
• $I$ is closed under taking barycenters. That is, if $\mu : [0,1] \to I$ is a measurable map and $\intop_0^1 \Vert \mu_t \Vert dt < \infty$ then $\intop_0^1 \mu_t dt \in I$.

Now assume that with every measure $\mu \in I$ we associate a $\mu$-equivalence class of a function: $f_\mu \in L^0(|\mu|)$. Call the family $(f_\mu, \mu \in I)$ consistent if:

• If $\nu \ll \mu$ then $f_\nu = f_\mu$ $\nu$-almost everywhere.
• For any measurably parametrized family $\mu : [0,1] \to I$ as above, the equivalence classes of functions $f_{\mu_t}$ are related to $f_\nu, \nu := \intop_0^1 \mu_t dt$ in the natural way: there is a Borel function $\tilde f : X \to \mathbb{R}$, such that $\tilde f = f_{\nu}$ $\nu$-almost everywhere and $\tilde f = f_{\mu_t}$ $\mu_t$-almost everywhere for Lebesgue-almost all $t$.

The latter condition allows to recover $f_{\mu_t}$ for Lebesgue-almost all $t$ from $f_{\intop \mu_t dt}$ and vice versa. Note that nothing really depends on the choice of the version $\tilde f$. The point is merely its existence, which imposes a strong compatibility requirement on $f_{\mu_t}$, even when $\mu_t$ are singular to each other.

Question: Obviously, if there is a universally measurable function $F : X \to \mathbb{R}$ that is a version of $f_\mu$ for all $\mu \in I$ simultaneously then the family $(f_\mu)$ is consistent. Is the converse true? If not, should I replace $I \in \mathcal{B}(M)$ by some better regularity condition in order for the answer to be positive?

Actually, I don't know the answer even for the ideal of all atomless measures.

My question is motivated by a general measure-theoretic problem that one frequently encounters in probability: the need to work with uncountably many mutually singular measures at once, and with functions that are initially only defined almost everywhere, not pointwise. Whenever such a situation arises, it seems like a common practice to resolve the problem in ad hoc ways, imposing requirements like continuity or quasi-continuity to pin down the "good enough" pointwise versions of the relevant functions. Broadly put, my question is whether there is a chance for a purely measure-theoretic (i.e. non-topological) way to construct a pointwise version of a function from its almost-everywhere "shadows" w.r.t. different measures.

Let $(X, \mathcal B(X))$ be a standard Borel space, and let $M$ be the Banach lattice of (signed, finite) Borel measures on it. We equip $M$ with the standard $\sigma$-algebra $\mathcal B(M)$ generated by the evaluation maps $\mu \mapsto \mu[A], A \in \mathcal B (X)$.

Let $I \subset M$ be norm-closed ideal (equivalently: a subspace that, together with every measure $\mu \in I$ contains all measures that are $\mu$-absolutely continuous). Assume that, furthermore,

• $I$ is a $\mathcal{B}(M)$-measurable set (note that $\mathcal{B}(M)$ is much smaller than the Borel $\sigma$-algebra of the norm topology, so this doesn't follow from being norm-closed).
• $I$ is closed under taking barycenters. That is, if $\mu : [0,1] \to I$ is a measurable map and $\intop_0^1 \Vert \mu_t \Vert dt < \infty$ then $\intop_0^1 \mu_t dt \in I$.

Now assume that with every measure $\mu \in I$ we associate a $\mu$-equivalence class of a function: $f_\mu \in L^0(|\mu|)$. Call the family $(f_\mu, \mu \in I)$ consistent if:

• If $\nu \ll \mu$ then $f_\nu = f_\mu$ $\nu$-almost everywhere.
• For any measurably parametrized family $\mu : [0,1] \to I$ as above, the equivalence classes of functions $f_{\mu_t}$ are related to $f_\nu, \nu := \intop_0^1 \mu_t dt$ in the natural way: there is a Borel function $\tilde f : X \to \mathbb{R}$, such that $\tilde f = f_{\nu}$ $\nu$-almost everywhere and $\tilde f = f_{\mu_t}$ $\mu_t$-almost everywhere for Lebesgue-almost all $t$.

The latter condition allows to recover $f_{\mu_t}$ for Lebesgue-almost all $t$ from $f_{\intop \mu_t dt}$ and vice versa. Note that nothing really depends on the choice of the version $\tilde f$. The point is merely its existence, which imposes a strong compatibility requirement on $f_{\mu_t}$, even when $\mu_t$ are singular to each other.

Question: Obviously, if there is a universally measurable function $F : X \to \mathbb{R}$ that is a version of $f_\mu$ for all $\mu \in I$ simultaneously then the family $(f_\mu)$ is consistent. Is the converse true?

My question is motivated by a general measure-theoretic problem that one frequently encounters in probability: the need to work with uncountably many mutually singular measures at once, and with functions that are initially only defined almost everywhere, not pointwise. Whenever such a situation arises, it seems like a common practice to resolve the problem in ad hoc ways, imposing requirements like continuity or quasi-continuity to pin down the "good enough" pointwise versions of the relevant functions. Broadly put, my question is whether there is a chance for a purely measure-theoretic (i.e. non-topological) way to construct a pointwise version of a function from its almost-everywhere "shadows" w.r.t. different measures.

Let $(X, \mathcal B(X))$ be a standard Borel space, and let $M$ be the Banach lattice of (signed, finite) Borel measures on it. We equip $M$ with the standard $\sigma$-algebra $\mathcal B(M)$ generated by the evaluation maps $\mu \mapsto \mu[A], A \in \mathcal B (X)$.

Let $I \subset M$ be a norm-closed ideal (equivalently: a subspace that, together with every measure $\mu \in I$ contains all measures that are $\mu$-absolutely continuous). Assume that, furthermore,

• $I$ is a $\mathcal{B}(M)$-measurable set (note that $\mathcal{B}(M)$ is much smaller than the Borel $\sigma$-algebra of the norm topology, so this doesn't follow from being norm-closed).
• $I$ is closed under taking barycenters. That is, if $\mu : [0,1] \to I$ is a measurable map and $\intop_0^1 \Vert \mu_t \Vert dt < \infty$ then $\intop_0^1 \mu_t dt \in I$.

Now assume that with every measure $\mu \in I$ we associate a $\mu$-equivalence class of a function: $f_\mu \in L^0(|\mu|)$. Call the family $(f_\mu, \mu \in I)$ consistent if:

• If $\nu \ll \mu$ then $f_\nu = f_\mu$ $\nu$-almost everywhere.
• For any measurably parametrized family $\mu : [0,1] \to I$ as above, the equivalence classes of functions $f_{\mu_t}$ are related to $f_\nu, \nu := \intop_0^1 \mu_t dt$ in the natural way: there is a Borel function $\tilde f : X \to \mathbb{R}$, such that $\tilde f = f_{\nu}$ $\nu$-almost everywhere and $\tilde f = f_{\mu_t}$ $\mu_t$-almost everywhere for Lebesgue-almost all $t$.

The latter condition allows to recover $f_{\mu_t}$ for Lebesgue-almost all $t$ from $f_{\intop \mu_t dt}$ and vice versa. Note that nothing really depends on the choice of the version $\tilde f$. The point is merely its existence, which imposes a strong compatibility requirement on $f_{\mu_t}$, even when $\mu_t$ are singular to each other.

Question: Obviously, if there is a universally measurable function $F : X \to \mathbb{R}$ that is a version of $f_\mu$ for all $\mu \in I$ simultaneously then the family $(f_\mu)$ is consistent. Is the converse true? If not, should I replace $I \in \mathcal{B}(M)$ by some better regularity condition in order for the answer to be positive?