You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
abstract: 'The *Filter Dichotomy* says that every uniform nonmeager filter on the integers is mapped by a finite-to-one function to an ultrafilter. The consistency of this principle was proved by Blass and Laflamme. A *medial limit* is a universally measurable function from $\mathcal{P}(\omega)$ to the unit interval $[0,1]$ which is finitely additive for disjoint sets, and maps singletons to $0$ and $\omega$ to $1$. Christensen and Mokobodzki independently showed that the Continuum Hypothesis implies the existence of medial limits. We show that the Filter Dichotomy implies that there are no medial limits.'
author:
- 'Paul B. Larson[^1]'
title: The Filter Dichotomy and medial limits
---

2000 Mathematics Subject Classification 03E35; 28A20

Universally measurable sets
===========================

A *measure* on a set $X$ is a function $\mu$ whose domain is some $\sigma$-algebra of subsets of $X$, with codomain $[0, \infty]$, such that $\mu$ is countably additive for disjoint families (by default, measures are countably additive, but we deal also with *finitely additive measures*, which are finitely additive but not necessarily countably additive). A set is said to be *measurable* with respect to $\mu$ if it is in the domain of $\mu$. A *Borel measure* is a measure on a topological space whose domain contains the Borel subsets of the space. A measure is *complete* if all subsets of sets of measure $0$ are in the domain of the measure (and thus have measure 0). The *completion* of a measure is the smallest complete measure extending it. If $\mu$ is a Borel measure on a topological space $X$, and $\mu^{*}$ is the completion of $\mu$, then a set $A \subseteq X$ is in the domain of $\mu^{*}$ if and only if there is a set $B$ in the domain of $\mu$ such that the symmetric difference $A \bigtriangleup B$ is contained in a set of $\mu$-measure 0 (see 212C of [@F2]). A measure $\mu$ on a set $X$ is a *probability* measure if $\mu(X) = 1$, *finite* if $\mu(X)$ is finite, $\sigma$-*finite* if $X$ is a countable union of sets of finite measure, and *atomless* if singletons have measure 0.

A subset of a topological space is said to be *universally measurable* if it is measurable with respect to every complete $\sigma$-finite Borel measure on the space (see [@Ke; @Ni], and 434D of [@F4], for instance). The collection of universally measurable sets does not change if one replaces “$\sigma$-finite" with “finite" or requires the measures to be atomless (see 211X(e) of [@F2]).

We will make use of the following standard observation.

\[immeas\] If $f \colon X \to Y$ is a homeomorphism between topological spaces, then any Borel measure $\mu$ on $Y$ induces a measure $\mu^{*}$ on $X$ defined by letting $\mu^{*}(B) = \mu(f[B])$ for any Borel $B \subseteq X$. It follows that, in this context, for any universally measurable $A \subseteq X$, $f[A]$ is a universally measurable subset of $Y$.

Medial limits
=============

A function between topological spaces is *universally measurable* if all preimages of open sets (equivalently, Borel sets) are universally measurable. For our purposes, a *medial limit* is a universally measurable function from $\mathcal{P}(\omega)$ to $[0,1]$ which is finitely additive for disjoint sets, and maps singletons to $0$ and $\omega$ to $1$ (i.e., a universally measurable finitely additive measure on $\mathcal{P}(\omega)$ giving $\omega$ measure $1$ and singletons measure $0$). Equivalently, a *medial limit* is a function $f \colon \mathcal{P}(\omega) \to [0,1]$, finitely additive for disjoint sets and mapping singletons to $0$ and $\omega$ to $1$, such that for every complete, $\sigma$-finite Borel measure $\mu$ on $\mathcal{P}(\omega)$ there is a Borel function $b \colon \mathcal{P}(\omega) \to [0,1]$ such that $\{ x \subseteq \omega \mid f(x) \neq b(x)\}$ is $\mu$-null. Medial limits (in various forms) appear the following publications, among others : [@MaPrWe; @Ta85; @Fi; @Ke91; @Co; @La; @JaKeLo; @Ka]. Christensen and Mokobodzki (see [@Ch; @Me]) independently showed that medial limits exist under the assumption that the Continuum Hypothesis holds. This assumption was weakened to Martin’s Axiom by Normann [@No]. As far as we know, the weakest hypothesis known to be sufficient is the statement that the reals are not a union of fewer than continuum many meager sets (i.e., that the covering number for the meager ideal is the continuum). This was apparently known to Mokobodzki in the 1970’s (see also 538S of [@F5]). The term “medial limit" is often used for the corresponding linear functional on $\ell^{\infty}$ (see [@Me]; 538Q of [@F5]).

Godefroy and Talagrand [@GoTa] proved in 1977 that if $f$ is a medial limit, then the filter $\{ x\subseteq \omega \mid f(x) = 1\}$ does not have the property of Baire. The next two theorems present their proof of this fact, using terminology from Farah’s [@Fa]. Note that a filter on $\omega$ has the property of Baire if and only if its corresponding ideal does.

An ideal $I$ on $\omega$ is said to be *c.c.c. over Fin* if all almost disjoint families (i.e., families of infinite subsets of $\omega$ which pairwise have finite intersection) disjoint from $I$ are countable (this is Definition 3.3.1 of [@Fa]). We say that an ideal on $\omega$ is *uniform* if it contains all finite subsets of $\omega$, and similarly that a filter is uniform if its corresponding ideal is.

\[medcc\] If $f$ is a medial limit and $I = \{ x \subseteq \omega \mid f(x) = 0\}$, then $I$ is a universally measurable uniform c.c.c. over Fin ideal.

That $I$ is a universally measurable uniform ideal follows from the definition of medial limit. To see that $I$ is c.c.c. over Fin, let $A$ be an almost disjoint family disjoint from $I$, and suppose that $A$ is uncountable. Then there is a positive integer $n$ such that the set of $x \in A$ such that $f(x) \geq 1/n$ is uncountable. Since $f$ is finitely additive for almost disjoint sets, and $f(\omega) = 1$, any such set can have size at most $n$.

The following is Lemma 3.3.2(c) of [@Fa].

\[ccnb\] A uniform c.c.c. over Fin proper ideal on $\omega$ cannot have the property of Baire.

Let $I$ be a uniform c.c.c. over Fin proper ideal on $\omega$. If $I$ is somewhere comeager, than there are two members of $I$ whose union is cofinite. To see this, suppose that $s\subseteq \omega$ is finite and $D_{n}$ $(n \in \omega)$ are dense open subsets of $\mathcal{P}(\omega)$ such that all $x \in [s] \cap \bigcap_{n \in \omega} D_{n}$ are in $I$, where $[s] = \{ x \subseteq \omega \mid x \cap (max(s) + 1) = s\}$. It is relatively straightforward to build sets $x$, $y$ in $[s] \cap \bigcap_{n \in \omega} D_{n}$ whose union is $\omega\setminus ((max(s) + 1)\setminus s)$. Similarly, if $I$ is meager, then there is a perfect set of subsets of $\omega$ which are almost disjoint and all not in $I$. The construction of such a perfect set is similar, using a collection of dense open sets $D_{n}$ $(n \in \omega)$ such that $\bigcap_{n \in \omega}D_{n}$ is disjoint from $I$.

The Filter Dichotomy
====================

The *Filter Dichotomy* is the statement that for each nonmeager filter $F$ on $\omega$, there is a finite-to-one function $h \colon \omega \to \omega$ such that $\{ h[x] \mid x \in F\}$ is an ultrafilter.

Blass and Laflamme showed [@BlLa] that the Filter Dichotomy holds in models previously considered by Miller [@Mi] and Blass and Shelah [@BlSh87; @BlSh89].

\[umeasmeager\] The Filter Dichotomy implies that universally measurable uniform filters on $\omega$ are meager.

Let $F$ be a nonmeager universally measurable uniform filter on $\omega$, and let $h \colon \omega \to \omega$ be finite-to-one such that $\{ h[x] \mid x \in F\}$ is an ultrafilter. Let $$S = \{ \bigcup_{n \in Z}h^{-1}[\{n\}] \mid Z \subseteq \omega\},$$ and let $G \colon \mathcal{P}(\omega) \to \mathcal{P}(\omega)$ be defined by $G(x) = h[x]$. Then:

1.  $S$ is a perfect subset of $\mathcal{P}(\omega)$.

2.  $F \cap S$ is a universally measurable subset of $S$.

3.  $G {\mbox{\logic\char'026}}S \colon S \to \mathcal{P}(\omega)$ is a homeomorphism.

4.  $G[F \cap S] = G[F]$ is not Lebesgue measurable.

This gives a contradiction, by Remark \[immeas\].

The first and third items above are easy, and the fourth follows from the fact that nonprincipal ultrafilters are not Lebesgue measurable ([@Si]; to see that $G[F]$ has to be nonprincipal, note that $F$ is uniform and $h$ is finite-to-one). To see the second item above, suppose that $\mu$ is a finite Borel measure on $S$. Define a measure $\mu^{*}$ on $\mathcal{P}(\omega)$ by letting $\mu^{*}(A) = \mu(A \cap S)$ for all Borel $A \subseteq \mathcal{P}(\omega)$. Since $F$ is universally measurable, there exist Borel sets $B$ and $N$ such that $F \bigtriangleup B \subseteq N$ and $\mu^{*}(N) = 0$. Then $\mu(N \cap S) = 0$, and $(F \cap S) \bigtriangleup (B \cap S) \subseteq (N \cap S)$.

Putting together the Blass-Laflamme result with Lemmas \[medcc\] and \[ccnb\] and Theorem \[umeasmeager\], we have the following.

\[nomed\] If ZFC is consistent, then so is ZFC + “there exist no medial limits."

Talagrand [@Ta80] proved that a filter on $\omega$ is meager if and only if there exists a finite-to-one function $h \colon
\omega \to \omega$ such that $\{ h[x] \mid x \in F\}$ is the set of cofinite subsets of $\omega$. The Filter Dichotomy can then be restated as: for every uniform filter $F$ on $\omega$ there is a finite-to-one function $h \colon \omega \to \omega$ such that $\{ h[x] \mid x \in F\}$ is either the cofinite filter or a nonprincipal ultrafilter. The condition that $h$ is finite-to-one is used in the proof of Theorem \[umeasmeager\] only to get the image ultrafilter to be nonprincipal, and for this one needs only that $h[x]$ is infinite for all $x \in F$. Blass has pointed out to us that the finite-to-one condition can be relaxed in the other (meager) case as well, via the following argument.

Suppose that $f \colon \mathcal{P}(\omega) \to [0,1]$ is finitely additive for disjoint sets, sends singletons to $0$ and $\omega$ to $1$. Let $F = \{ x \subseteq \omega \mid f(x) = 1\}$, and let $h \colon \omega \to \omega$ be such that $f(h^{-1}[\{n\}]) = 0$ for all $n \in \omega$. Then $\{ h[x] \mid x \in F\}$ is not the cofinite filter on $\omega$.

First note that $y \mapsto f(h^{-1}[y])$ defines a function from $\mathcal{P}(\omega)$ to $[0,1]$ which is finitely additive for disjoint sets, sends singletons to $0$ and $\omega$ to $1$. Furthermore, $\{ h[x] \mid x \in F\} = \{ y \subseteq \omega \mid f(h^{-1}[y]) = 1\}$, so it suffices to consider the case where $h$ is the identity function, and to show that $F$ is not the cofinite filter. Split $\omega$ into two infinite pieces, and let $A_{0}$ be the one with smaller measure with respect to $f$, or either piece in case of a tie. Let $B_{0}$ be the other piece minus its first element. Then $f(A_{0}) \leq 1/2$, $f(B_{0}) \geq 1/2$ and $| \omega \setminus (A_{0} \cup B_{0})| = 1$ (note that finite sets have $f$-measure $0$). Split $A_{0}$ into two infinite pieces, let $A_{1}$ be the smaller one with respect to $f$, and let $B_{1}$ be the larger one minus its first element. Then $f(A_{1}) \leq 1/4$, $f(B_{0} \cup B_{1}) \geq 3/4$ and $| \omega \setminus (A_{1} \cup B_{0} \cup B_{1})| = 2$. Continue in this way, defining infinite sets $A_{n}$ and $B_{n}$ for each $n \in \omega$ such that $f(A_{n}) \leq 2^{-n-1}$, $f(B_{0} \cup \ldots \cup B_{n}) \geq 1 - 2^{-n-1}$ and $|\omega \setminus (A_{n} \cup B_{0} \cup \ldots \cup B_{n})| = n+1$. Then $\bigcup_{n \in \omega}B_{n}$ has $f$-measure $1$, but is not cofinite.

From this it follows that the nonexistence of medial limits follows from the following weak form of the Filter Dichotomy: for every uniform filter $F$ on $\omega$ there exists a function $h \colon \omega \to \omega$ such that $h^{-1}[\omega \setminus \{n\}] \in F$ for all $n \in \omega$ and $\{ h[x] \mid x \in F\}$ is either the cofinite filter or a nonprincipal ultrafilter. Since our proof uses only the universal measurability of the filter of measure $1$ sets, we have that this weak form of the Filter Dichotomy implies that whenever $f$ is a finitely additive measure on $\mathcal{P}(\omega)$ such that $f(\omega) = 1$, the filter $\{ x \subseteq \omega \mid f(x) = 1\}$ is not universally measurable.

Another proof that the Filter Dichotomy implies the nonexistence of medial limits is given by the logical combination of previously known facts. We include the following definitions for completeness, the second of which is Definition 538A of [@F5]. We refer the reader to [@Ke; @F5] for the definition of *probability space*.

Given two filters $F$ and $G$ on $\omega$, $F$ is said to be *Rudin-Keisler* above $G$ ($F \geq_{RK} G$) if there is a a function $h \colon \omega \to \omega$ such that $\{h[x] \mid x \in F\}$ generates the filter $G$.

A filter $\mathcal{F}$ on $\omega$ is said to satisfy the *Fatou property* if for any probability space $(X, \Sigma, \mu)$, if $\langle E_{n} : n \in \omega\rangle$ is a sequence in $\Sigma$, and $X = \bigcup_{A \in \mathcal{F}} \bigcap_{n \in A} E_{n}$, then $\lim_{n \to \mathcal{F}} \mu(E_{n})$ is defined and equal to $1$.

The Filter Dichotomy can be restated as saying that every uniform nonmeager filter on $\omega$ is Rudin-Keisler above a nonprincipal ultrafilter on $\omega$. The nonexistence of medial limits under the Filter Dichotomy then follows from the following facts, where $f$ is a supposed medial limit and $F$ is the filter $\{ x \subseteq \omega \mid f(x) = 1\}$.

-   The filter $F$ is uniform and nonmeager. ([@GoTa])

-   The filter $F$ has the Fatou property. (Proposition 538Rd of [@F5])

-   Whenever $G$ and $H$ are filters on $\omega$ such that $G \geq_{RK} H$, and $G$ has the Fatou property, then $H$ has the Fatou property. (Proposition 538Ob of [@F5])

-   No nonprincipal ultrafilter on $\omega$ has the Fatou property (Exercise 538Xn of [@F5])

The Semifilter Trichotomy
=========================

A *semifilter* on $\omega$ is a proper subset of $\mathcal{P}(\omega)$ which is closed under finite changes and supersets. The proof of Theorem \[umeasmeager\] uses only these properties of the filter $F$, and thus gives the following.

\[threetwosem\] If $F$ is a universally measurable semifilter and $h \colon \omega \to \omega$ is a finite-to-one function, then $\{ h[x] \mid x \in F\}$ is universally measurable.

The following statement was shown by Laflamme [@Laf] to hold in the models of the Filter Dichotomy mentioned above.

The *Semifilter Trichotomy* is the statement that for every semifilter $F$ on $\omega$ there is a finite-to-one function $h \colon \omega \to \omega$ such that $\{ h[x] \mid x \in F\}$ is either the cofinite filter, a nonprincipal ultrafilter or the set of all infinite subsets of $\omega$.

Talagrand’s proof of his result mentioned after Corollary \[nomed\] applies to semifilters, showing that a semifilter $F$ on $\omega$ is meager if and only if there is a finite-to-one $h \colon \omega \to \omega$ such that $\{h[x] \mid x\in F\}$ is the cofinite filter. Blass has pointed out to us that the third case of the Semifilter Trichotomy occurs if and only if the semifilter $F$ is comeager. We include a proof for the sake of completeness.

\[blassco\] Suppose that $F$ is a semifilter on $\omega$. Then $F$ is comeager if and only if there exists a finite-to-one function $h \colon \omega \to \omega$ such that $\{ h[x] \mid x \in F\}$ is the set of all infinite subsets of $\omega$.

In each direction we will use the following characterization : a set $Z \subseteq \mathcal{P}(\omega)$ is comeager if and only if there exist a division of $\omega$ into finite sets $I_{k}$ $(k \in \omega)$ and a set $y \subseteq \omega$ intersecting each $I_{k}$ such that every $x \subseteq \omega$ for which $x \cap I_{k} = y \cap I_{k}$ for infinitely many $k$ is in $F$ (see [@MaPrWe], for instance).

For the forward direction of the theorem, applying the assumption that $F$ is comeager we can let $h$ be the function which maps $I_{k}$ to $k$, where the set of $I_{k}$’s is as above. For the reverse direction, we can let $I_{k} = h^{-1}[\{k\}]$ and let $y = \omega$. Then if $x$ is a subset of $\omega$ with the property that $x \cap I_{k} = y \cap I_{k} = I_{k}$ for infinitely many $k$, there exists a $z \in F$ such that $h[z] = \{ k \in \omega \mid
x \cap I_{k} = I_{k}\}$. Then $z \subseteq x$, which implies that $x \in F$.

Putting together Theorems \[threetwosem\] and \[blassco\] with Talagrand’s characterization of meager semifilters, we have the following.

The Semifilter Trichotomy implies that universally measurable semifilters on $\omega$ have the property of Baire.

It is an open question (asked in [@MaPrWe]) whether consistently all universally measurable sets of reals have the property of Baire.

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Department of Mathematics\
Miami University\
Oxford, Ohio 45056\
United States\
<PRESIDIO_ANONYMIZED_EMAIL_ADDRESS>

[^1]: Supported in part by NSF grant DMS-0801009.