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3 votes
2 answers
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An uncountable measurable subset of $\Bbb R$ containing no nonempty perfect set

$\newcommand\R{\Bbb R}$Assuming the axiom of choice, is there an uncountable Lebesgue-measurable subset $S$ of $\R$ that contains no nonempty perfect set? Of course, such a set $S$, if it exists, ...
Iosif Pinelis's user avatar
3 votes
0 answers
93 views

What axioms are needed to show that the range of a finitely additive diffuse measure on $\mathbb N$ is not closed?

The other day I learned of a small error in the book Theory of Charges: A Study of Finitely Additive Measures. Example 11.4.1 goes as follows. Let $\mu_0$ be a finitely additive probability measure ...
aduh's user avatar
  • 869
11 votes
1 answer
500 views

Uncountable families of measurable sets with pairwise positive intersections

Let $(X,\mathcal{B},\mu)$ be an arbitrary finitely additive probability measure space, let $a>0$ and let $(A_i)_{i\in I}$ be an uncountable family of subsets with measure $\geq a$. Is there an ...
Saúl RM's user avatar
  • 10.6k
5 votes
1 answer
620 views

Non-atomic probability measures on N

One can intuitively imagine picking a random natural number and ask to what extent the intuition can be axiomatized. Using the axiom of choice, there is a total finitely additive (monotonic) averaging ...
Dmytro Taranovsky's user avatar
3 votes
0 answers
90 views

Existence of symmetric total measures

Is it consistent that there is a total finitely additive measure $μ$ on $ℝ$ extending the Lebesgue measure such that for every Borel Lebesgue-measure-preserving bijection $f$ of $ℝ$, $∀α∈Ord \, ∀s∈Ord^...
Dmytro Taranovsky's user avatar
12 votes
2 answers
622 views

Countable set meeting uncountable family of positive measure sets

Suppose $\mu:\mathcal{P}([0, 1]) \to [0.1]$ is a probability measure and $\{A_i: i < \omega_1\}$ is a family of subsets of $[0, 1]$ such that $\mu(A_i) \geq 1/2$ for every $i < \omega_1$. Can we ...
Adam's user avatar
  • 129
3 votes
0 answers
117 views

Which sigma-ideals in a sigma-algebra are contained in an ideal of null sets?

Let $X$ be a Polish space and $\mathcal{B}(X)$ be the $\sigma$-algebra of Borel subsets of $X$. Given a Borel probability measure $\mu$ on $X$, we write $\mathcal{N}(\mu) := \{ B \in \mathcal{B}(X) : \...
Stefan Schrott's user avatar
4 votes
0 answers
198 views

When a null uncountable set can be image of some increasing function with discontinuities on a dense countable set

Consider the following result: A: Let $f:D \to \mathbb R$ be an increasing function with discontinuities on a dense countable subset of $D$ such that the jump values sum to $\mu(D)$, where $D$ is a ...
Amir's user avatar
  • 303
4 votes
1 answer
204 views

How probability-rich is the $\sigma$-algebra generated by a sequence of sets? (Sierpiński's theorem on non-atomic measures without using the AoC.)

$\newcommand\F{\mathcal F}\newcommand\si{\sigma}\newcommand\Om{\Omega}\newcommand\ep{\varepsilon}$Let $p\in(0,1)$ and let $(\Om,\F,P)$ be a probability space. Let $(A_n)$ be a sequence in $\F$ such ...
Iosif Pinelis's user avatar
1 vote
1 answer
112 views

Smallest ensemble of sets stable by any intersections and finite union

Let $E$ be a set, and let $S$ be any set of subsets of $E$, such that $S$ contains the empty set. Can you identify the smallest set of subsets $T$ of $E$ such that $T$ contains all elements of $S$, ...
Guest2024bis's user avatar
6 votes
1 answer
290 views

Subset of the reals with zero inner measure and "full" outer measure in $\mathsf{ZF}+\mathsf{DC}$

Working in $\mathsf{ZF}+\mathsf{DC}$ (that is, we are allowed to use Dependent Choice but not full choice), suppose that there exists a non-measurable subset of the unit interval $[0,1]$ (just non-...
David Fernandez-Breton's user avatar
8 votes
1 answer
1k views

Worst of both worlds?

It's well known that $\mathsf{AC}$ implies the existence of non-measurable sets. And it's also true that, if all sets are measurable, then $|\mathbb{R}/\mathbb{Q}| > |\mathbb{R}|$. But is there a ...
Zemyla's user avatar
  • 309
14 votes
3 answers
2k views

Every function on reals a sum of two surjective real functions?

From this question, and the answer thereof, we can see that every real valued function on reals is a sum of two injective functions. Is the same true if we replace injectivity by surjectivity. For ...
vidyarthi's user avatar
  • 2,089
7 votes
1 answer
650 views

What can be the measure of a Vitali set?

Suppose the continuum $\mathfrak{c}$ is real-valued measurable, i.e., there exists a countably additive probabilistic measure on $\mathfrak{c}$ that measures all subsets. Then by the construction on p....
new account's user avatar
8 votes
3 answers
696 views

Is "the purely probabilistic version of Freiling's axiom of symmetry" disprovable in ZFC?

I'm trying to pinpoint the "intuitive argument" for Freiling's Axiom of Symmetry. It's meant to be a "probabilistic" argument, so thinking about what seems to me to be the ...
Julian Newman's user avatar
25 votes
2 answers
2k views

Writing a function on $\mathbb{R}$ as a sum of two injections

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function. It is well-known that, using transfinite recursion with a well-ordering of $\mathbb{R}$, one can construct two injective functions $g,h: \...
Burak's user avatar
  • 4,265
2 votes
2 answers
227 views

Weak convergence of measures on continuous function spaces

Let $S$ be the unit sphere of $C[0,1], \|\cdot\|_{\infty})$, let $(B_{t})_{t}$ the brownian motion. I would like to show that the measure $\mu_r$ defined on $\mathbb{B}(S)$ by $\mu_r(A):=P\Big(\frac{...
Paul's user avatar
  • 21
1 vote
1 answer
258 views

What is the measure of two sets which partition the reals into subsets of positive measure?

This is a follow up to this question, where I wish to partition the reals into two sets $A$ and $B$ that are dense (with positive measure) in every non-empty sub-interval $(a,b)$ of $\mathbb{R}$. (In ...
Arbuja's user avatar
  • 63
1 vote
0 answers
154 views

Polish spaces and analytic sets

Can we conclude that an analytic subset $A$ of a Polish space $X$ is also Polish? Let $\mathcal{M}(R^d)$ denotes the family of Borel probability measures on $R^d$ equipped with the Lévy-Prokhorov ...
B-S's user avatar
  • 39
10 votes
3 answers
1k views

How to prove that the Lebesgue $\sigma$-Algebra is not countably generated?

How to prove that there can't exist a countable set $\{A_1,A_2,\dots\}\subset \mathcal{L}(\mathbb{R})$ (where $\mathcal{L}(\mathbb{R})$ denotes the family of all Lebesgue measurable sets) such that $\...
Joris Wk's user avatar
  • 243
10 votes
1 answer
1k views

Within ZFC, is $2^{\aleph_0}<2^{\aleph_1}$ provable/independent?

So, I ask whether from the ZFC axioms one can prove X that every uncountable set has strictly more than continuum many subsets, or whether X is independent of the ZFC axioms. Note that (within ZFC) ...
TaQ's user avatar
  • 3,584
7 votes
1 answer
381 views

Consistency of a strong Fubini type theorem for measure zero sets

Is the following statement (†) consistent with ZFC? If $E \subseteq [0,1]^2$ is such that $E_x := \{y\in[0,1] : (x,y)\in E\}$ has measure zero for almost all $x$, then $E^y := \{x\in[0,1] : (x,y)\in ...
Gro-Tsen's user avatar
  • 32.5k
5 votes
0 answers
216 views

Applications of Baire's theorem on functions of first class

I found the following theorem on page 32 of John Oxtoby's Measure and Category. Theorem 7.3. If $f$ can be represented as the limit of an everywhere convergent sequence of continuous functions, then $...
i like math's user avatar
8 votes
2 answers
960 views

Is there a measure theory for proper classes?

This question is naive, but I didn't get an answer at MSE: Is it straightforward to extend measure theory to proper classes? Of course when one tries to define measures on "large sets" ...
aduh's user avatar
  • 869
4 votes
1 answer
275 views

Supremum of infimum of measure of members of a free ultrafilter

For a set $A\subseteq \omega$ we let the upper density of $A$ be defined as $d^+(A) := \lim\sup_{n\to\infty}\frac{|A\cap(n+1)|}{n+1}$. Let $\text{FrU}(\omega)$ be the collection of free ultrafilters ...
Dominic van der Zypen's user avatar
2 votes
1 answer
197 views

Large cardinals and measurability in $L(A)$

Under suitable large-cardinal assumptions, in the inner model $L(\mathbb R)$ one can have $\omega_1$ and $\omega_2$ measurable (this follows from determinacy). I was wondering if it is possible to ...
Curious_poster's user avatar
6 votes
1 answer
191 views

Steinhaus number of a group

$\newcommand\Sn{\mathit{Sn}}$A subset $A$ of a group $X$ is called algebraic if $A=\{x\in X: a_0xa_1x\dotsm xa_n=1\}$ for some elements $a_0,a_1,\dotsc,a_n\in X$. Let $\mathcal A_X$ be the family of ...
Taras Banakh's user avatar
  • 41.8k
4 votes
1 answer
351 views

$\sigma$-algebra generated by analytic sets

The Borel $\sigma$-algebra $\cal B$ on real numbers has many good properties. For instance, all continuous functions are $\cal B/\cal B$-measurable. On the other side, not only $\cal B$ is not ...
Giafazio's user avatar
  • 205
4 votes
1 answer
259 views

Reference request: large generalized probability measures

I'm interested in references relevant to the following: what is the right generalization, if there is one, of a probability measure that takes on values in an structure of more than continuum size? I'...
Beau Madison Mount's user avatar
15 votes
2 answers
2k views

Is "There exists an unbounded non-measurable set but no bounded non-measurable set" consistent with $\mathsf{ZF}$?

This is a follow-up to this question. We say that a set $A \subseteq \mathbb{R}$ is bounded if there exists a finite interval $(a,b)$ such that $A \subseteq (a,b)$. Working in $\mathsf{ZFC}$, the ...
Clement Yung's user avatar
  • 1,372
7 votes
1 answer
393 views

Models with fixed cardinality of non-Lebesgue measurable sets

In the usual $\mathsf{ZFC}$, we know that there are $2^\mathfrak{c}$ many subsets of $\mathbb{R}$ that are not (Lebesgue) measurable. On the other hand, the Solovay model also provides us a model of $\...
Clement Yung's user avatar
  • 1,372
6 votes
2 answers
276 views

Extending contents

Suppose $\mu$ is a finitely additive measure on $X$ (aka “content”) with $\mu(X) < \infty$, defined on an algebra of sets $\mathcal A$. Let $$\mu^*(Y) = \inf \{ \mu(E) : E \in \mathcal A \wedge E \...
Monroe Eskew's user avatar
  • 18.6k
2 votes
0 answers
320 views

Showing that the procedure to generate an algebra does not generate a $\sigma$-algebra

A while ago I asked this question on mathstackexchange: Let $\mathscr{C}\subset \mathscr{P}(\Omega)$ be a class of subsets of a nonempty set $\Omega$ containing $\Omega$ and $\varnothing$. Define $\...
Eduardo's user avatar
  • 757
0 votes
1 answer
169 views

Haar measure on ${\cal P}(\omega)$

First, we note that there is a natural bijection ${\cal P}(\omega) \to \{0,1\}^\omega$ and endow the latter with the product topology (where $\{0,1\}$ carries the discrete topology). So we get a ...
Dominic van der Zypen's user avatar
4 votes
1 answer
746 views

Can all uncountable (but small) families of sets with positive measure have an uncountable subfamily with an intersection of positive measure?

My general question was is it consistent that any uncountable family of less than $\mathrm{non}(\mathcal{N})$ sets, each with positive measure, has an uncountable subfamily $\mathcal{F}$ s.t. $\bigcap ...
mtg's user avatar
  • 135
1 vote
0 answers
155 views

$f:Y\to X$ continuous with $f^{-1}(x)$ compact for $x\in X$, does there exist a Borel measurable map $g:X\to Y$?

Let $X,Y$ be Polish, metric spaces. $f:Y\to X$ is a continuous, surjective map and for any $x\in X$, $f^{-1}(x)\subset Y$ is compact. Is it true that there is a injective, Borel measurable map $g:X \...
mathmetricgeometry's user avatar
3 votes
1 answer
331 views

Loeb measures and non-standard hull of Banach spaces

$\DeclareMathOperator\Fin{Fin}$I am trying to understand the notion of "liftings" of Loeb measurable functions to internal, internally measurable functions, and its connection to non-...
BharatRam's user avatar
  • 949
4 votes
1 answer
718 views

Is every element of $\omega_1$ the rank of some Borel set?

It is well known that we can obtain the $\sigma$-algebra of Borel subsets of $2^{\omega}$ in the following way: Let $B_0$ be the collection of all open subsets of $2^{\omega}$. For $\alpha=\beta+1$, ...
Hannes Jakob's user avatar
  • 1,799
15 votes
2 answers
530 views

Nontrivial signed measure on Lebesgue measurable sets being trivial on Borel sets

Let $\mathfrak{L}(\mathbb{R})$ be the collection of Lebesgue measurable sets and $\mathfrak{B}(\mathbb{R})$ be the Borel sets. Question: Is there a nontrivial signed measure on $\mathfrak{L}(\mathbb{R}...
Zhang Yuhan's user avatar
14 votes
1 answer
918 views

Was Cantor aware of Lebesgue theory of integration?

Georg Cantor died in 1919, more than ten years after appearance of the Lebesgue theory of measure and integration at the beginning of the twentieth century. Lebesgue theory has a deep connection with ...
XIII's user avatar
  • 747
5 votes
1 answer
314 views

Do Borel subsets of the plane with null sections have Borel projections?

This might be a very easy question, and it might be better for mathstackexchange in which case I apologize. I'm stuck on something an anonymous referee wrote to me about a paper of mine and I'm hoping ...
Corey Bacal Switzer's user avatar
8 votes
4 answers
775 views

Self-contained formalization of random variables?

I have not been able to find any formalization of random variables that supports construction of new random variables dependent on previously constructed ones. In what I have found, a random variable $...
user21820's user avatar
  • 2,912
6 votes
2 answers
1k views

Foundational results dependent on/equivalent to the continuum hypothesis or its negation?

I remember at a certain point early in my mathematical studies learning that the Axiom of Choice is equivalent to the following statement on Cartesian products: If $\{ X_i \}_{i \in I}$ is any ...
Rivers McForge's user avatar
9 votes
1 answer
831 views

Baire category theorem for uncountable unions

Any compact Hausdorff space $X$ is a Baire space: if the set $X$ is a meager set (meaning a countable union of nowhere dense subsets, also known as a set of first category), then $X$ is empty. I am ...
Dmitri Pavlov's user avatar
1 vote
0 answers
150 views

Is there difference in notion of measurability in classical versus constructive?

Are there notions of measure and examples of sets measurable in that measure in classical logic but not in constructive logic (I think there cannot be counterexamples in other direction)? Are there ...
VS.'s user avatar
  • 1,826
2 votes
1 answer
156 views

Atomless, c-additive measures in ZFC

This is a follow-up question to this one. Is there a ZFC example of an atomless measure that is $2^\omega$-additive, meaning, fewer than continuum many null sets have measurable union that is null?
Kant's user avatar
  • 55
3 votes
1 answer
207 views

Finitely additive, $\kappa$-additive atomless measures in ZFC

Under Martin's Axiom (and non-CH) the Lebesgue measure is $2^\omega$-additive in the sense that unions of fewer than continuum ($2^\omega$) many null sets are measureable and null. In ZFC we may ...
Kant's user avatar
  • 55
5 votes
1 answer
330 views

Function whose graph is a Borel relation

Suppose $f\colon\mathbb{R}^{\omega}\longrightarrow\mathbb{R}$ is a function such that $$G(f):=\{(x,y)\in\mathbb{R}^{\omega}\times\mathbb{R}\mid f(x)=y\}$$ is a Borel set. Does it necessarily follow ...
Hannes Jakob's user avatar
  • 1,799
1 vote
1 answer
132 views

Lambda system generated by a non-atomic collection

Consider a probability space $(X,\Sigma,P)$. Let say that a collection $\mathcal{B}\subseteq\Sigma$ is non-atomic if for every $E\in\mathcal{B}$ and $\alpha\in(0,P(E))$ there exists $F\in\mathcal{B}$ ...
Johann's user avatar
  • 11
2 votes
1 answer
348 views

On the hereditary Lindelof topological spaces

I received the following interesting point in (1). I could not find any reference or clear proof. Any suggestion? Theorem. A topological space $X$ is hereditary Lindelof if and only if for any ...
ABB's user avatar
  • 4,058

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