13

Define $f:X\to X\times X$ by $f(x)=(x,x)$. Then $f$ is continuous, and $f^{-1}(A\times A)=A$. If $A\times A$ is Borel, it follows that $A$ is Borel.


10

Observe that $\{x\,:\,f(x) \geq a\} = \bigcap_{n=1}^{\infty} \left\{x \,:\,f(x) \gt a-\frac{1}{n}\right\}$, and the latter is a countable intersection of open sets by definition of lower semicontinuity. Added later: See also this related thread for an argument using a different definition of (upper) semicontinuity.


10

Your idea is sound, but it requires more work. As pointed out, you have only described so far a very small subcollection of the Borel sets. Instead, show that you can associate to each Borel set a code that keeps track of the "history" of its construction starting from basic open sets, and then count the number of such codes. There is a lot of leeway on how ...


9

The answer is no (except in the trivial case $n=1$), by a counting argument. Let $C$ be the closed unit ball centered at the origin, and let $O$ be the open unit ball centered at the origin. Any set $X$ with $$O\subseteq X\subseteq C$$ is convex; this is just because no line can meet the surface of a ball in three points. But since the surface of the ball (...


9

This is an extraordinarily strong property: it is equivalent to $X$ having only finitely many open sets. Indeed, it is easy to see that if $X$ has finitely many open sets, then it has only finitely many Borel sets, and so your condition holds trivially. Conversely, suppose $X$ has infinitely many open sets. Then in particular $X$ has infinitely many Borel ...


8

As shown in this question there are $\mathfrak c$ Borel sets, which can be extended to the fact that there are $\mathfrak c$ Borel sets of positive measure. Each one with positive measure must have continuum many points. Put them in a well order in type $\mathfrak c$, so each one has $\lt \mathfrak c$ predecessors. Take each set in turn and find two ...


6

Any separable infinite-dimensional Banach space is a counterexample, for example $L_1$. The compact sets are meagre so they generate a sub-$\sigma$-algebra of the $\sigma$-algebra of sets that are either meagre or comeagre. But the Borel $\sigma$-algebra contains the unit ball, which is neither meagre nor comeagre.


6

Yes. Because countable choice is enough to prove that every Borel set has a code, and there are only continuum many codes. To see that this is the case, first notice—as you did—that countable choice is enough to ensure that the Borel hierarchy is of height $\omega_1$. Next, by induction we can prove every $\bf\Sigma^0_\alpha$ set has a code, simply by ...


6

This is an excellent question! Note first that, just from cardinal arithmetic considerations, whether we obtain all sets is independent (and therefore so is whether we obtain all Lebesgue measurable sets). The right context to study this question is descriptive set theory, and indeed the problem was considered early on. For example, Sierpiński proved in ...


6

The image of a Jordan curve is a compact set, so that its complement is open. The interior of the Jordan curve is one of the two connected components of that complement, and therefore open as well. Every open set is a Borel set.


5

Let $A$ be a subset of $\mathbb R$ that is not Borel. I'll produce a net of Borel functions on $\mathbb R$ whose limit is the characteristic function of $A$. The index set $I$ for my net is the set of finite subsets of $A$, and the partial ordering of $I$ is the inclusion relation $\subseteq$. This partial order is directed because the union of any two ...


5

Statement (3) is not equivalent to saying that every bounded Borel function is a pointwise limit of a uniformly bounded sequence of continuous functions. Rather, it is equivalent to every bounded Borel function being either (1) a pointwise limit of a uniformly bounded sequence of continuous functions, or (2) a pointwise limit of a uniformly bounded sequence ...


5

We want to show that $$\mathscr{B} (\mathbb{R}^n) \subset \mathscr{A}:= \{A \subset \mathbb{R}^n : f(A) \in \mathscr{B}(\mathbb{R}^n) \}$$ By the definition of Borel $\sigma$-algebra it is enough to show that $\mathscr{A}$ is a $\sigma$-algebra and it contains all open sets. Claim 1: $\mathscr{A}$ is a $\sigma$-algebra. Proof: Indeed $f(\emptyset)=\...


5

The system of sets you've described is the $\omega_1$-Borel hierarchy. I don't know who first introduced it, but Arnie Miller has a bunch of work on it and related topics - see e.g. this article and this set of notes. This is well outside my comfort zone, so let me just say a brief bit about one of the very basic questions (cribbing horribly from page $2$ ...


5

Yes - this is called the perfect set property. The analytic (= continuous image of Borel) sets also have this property; for more complicated sets (e.g. complements of analytic sets), the perfect set question is undecidable from the usual axioms of set theory. While it's trivially true for open sets, already for closed sets it takes some work (the easiest ...


4

One problem with the "general definition" is that a continuous function might not be measurable. There is, for example, a continuous injective function $f$ from $\mathbb R$ to $\mathbb R$ that is a homeomorphism from a closed set $S$ of positive measure to the Cantor set $E$ (which has measure $0$). For any $V \subseteq S$, $V = f^{-1}(f(V))$, but $f(V) \...


4

The answer in general depends on $X$. For instance, if we take $X$ to be the metric space with only one point, then each $\Sigma^0_\alpha$ class is closed under arbitrary union. More generally, in any discrete metric space every set is $\Sigma^0_1$, so each $\Sigma^0_\alpha$ class is closed under arbitrary unions. However, as long as $X$ is sufficiently "...


4

Yes there is a natural and useful construction of $\sigma$-algebras as a "limit". But not as the limit of a sequence, rather the limit of an uncountable collection, indexed by the countable ordinals (something logicians might call an $\omega_1$-sequence, but which the rest of us don't usually call a "sequence", since it's not countable). First, given $A\...


4

The phrase means just this; if $$a_k 10^k + a_{k-1} 10^{k-1} + \dots + a_0 + a_{-1} 10^{-1} + a_{-2} 10^{-2} + \cdots$$ is the decimal expansion of a real number, then the set $$ \{ n : a_n = 5, -\infty < n \leq k \} $$ is an infinite set. Pulled up from the comments below: to elaborate, the given condition means that there is some infinite subsequence ...


4

There are many ways to enumerate the rational numbers. Since you did not specify an enumeration, let me pick one for you. For each positive integer $m$, choose a rational number $q_{2m}$ so that $|q_{2m}-\sqrt2|\lt\frac1{2^{3m}}$. In this way we get a sequence $q_2,q_4,q_6,\dots$ of rational numbers converging to $\sqrt2$. Enumerate the rest of the rational ...


4

Start with a set $A$ of measure zero. Choose some insane subset $B\subseteq A$ (e.g. a Bernstein set if $A$ is closed, or a Vitali set if it makes sense). Then $B$ is Lebesgue measurable, but not analytic. Also, somewhat trivially (relative to the fact that analytic sets are measurable), every coanalytic set is Lebesgue measurable (in particular, a ...


4

One can find proofs in many set theory texts for the stronger result that, for each countable ordinal level of the Borel hierarchy, there exist Borel sets not belonging to that level. Look in book indexes for "universal set". You can also google Borel + "universal set". However, I've found very few published proofs that limit themselves to the finite levels ...


4

Following Nate Eldredge, I'll write "$(a,b)_{\mathbb{Q}}$" for $(a,b)\cap\mathbb{Q}$ (with $a<b$ reals; note that we don't need $a,b\in\mathbb{Q}$ themselves for this to make sense). And I'll reserve "$(a,b)$" for the full interval of real numbers, as usual. Now the key point is that each $(a,b)_\mathbb{Q}$ is countable. This implies that every element $...


4

$G_\delta$ is not the complement of $F_\sigma$. The complement of an $F_\sigma$ set is a $G_\delta$ set.


3

As pointed out by Zestylemonzi, the set of irrational numbers has an infinite measure and does not contain any closed or non-empty open interval because any such interval contains rational numbers.


3

Under the extra condition that $m(\mathbb R)=n(\mathbb R)<\infty$ this problem will evidently not arise. For $k=1,2,\dots$ define measures $n_k,m_k$ by $n_k(A):=n(A\cap(-k,k))$ and $m_k(A):=m(A\cap(-k,k))$ for $A\in B$. Then applying your method we find $n_k=m_k$ for every $k$. If $E\in L$ then $n(E^c)=\lim_{k\to\infty}n_k(E^c)=\lim_{k\to\infty}m_k(E^...


3

For any subset $\Omega'$ it's fairly easy to see that $$\mathcal B(\Omega', \mathcal T') = \{ A \cap \Omega': A \in \mathcal B(\Omega, \mathcal T)\}$$ Proof: the right hand side is clearly a $\sigma$-algebra on $\Omega'$ and it contains all of $\mathcal{T}'$ as all subspace open sets are of the form $O \cap \Omega'$ with $O \in \mathcal{T} \subset \...


3

Well i'd say it depends of the context but one reason that come to my mind is that the borel $\sigma$-algebra is simpler (and smaller) than the Lebesgue $\sigma$-algebra $\mathscr{M}(\mathbf{R})$. For a lot of things the seting of borel functions or borel sigma alegbra is enough for what you want to do, using the Lebesgue sigma algebra would only make the ...


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