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13 votes
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Proving Caratheodory measurability if and only if the measure of a set summed with the measure of its complement is the measure of the whole space.

Step 0: The inequality $\mu^{\ast}(A) \leqslant \mu^{\ast}(A\cap E) + \mu^{\ast}(A \cap E^C)$ for all $A \subset X$ follows directly from the subadditivity of outer measures. Step 1: For a ...
Daniel Fischer's user avatar
6 votes
Accepted

How does the frequentist definition of probability work with non-measurable sets?

Initial answer: I don't have a rigorous answer for you, but let me point out that every subset of $[0,1]$ (measurable or not) has a rigorously defined inner and outer Lebesgue measure, both of which ...
pre-kidney's user avatar
  • 30.3k
6 votes
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Let $E_1 \subset E_2$ both be compact and $m(E_1) = a, m(E_2) = b$. Prove there exists a compact set $E$ st $m(E) = c$ where $a < c < b$.

Let $E_1, E_2$ be two compact sets in $\mathbb R^d$. Let $f : \mathbb R \to \mathbb R$ be defined by $$ f(t) = m(E_1 \cup (E_2 \cap \{ x_1+ \cdots + x_d \le t\})),$$ where $x_1, \cdots, x_d$ are the ...
Arctic Char's user avatar
  • 16.2k
5 votes
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Proving the *Caratheodory Criterion* for *Lebesgue Measurability*

Note: I just realized that your statement of Carathéodory's criterion doesn't agree with the usual one, where one tests against any possible test $A$. If we show that it is possible to test using open ...
Miguel's user avatar
  • 1,644
5 votes
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Prove that $f\in L^1(A)\Leftrightarrow \sum_{n}^{\infty}m(\{ x\in A : f(x)\geq n \}) < \infty$

As stated in the comment you have that $$ f^{-1}([n,\infty ))= \bigcup_{k\geqslant n}f^{-1}([k,k+1)) $$ Therefore $$ \begin{align*} \sum_{n\geqslant 0}m(f^{-1}([n,\infty ))&= \sum_{n\geqslant 0}m\...
Masacroso's user avatar
  • 30.7k
5 votes
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Let $f$ be measurable then prove a certain set is measurable.

we have that the function $\phi:\mathbb{R}^2\to \mathbb{R}$, where $\phi(x,y)=x-y$ is measurable since it is continuous. So is the function $F(x,y)=(f(x),f(y))$ (from $\mathbb{R}^2\to \mathbb{R}^2$), ...
alphaomega's user avatar
  • 2,236
5 votes
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Is it possible to write a metric a space as a countable disjoint union of compact sets?

The key assumption here is that $\mu$ is a Radon measure, meaning it is inner regular with respect to compact sets. Without this assumption, this is not true, not even if $\mu$ is finite (for instance,...
tomasz's user avatar
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5 votes
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Vitali set proof: Why is $1 \leq \sum_{k \in \mathbb{N}}m(\mathcal{N}_k) \leq 3$ impossible?

Since $m(\mathcal N_k)=m(\mathcal N)$ for each $k$, both cases lead to a contradiction: If $m(\mathcal N)=0$, then the infinite sum equals $0$ as well, yielding $1\leq 0\leq 3$, which is absurd. If $...
Zuy's user avatar
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5 votes

commonness of non measurable sets

This answer tries to walk a fine line - I think this is a very natural and simple-to-pose question which nonetheless benefits from quite technical material, and that makes exposition difficult. I've ...
Noah Schweber's user avatar
4 votes
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Can an integral defined by a density with respect to folner sequences of $A$ exist on a function defined on subsets of $A$?

I will try and give a high level answer. Given a Folner sequence $(F_n)$ on a countable group $G$ and a subset $A\subset G$, we can try and define how big A is "according to the Folner sequence" as ...
pseudocydonia's user avatar
4 votes

Measurable functions : $f(A) \in \mathcal{B}$

Inverse images behave much more nicely than direct images. Indeed, one has the identities $$ f^{-1}(B\cap C)=f^{-1}(B)\cap f^{-1}(C) $$ and $$ f^{-1}(B^c)=(f^{-1}(B))^c $$ where ${}^c$ is the ...
Reveillark's user avatar
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4 votes
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The function that generates a measurable graph is measurable

The mapping $$X \ni x \mapsto h(x):=(x,a) \in X \times (0,\infty)$$ is measurable for any fixed $a>0$. Thus, $h^{-1}(U_f) \in \mathcal{S}$. Noting that $$h^{-1}(U_f) = \{x \in X; a<f(x)\}=f^{-1}...
saz's user avatar
  • 121k
4 votes
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Show that $\mathcal A_1$ $\cap$ $\mathcal A_2$ is also a $\sigma$-algebra

As mentioned by @Reveillark, you shall need the definition of intersection of sets. Based on it, we can proceed as follows. Let $\Omega$ be a nonempty set and $\mathcal{A}_{1}$ and $\mathcal{A}_{2}$ ...
user0102's user avatar
  • 21.6k
4 votes
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Borel algebra always strictly smaller than its completion?

I'm going to stick with the Lebesgue-Stieltjes part of the question. Hopefully Folland had something simpler in mind, but here's one way to show that the domain of the completion has cardinality $2^{\...
Michael Jesurum's user avatar
4 votes
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Proving that a trivial product sigma algebra is the product of sigma algebras

Your argument for the one inclusion is correct. Let me propose an argument proving both inclusions at once, which might seem a bit abstract, but highlights what's going on. Some details are left for ...
Thorgott's user avatar
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4 votes
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Are nonmeasurable sets whose sections are null always contained in a null measurable set?

No. For instance, let us consider Lebesgue measure on $\mathbb{R}^2$. It can be shown that if $A\subseteq\mathbb{R}^2$ is measurable with positive measure, then there are $2^{\aleph_0}$ different ...
Eric Wofsey's user avatar
3 votes

Why is $\{\{c\}:c \in \mathbb{R} \}$ not a generator of the Borel $\sigma$ field in $\mathbb{R}$?

The set $\mathcal{A}$ of all subsets of $\Bbb R$ that are either countable or have a countable complement, forms a $\sigma$-algebra. It clearly contains $\mathcal{S}:=\{\{c\}: c \in \Bbb R\}$, all ...
Henno Brandsma's user avatar
3 votes
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Let $m^\ast(E)$ < ∞.If for every interval (a, b) we have that $b-a$=$m^\ast((a,b)∩E) + m^\ast((a,b) ∩ E^c)$ then $E$ is lebesgue measurable

Note that the assumption is saying that for any interval $I = (a,b)$ we have that $$m^*(I) = m^*(I \cap E) + m^*( I \cap E^c)$$ Sub additivity gives us that $ m^*(A) \leq m^*(A \cap E) + m^*( A \cap ...
Dionel Jaime's user avatar
  • 3,908
3 votes
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Continuous map from subset of $\mathcal{C}$ (Cantor) to non-measurable set.

You are absolutely correct, for the reason you gave, and the "proof" sketched is false. What we know is that $f(C)$ is a compact set of positive measure. And any set of positive measure contains a ...
zhw.'s user avatar
  • 106k
3 votes
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Show that there exists an open interval $(a,b)$ such that $m(E\cap(a,b)) > \frac{1}{2}m(a,b) > 0$ when $m(E) > 0$

As the above comments point out, your proof is not correct. Here's an approach to the problem. Note that for any given $\epsilon>0$, there exists a finite family of disjoint intervals $I_j=(a_j,...
Myunghyun Song's user avatar
3 votes

Lipshitz function and measurable sets

Let $E $ be measurable. By inner regularity, $$m (E)=\sup\{m (K):\ K\subset E, \ K \text { compact }\}. $$ So we can write $$E=E_0\cup\bigcup_nK_n, $$ an increasing union of compacts, and $E_0$ a null ...
Martin Argerami's user avatar
3 votes
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Why is the set $\{\omega\in \Omega: \exists t \geq 0: -\infty < D^-(t,\omega) \leq D^+(t,\omega) < + \infty\}$ measurable?

Billingsley doesn't assert that $E$ is measurable. He does show that $E$ has outer measure $0$, and so will be measurable provided the measure space $(\Omega,\mathcal F,P)$ is complete. A discussion ...
John Dawkins's user avatar
  • 26.7k
3 votes
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Borel-Cantelli Lemma Proof Verification

We have $E= \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty E_k$ (why?) Hence, $E$ is trivially measurable by the axioms of $\sigma$-algebra. The assumption $\sum_k m(E_k) < \infty$ is not necessary.
J. De Ro's user avatar
  • 21.6k
3 votes
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Let X be a set. How many $\sigma$-algebras of subsets of X contain exactly $5$ elements?

The answer is none. The reason is the following theorem: Theorem. If $\mathcal{F}$ is a $\sigma$-algebra on $X$ that has a finite number of elements, then $\mathrm{card}(X)=2^n$ for some $n$. I'll ...
Alonso Delfín's user avatar
3 votes
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Theorem 0.7. Walters' book of Ergodic Theory

Hint: The proof is rather lengthy, but the idea is simple. Consider the collection of subsets: $$ {\cal C} = \{ S \in {\cal B} \ | \ \forall \epsilon>0, \exists A\in {\cal A} : \mu(S\Delta A)<\...
H. H. Rugh's user avatar
  • 35.4k
3 votes
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union of two spaces

If $x \notin E_1$ and $x \notin E_2$, then $|f(x) - h(x)| \le |f(x) - g(x)| + |g(x) + h(x)| < r + t$, so $x \notin \{y \in X \mid |f(y) - h(y) \ge r+t\}$. This proves $$E_1^c \cap E_2^c \subseteq \{...
angryavian's user avatar
  • 90.9k
3 votes
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Prove that $\Omega$ equals the $\sigma$-algebra $\sigma(\Sigma,S)$

Hint: You have to first verify that $\Omega$ is a $\sigma-$ algebra. Then observe that it contains $\Sigma$ and the set $S$. It follows that it contains $\sigma (\Sigma, S)$.
Kavi Rama Murthy's user avatar
3 votes

Lebesgue measurable function and set, $\forall t\in (0,1), $there exists $ E_t \subset E$ such that $\int_{E_t} f(x) dx = t \int_{E} f(x) dx$.

Since your $f$ can take both positive and negative values on $E$, I'll break up the solution to two steps. Step 1: Reduce to the case where $f$ on its domain of integration is a nonnegative measurable ...
KCd's user avatar
  • 47.5k
3 votes
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Lebesgue measurable function and set, $\forall t\in (0,1), $there exists $ E_t \subset E$ such that $\int_{E_t} f(x) dx = t \int_{E} f(x) dx$.

For your statement to make sense, $f$ must be assumed to be not only measurable but integrable (but $f$ is not necessarily of constant sign, and its integral $A\in\Bbb R$ does not need to be $>0$). ...
Anne Bauval's user avatar
  • 38.8k
3 votes
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Show that $S$ has Lebesgue measure zero.

For $k,m \in \mathbb{N}$ let $$ A_{m,k}:=\{x \in \mathbb{R} \mid \forall n \ge m: ~ |\sin(n!\pi x)| \le \frac{1}{k+1}\}. $$ Each $A_{m,k}$ is closed, hence measurable and $$ S=\bigcap_{k\in \mathbb{N}}...
Gerd's user avatar
  • 7,419

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