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Questions tagged [outer-measure]

Outer measure on $X$ is a function $ \phi : 2^X \rightarrow [0,\infty]$ defined on all subsets on $X$ that satisfies the following : (1) $\phi ( \emptyset )=0$ (2) Monotonicity : $A\subset B$ implies $\phi (A)\leq\phi (B)$ and (3) Countable subadditivity : $\phi (\bigcup_{i=1}^\infty A_j) \leq \sum_{j=1}^\infty \phi (A_j) $

Outer measure on $X$ is a function $ \phi : 2^X \rightarrow [0,\infty]$ defined on all subsets on $X$ that satisfies the following :

  1. $\phi ( \emptyset )=0$

  2. Monotonicity : $A\subset B$ implies $\phi (A)\leq\phi (B)$

  3. Countable subadditivity : $\phi (\bigcup_{i=1}^\infty A_j) \leq \sum_{j=1}^\infty \phi (A_j) $

This allows us to define the concept of measurability as follows : a subset $E$ of $X$ is $\phi$-measurable iff for every subset $A$ of $X$ $$ \phi(A) = \phi(A \cap E) + \phi(A \cap E^c)$$

587 questions
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Definitions of Measurability: Outer-inner measure convergence vs. Caratheodory criterion

If we are looking over subsets of $\mathbb R$ and considering the outer measure defined exactly as $$\mu^*(A) = \inf\left\{ \sum_{k=1}^\infty \ell(I_k) \text{ where the $I_k$ are open intervals such that } A\subset \bigcup_{k=1}^\infty I_k…
D.R.
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Why is the inner measure problematic?

Currently the Lebesgue measure is defined by the outer measure $\lambda^*(A)$ by the criterion of Carathéodory: A set $A$ is Lebesgue measurable iff for every set $B$ we have $\lambda^*(B)=\lambda^*(B\cap A) + \lambda^*(B\cap A^C)$. However, before…
Stephan Kulla
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Problems with Set Function

(Questions are at the bottom of the post. I’ve added two more doubts since the recent answer.) Introduction Consider $f:A\to\mathbb{R}$ where $A\subseteq[a,b]$, $a,b \in \mathbb{R}$ and $S$ is a fixed subset of $A$. Before mentioning my set…
Arbuja
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Constructing an outer measure on a set whose measurable sets are exactly a given sigma algebra on the set.

Consider an arbitrary set $X$ and an arbitrary $\sigma$-algebra $\mathcal{M}$ on $X$. My question is that can one construct an outer measure on the set $X$ whose measurable sets is exactly the collection $\mathcal{M}$. I tried to find an answer…
Satwata Hans
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On the abscence of the Inner Measure in introductory texts on Measure Theory

Letting $\mu$ and $\mu^*$ be the Lebesgue and outer Lebesgue measure repsectively, the inner Lebesgue measure can defined as $$\mu_*:S\mapsto \sup\left\{\mu(K) : \text{$K$ is a compact subset of $S$}\right\}$$ or, on bounded sets, as $$S\mapsto…
Sam
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Proving that the outer measure of a closed interval $[a,b]$ is $b-a$

In Sheldon Axler's book, Measure Integration, and Real Analysis, he defines outer measure of a set as $|A| = \inf\big\{\sum_{k=1}^\infty \ell(I_k): I_1, I_2, \dots \text{are open intervals such that} A\subset \bigcup_{k=1}^\infty I_k\big\}$, where…
Subhasish Mukherjee
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Do we need intervals to define the Lebesgue measure?

The Lebesgue measure is conventionally defined as $$\mu(X) = \inf\{\sum_{n \in \mathbb{N}}(b_n-a_n) | X \subseteq \bigcup_{n \in \mathbb{N}}(a_n,b_n) \}$$ Which can be thought of intuitively as covering the set with translated and stretched copies…
Zoe Allen
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Prove that $\mu^* (F \setminus E) = 0.$

Let $\mathcal A$ be an algebra of subsets of a set $X$ and $\mathcal S (\mathcal A)$ be the $\sigma$-algebra of subsets of $X$ generated by $\mathcal A.$ Let $\mu : \mathcal A \longrightarrow [0,+ \infty]$ be a measure on $\mathcal A.$ Let $\mu^*$…
math maniac.
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Is there a nonmeasurable subset of $\mathbb{R}^2$ that is $1-$dimensional Hausdorff measurable?

For $n\in\mathbb{N}^*$ and $s\in\mathbb{R}_{\ge 0}$, the $s-$dimensional Hausdorff measure $H^s$ is an outer measure over $\mathbb{R}^n$, and the $\sigma-$algebra of $s-$dimensional measurable subsets of $\mathbb{R}^n$ is given by Carathéodory's…
Jianing Song
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Are there other important measure spaces which are not obtained directly from an outer measure or from the Caratheodory extension theorem?

I have just started to study measure theory and I have a question. But before presenting it, I will provide the context from which it comes. Given a nonempty set $\Omega$, we say that a set function $\mu$ defined on a algebra…
user0102
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Is the Carathéodory measurability criterion optimal in some sense?

If $m$ is an outer measure defined on a set $X$, we say that a subset $E$ of $X$ is Carathéodory-measurable with respect to $m$ if for all subsets $A$ of $X$, we have $m(A)=m(A\cap E) + m(A\cap E^c)$. And if $M$ is the set of all…
Keshav Srinivasan
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Completion and outer-measure extension of sigma finite measure

I need some help with the proof of the following. Let $(X,\mathcal{A},\mu)$ be a measure space with $\mu$ being $\sigma$-finite, $\mu^*$ be the outer measure given by the formula $\mu^*(E)=\inf\{\sum_{n}\mu(A_n): E\subset\bigcup A_n,…
Just dropped in
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A descending sequence of sets with outer measure 1

Show that there is a sequence of sets $A_n\subseteq [0,1]$, with outer measure 1 ($\mu^*(A_n)=1$ for every $n$), so that $A_1\supseteq A_2\supseteq A_3\supseteq...$ and $\bigcap_{n=1}^{\infty}A_n=\varnothing$. From these conditions, the sets mustn't…
35T41
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If the measure of union = sum of outer measures, then the sets are measurable

I am trying to show that for $A$, $B \; \subset \mathbb{R}$ $A \cup B$ is Lebesgue measurable such that \begin{align} \infty > \mu(A \cup B) = \mu^{*}(A) + \mu^{*}(B), \end{align} Then $A$ and $B$ are Lebesgue measurable. Thus far, I have tried…
DRich
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Prove that $\mu_*\leq\mu^*$ where the definition of an inner and outer measure is induced by a measure

I know that similar question has been asked and answered here, here, and here. But I am looking for a different proof based on different definitions. We have the following definition of an outer and inner measure indeuced by a…
Beerus
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