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 $X$ $$ \phi(A) = \phi(A \cap E) + \phi(A \cap E^c)$$

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Find the outer measure of $[-1,2]\cup \{3\}$.

As we know, the outer measure $m^*$ of union of two sets $A$ & $B$ is given by $m^*(A\cup B)=m^*(A)+m^*(B)-m^*(A\cap B)$. If $A$ and $B$ are disjoint then $m^*(A\cap B)=0$. Therefore, $m^*(A\cup B)=m^*(A)+m^*(B)$. Here, $[-1,2]$ and {3} are…
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Continuous image of set having finite Lebesgue outer measure .

Let $f:\Bbb R\rightarrow\Bbb R$ be a continuous function and $E\subseteq\Bbb R$ having finite outer measure. Then can I say that image of $E$ also has finite outer measure ? If I take linear maps it works . In general I have no idea . Please…
neelkanth
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Rearranged sequences to prove countable subadditivity of outer measure

I'm working through the countable subadditivity of outer measure proof in Axler (Measure, Integration & Real Analysis), and I'm finding it difficult to understand the purpose or intuition of the rearrangement of sequences. Suppose $A_1$, $A_2$,…
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Outer measure and g_delta set

I think that my logic is not wrong, but I can't prove this problem completely. How can I correct my solution if my logic is wrong? If there is nothing wrong, how can I complete proof? Oh, Korean word [풀이] was not edited... 풀이 means solution.
Mvaldi
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doesn't exist a closed subset $F$ of $[0,1]$ such that $F\subset\mathbb{R}\setminus\mathbb{Q}$, and $|F|=1$ Axler Measure, Integration & Real Analysis

I am reading "Measure, Integration & Real Analysis" by Sheldon Axler. The following exercise is Exercise 13 on p.24 in Exercises 2A in this book. Suppose $\epsilon>0$. Prove that there exists a subset $F$ of $[0,1]$ such that $F$ is closed, every…
tchappy ha
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Determine the outer measure induced by a given set function (which counts the number of integers in a subset of $\mathbb{R}$)

On the collection $\mathcal{S}$ of all subsets of $\mathbb{R}$, define the set function $\mu: \mathcal{S} \rightarrow \mathbb{R}$ by setting $\mu(A)$ to be the number of integers in $A$. Determine the outer measure induced by $\mu$ and the…
cxxu96
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The outer measure of 2 disjoint bounded sets

When $d(A,B)>0$, we have $$m^*(A)+m^*(B)=m^*(A\cup B)$$I want to examine whether the identity holds if we only have $A\cap B=\varnothing$. My attempt: The proof of the origin theorem requires a $\delta$-distance to separate these 2 sets. So the only…
Isomorphism
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Prove that there exists a bounded set $A$ such that $m^*(F)\le m^*(A)-1$ for every closed set $F\subset A$

Prove that there exists a bounded set $A\subset\mathbb{R}$ such that $m^*(F)\le m^*(A)-1$ for every closed set $F\subset A$. Here $m^*$ is the outer mesure. I used the bounded and non measurable Vitali set $V$, and defined $A=tV$, where $t>0$, but…
AlephZero
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Outer Measure: Theorem) for an interval $I$, $|I|_e=v(I)$?

I do not understand a theorem. So can someone explain for me? Definition closed n-dimensional intervals are $$I=\{\mathbf{x}:a_j\le x_j\le b_j,~~j=1, \cdots, n\}$$ their volumes are $$v(I)=\prod_{j=1}^n(b_j-a_j)$$ A countable collection $S$ of…
Danny_Kim
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