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Hot answers tagged outer-measure

12

YES. First construct a "Cantor like set" with positive measure $C$ - otherwise known as fat Cantor set. Then define the product $C\times [0,1]\subset \mathbb R^2$ and finally to make it connected define $$T=C\times [0,1]\cup [0,1]\times\{0\}.$$ What is a Cantor like set? Take $C_0=[0,1]$, then $C_1=[0,1]\setminus [a_1^0,b_1^0]$, then $C_2=C_1\setminus([... 8 The answer "yes" follows from results in the article "On Two Halves Being Two Wholes" by Andrew Simoson in The American Mathematical Monthly Vol. 91, No. 3 (Mar., 1984), pp. 190-193. Some readers might not have access, but here is the JSTOR link: http://www.jstor.org/stable/2322357 The article was cited in the article i707107 linked to in a comment. In ... 4 The family$\mathcal F$of nonempty open subsets of$[0,1]$of measure$< 1$has cardinality$c$, so it can be well-ordered so that each member of$\mathcal F$has fewer than$c$predecessors. Since the complement in$[0,1]$of any member of$\mathcal F$has cardinality$c$, we can choose by transfinite induction, for each$U \in \mathcal F$, a countably ... 4 Actually, it's the other way around:$\mu^*(A)\le \mu(A)$. This is because$A$itself covers$A$. Specifically, let$A_0=A$and$A_i=\emptyset$for$i>0$. Then$A\subseteq\bigcup A_i$. Now it's easy to see that there are two possibilities for$\mu(\emptyset)$, by finite additivity: either$0$or$\infty$. If$\mu(\emptyset)=0$, then$\sum\mu(A_i)=\mu(A)$... 4 Here's a proof by contrapositive: suppose that$C \subseteq \Omega$and there is an$A \in \mathcal A$with$\mu^*(A \cap C) < \mu(A)$. Noting that$C \subset A^c \cup (A \cap C)$, we find that $$\mu^*(C) \leq \mu^*(A \cap C) + \mu^*(A^c) = \mu^*(A \cap C) + \mu(A^c) < \mu(A) + \mu(A^c) = 1$$ So,$\mu^*(C)<1$. We have reached the desired ... 4 If$C$is the Cantor set and$A=B=\{e^{x}:x\in C\}$then$A$and$B$have measure$0$but$AB$contains all numbers of the form$e^{x+y}$with$x,y \in C$. Since$C+C=[0,2]$it follows that$AB$does not have measure$0$. [I have used the following facts:$e^{x}$is absolutely continuous on$[0,1]$and any absolutely continuous function maps sets of measure$...

3

From the article provided by @Jonas Meyer: http://www.jstor.org/stable/2322357 An alternative approach to this problem is possible. This approach is already provided by @Syang Chen at MathOverflow: https://mathoverflow.net/questions/59978/additive-subgroups-of-the-reals/59980 Let $f(x) = m^*(G\cap [0,x])$. Then, for any $x, y \in G \cap (0,\infty)$, we ...

3

Step 1: Let $E \subseteq X$ be an arbitrary set. Show that there exists $B \in S(R)$ such that $E \subseteq B$ and $\mu(B) = \mu^*(E)$. Step 2: Let $E \subseteq X$ be such that $\mu^*(E) < \infty$. Choose $B$ be as in step 1. Then \begin{align*} \mu_*(E) &= \sup \{\mu(A); A \subseteq E, A \in S(R)\} \\ &= \sup\{\mu(A); B \cap A^c \supseteq B \... 3 Generalize the construction of a Bernstein set. Enumerate A=\{x_\xi:\xi<2^\omega\}. Let \mathscr{F}=\{F_\xi:\xi<2^\omega\} be the family of uncountable closed subsets of A. Recursively construct distinct y_{\xi,n}\in A for \langle\xi,n\rangle\in 2^\omega\times\omega as follows. If \eta<2^\omega, and we have y_{\xi,n} for each \... 3 Ok lets look at it step by step f(x)=c_1 Now we need to check the \{x\in X| f(x)>c\} Basically this is the same as \{x\in X| c1>c\} Now we debate depending on variable c For c \ge c_1 We get \{x\in X| c_1>c\ge c_1\}=\emptyset since c_1>c_1 cannot happen. And as we know empty set is in sigma algebra by definition. Now for c <... 3 You can but the result is somewhat trivial. Since every open set is Lebesgue measurable, we have\inf\{m(E)\mid E \text{ is measurable},\; A \subseteq E\} \leq \inf\{m(E)\mid E \text{ is open}, \;A\subseteq E\} = m^*(A).$$This follows because the left infimum has a greater range. By monotonicity, m^{*}(A) \leq m^{*}(E) = m(E) holds for all measurable E... 3 For x\in A, let U_x=\{y\in X:d(y,x)<\frac12 d(A,B)\}. Clearly this set is open, so U:=\bigcup_{x\in A}U_x is open. Obviously A\subseteq U. If y\in U, then there exists x\in A such that$$d(x,y)<\frac12 d(A,B)\le\frac12\inf_{z\in B}d(x,z),$$so in particular y\notin B. Hence$$\mu^*(A\cup B)=\mu^*\big((A\cup B)\cap U\big)+\mu^*\big((A\cup ...

3

For each $m\in\mathbb N$, you know that there is a family $(U_{k,m})_{k\in\mathbb N}$ of open sets in $\mathbb{R}^n$ such that $A\subset\bigcup_{k\in\mathbb N}U_{k,m}$ and that $\sum_{k=1}^\infty l(U_{k,m})<\mu^*(A)+\frac1m$. Define $B_m=\bigcup_{k\in\mathbb N}U_{k,m}$. Then $A\subset B_m$ and$$\mu^*(A)\leqslant\mu(B_m)<\mu^*(A)+\frac1m.$$ Now, define, ...

3

Denote by $(X_n)_{n \in \mathbb{N}} \subseteq \mathcal{A}$ a sequence of increasing sets, $X_n \uparrow X$, such that $\mu(X_n)< \infty$ for all $n \in \mathbb{N}$. You have already shown that $$B \in \mathcal{M} \implies \exists F \in \mathcal{A}, F \supseteq B: \mu(F) = \mu^*(B). \tag{1}$$ Roughly speaking, the idea is to use $(1)$ for $B:=A^c \in \... 3 HINT:$P+n$form a partition of$\mathbb{R}$, so$A\cap (P+n)$form a partition of$A$. Now use$\mu(A\cap (P+n) = \mu((A-n) \cap P)$3 Fix$\epsilon>0$. For all$i\ge 1$,$\exists B_i\in \mathcal{A}$s.t.$\mu(B_i)\le \mu^*(A_i)+\epsilon2^{-i}$. Then $$\mu^*\left(\bigcup_{i\ge 1}A_i\right)\le \mu^*\left(\bigcup_{i\ge 1}B_i\right)\le \sum_{i\ge 1}\mu^*(A_i)+\epsilon.$$ It's clear that$\mu^*\mid_{\mathcal{A}}=\mu$(by monotonicity of$\mu$). If$A\in\mathcal{A}$, then for any$E\subset ...

3

What we want to show here is that for any countable collection $\lbrace A_{i} \rbrace$ of subsets of $X$, we have $$\mu^{*} \left( \bigcup_{i = 1}^{\infty} A_{i} \right) \leq \sum_{i=1}^{\infty} \mu^{*}(A_{i})$$ For #1, note that if $\bigcup_{i=1}^{\infty} A_{i} = \emptyset$, then each $A_{i} = \emptyset$. Clearly, $\mu^{*} \left( \bigcup_{i=1}^{\infty} A_{... 3 6)$\displaystyle\sum_{k=1}^{\infty}\frac{\epsilon}{2^{k}}=\epsilon\sum_{k=1}^{\infty}\dfrac{1}{2^{k}}=\epsilon\cdot\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}}=\epsilon$. 7) No, the sum is$\epsilon$, but now we have$0\leq m^{\ast}(C)<\epsilon$for every$\epsilon>0$, if$m^{\ast}(C)>0$, then put$\epsilon$to be$m^{\ast}(C)$, then we arrive to$m^{\ast}...

3

(This answer assumes that $I$ is countable) This is a case for the $\epsilon/2^n$ trick, in combination with the "give yourself an epsilon of room" trick. Once we've internalized these tricks the proof seems straightforward. Let $\epsilon > 0$. For each positive integer $n$, let $\{I_{nj}\}$ be a countable collection of intervals such that $A_n \subset \... 3 Partial answer. Here's a proof for$n=1$. We actually show that$E_\epsilon$is the union of finitely many pairwise disjoint balls. Since$\{x : G(x) > 0\}$is open, we may write it as$\{G > 0\} = \sqcup_{n=1}^\infty (a_n,b_n)$, a countable union of disjoint intervals (the proof is just to take maximal intervals in$\{G > 0\}$). Now, since$G$is ... 2 (3) The sum starts at$n=1$and$\frac{1}{2} + \frac{1}{2^2} + \cdots = 1$. Even if the sum really were$2\epsilon$, which would be an actual error in the proof, it would be trivial to fix (just carry forward the$2\epsilon$, having a$2\epsilon$i nthe final inequality is just as good. (2) If you know that$a<b$then also$a\le b$. Which kind of ... 2 In general, whenever we deal with measures (or integrals, or things of the sort), there is some kind of simple sets (or functions) which generate all other things, in some sense. You can see this, for example, when we define integrals: First define integrals for characteristic functions, then for simple functions, and then take some limits and define for ... 2 Another reason besides the excellent answer given is that every set has an outer measure, whereas not every set has a measure. If a set can't be proven to be measurable, it's common to investigate it with the outer measure. If the set turns out to be measurable, the outer measure results still apply because they agree 2 It is true when$\epsilon=1/2$:$m^\ast[a,b]\leq b-a+2\times 1/2$It is true when$\epsilon=1/4$:$m^\ast[a,b]\leq b-a+2\times 1/4$It is true when$\epsilon=1/10000000$:$m^\ast[a,b]\leq b-a+2\times 1/10000000$Since we can make$\epsilon$arbitrarily small, it must be true that$m^\ast[a,b]\leq b-a$. If$m^\ast[a,b]$were even a little bit bigger than$...

2

The important difference between the two is that one covering is allowed to be countable, while the other is finite. The fact that allowing countable coverings gives $\mathbb Q \cap [0,1]$ measure 0 is the same as for all other countable sets; that a finite covering has measure at least 1 is a good exercise.

2

You mentioned $m$, so I am assuming you are talking about the Lebesgue measure $m$ on $\mathbb{R}^k$. Given $\epsilon > 0$, the entire space can be written as the union of countably many disjoint $k$-cells $W_n$ such that $m(W_n) < \epsilon$ for all $n$. Since $m(E) < \infty$, we can choose compact $K\subset E$ such that $m(E-K) < \epsilon$. $K$ ...

2

$$\mathbb{Q}\cap[0,1] \subseteq \bigcup_{n=1}^k I_n$$ $$\overline{\mathbb{Q}\cap[0,1]} \subseteq \overline{\bigcup_{n=1}^k I_n}= \bigcup_{n=1}^k \overline{I_n}$$ $$1=m[0,1]=m(\overline{\mathbb{Q}\cap[0,1]})\le m(\bigcup_{n=1}^k \overline{I_n})\le \sum_{n=1}^k l(\overline{I_n})=\sum_{n=1}^k l(I_n)$$

2

For any set $A$; $m^*(A)+m_*([0,1]\setminus A)=1$ Also it is given that $m^*(A)+m^*([0,1]\setminus A)=1$ Hence we have $m^*([0,1]\setminus A)=m_*([0,1]\setminus A)\implies [0,1]\setminus A$ is measurable $\implies A$ is measurable

2

Since you aren't given that $f$ is measurable, you can't say that $\{x \in E : f(x) > \alpha\}$ is measurable for any $\alpha \in \mathbb{R}$. Instead, recall that continuous functions are measurable. So $f$ restricted to the set to $E \backslash D$ is measurable. Denote this restriction $g \doteq f \vert_{E \backslash D}$. Then we have for any \$\...

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