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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([... 10 I know the existence of a non-Lebesgue measurable sets (i.e. Vitali set) having outer Lebesgue measure$0$That's not true. Vitali sets have positive outer measure, and indeed that's how we prove they're not measurable (if they were, by the countable additivity of Lebesgue measure the interval$[0,1]would have to have infinite measure). It is indeed the ... 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 ... 6 Tao talks about this in his measure theory book, which last time I checked was still freely available. He says roughly that there is an asymmetry here because when measuring finite unions of boxes, the measure is subadditive and not superadditive. This leads to the fact that the Jordan inner measure $$m^{*, J}(U) := \sup \{m(A)\ :\ A \subset U\ \text{is a ... 5 If C is the Cantor set and A=B=\{e^{x}:x\in C\} then A and B have measure 0 but AB contains all numbers of the form e^{x+y} with x,y \in C. Since C+C=[0,2] it follows that AB does not have measure 0. [I have used the following facts: e^{x} is absolutely continuous on [0,1] and any absolutely continuous function maps sets of measure ... 5 You are arguing that$$ |[a, b]| \ge |(a, b)| \ge b-a \, , $$but the right inequality needs to be justified. We know that \ell((a, b)) = b-a, but it not obvious from the definition that |(a, b)| = \ell((a, b)). It is in fact easier to prove$$ |[a, b]| \ge b-a $$first, because any open covering \bigcup_{k=1}^\infty I_k of the compact interval [a, b]... 4 Ok lets look at it step by step f(x)=c_1 Now we need to check the \{x\in X| f(x)>c\} Basically this is the same as \{x\in X| c1>c\} Now we debate depending on variable c For c \ge c_1 We get \{x\in X| c_1>c\ge c_1\}=\emptyset since c_1>c_1 cannot happen. And as we know empty set is in sigma algebra by definition. Now for c <... 4 Generalize the construction of a Bernstein set. Enumerate A=\{x_\xi:\xi<2^\omega\}. Let \mathscr{F}=\{F_\xi:\xi<2^\omega\} be the family of uncountable closed subsets of A. Recursively construct distinct y_{\xi,n}\in A for \langle\xi,n\rangle\in 2^\omega\times\omega as follows. If \eta<2^\omega, and we have y_{\xi,n} for each \... 4 The family \mathcal F of nonempty open subsets of [0,1] of measure < 1 has cardinality c, so it can be well-ordered so that each member of \mathcal F has fewer than c predecessors. Since the complement in [0,1] of any member of \mathcal F has cardinality c, we can choose by transfinite induction, for each U \in \mathcal F, a countably ... 4 Actually, it's the other way around: \mu^*(A)\le \mu(A). This is because A itself covers A. Specifically, let A_0=A and A_i=\emptyset for i>0. Then A\subseteq\bigcup A_i. Now it's easy to see that there are two possibilities for \mu(\emptyset), by finite additivity: either 0 or \infty. If \mu(\emptyset)=0, then \sum\mu(A_i)=\mu(A)... 4 Here's a proof by contrapositive: suppose that C \subseteq \Omega and there is an A \in \mathcal A with \mu^*(A \cap C) < \mu(A). Noting that C \subset A^c \cup (A \cap C), we find that$$ \mu^*(C) \leq \mu^*(A \cap C) + \mu^*(A^c) = \mu^*(A \cap C) + \mu(A^c) < \mu(A) + \mu(A^c) = 1 $$So, \mu^*(C)<1. We have reached the desired ... 4 Denote by (X_n)_{n \in \mathbb{N}} \subseteq \mathcal{A} a sequence of increasing sets, X_n \uparrow X, such that \mu(X_n)< \infty for all n \in \mathbb{N}. You have already shown that$$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 \... 4 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}... 4 You already know that \mu^\star(A\triangle B)=0 implies that \mu^\star(A)=\mu^\star(B). Now apply this to A^c and B^c, together with the nice identity A\triangle B=A^c\triangle B^c, to conclude that \mu^\star(A\triangle B)=0 also implies that \mu^\star(A^c)=\mu^\star(B^c), which implies that \mu_\star(A)=\mu_\star(B) by subtracting both sides ... 4 The claim to prove is wrong, as shows the following example. Let X=[0,1], \mu be the Lebesgue measure on X, and \mathcal A=\mathcal S(\mathcal A) be a \sigma-algebra of Lebesgue measurable subsets of X. Let E\subset X be an arbitrary set such that E\not\in\mathcal A. For instance, E can be a Vitali set. Let F\in\mathcal A be a set such ... 4 Another extension result provides a different construction of measures. It starts with a \sigma-finite “measure” \nu that is only defined on an algebra \mathcal{A} of sets (meaning, it satisfies countable additivity \nu(\bigcup_n Q_n) = \sum_n \nu(Q_n) only in case the countable union belongs to \mathcal{A}). Then there is a unique measure \nu' ... 4 The answer is negative. In fact there is a set S\subset[0,1] of Lebesgue measure 0 such that S\cap(a,b) is uncountable for any 0\leqslant a<b\leqslant 1 (thus, for such an S, we have \mu^*(c,S)=1 and \mu_*(c,S)=0 for any 0<c<1). The idea to build S is to take the Cantor set and "fill in the holes" by its homothetic copies.... 4 Suppose that | A | + |( b, c ) \setminus A | = c − b and \epsilon>0 be any number. There exist countable covers \mathcal C_1 of A and \mathcal C_2 of ( b, c ) \setminus A by open subintervals of (b,c) such that S_1+S_2\le c-b+\epsilon, where S_i=\sum_{C\in\mathcal C_i} |C| for each i. It is easy to check that if |\bigcup\mathcal ... 4 Welcome to MSE! You mention the definition of measurability in terms of being "almost closed", however Axler actually gives a whole slew of equivalent definitions of measurable. These show up in the section 2D, and seeing as your exercise is 2D.12, it makes sense that you might want to use these definitions too! I've included a screenshot for ... 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 You mentionedm$, so I am assuming you are talking about the Lebesgue measure$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$... 3 Fix$\epsilon > 0$. Then there is a countable set$\{I_n \}$of intervals such that $$\sum_n l(I_n) < m^*(A \cup B) + \epsilon/2.$$ Write each open interval$I_n$as a finite union of open subintervals$\{ J_{i,n}\}_{i=1}^{N(n)}$such that$l(J_{i,n}) < \alpha$and$\sum_{i=1}^{N(n)}l(J_{i,n}) < l(I_n) + \frac{\epsilon}{2^{n+1}}$. Now, the ... 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 each$A_n$, let$B_n\supset A_n$a$m$-measurable set with$m(B_n)=m(A_n)$. Define an ascending sequence$\{C_n\}$of$m$-measurable sets as the following: $$C_n = \bigcap_{i=n}^\infty B_i,~~n\in\mathbb{N}.$$ Note that$A_n\subset C_n\subset B_n$for all$n\in\mathbb{N}$. Then $$m(\bigcup_{n=1}^\infty A_n)\leq m(\bigcup_{n=1}^\infty C_n) = \lim_{n\to\... 3 For x\in A, let U_x=\{y\in X:d(y,x)<\frac12 d(A,B)\}. Clearly this set is open, so U:=\bigcup_{x\in A}U_x is open. Obviously A\subseteq U. If y\in U, then there exists x\in A such that$$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 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 ...

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