# Tag Info

## New answers tagged measure-theory

1

Hint for $(1)$: What happens if you take the preimage of $[0,1]$? Hint for $(2)$: How many points does the set $\{x\in[0,1]:f(x)=a\}$ have?

0

Hint: $f$ is not measurable since $f^{-1}((0,1))=N-\{0\}$.

0

Dirac delta is not a function. But can be considered as limit of a sequence of measurable functions or as measure as done in probability theory.

1

Suppose the sigma algebra is countably infinite. Then the set $X$ on which the sigma algebra acts on must be infinite. (Why?) Let $A$ be an infinite measurable set. Suppose there does not exist a proper infinite subset of $A$. This is a contradiction to the assumption that the sigma algebra is countable infinite. (Why? Consider intersections and complements ...

1

In one dimension, $(0,1)\cup [1,2) = [0,2)$ is such an example: the perimeter of every interval is equal to $2$. In higher dimensions $n>1$, multiply the above example by a cube $(0,1)^{n-1}$. The perimeter of a rectangular box is the same regardless of the box being open, closed, or half-closed: indeed, the Caccioppoli definition of perimeter shows it ...

4

If it were $L^2$ then it would satisfy Cauchy-Schwarz, i.e. you would have $|f(0)| \leq C \| f \|_{L^2}$ for some $C$. Construct a sequence of functions $f_n$ such that $|f_n(0)|>n \| f_n \|_{L^2}$ to contradict this.

1

Assuming we've shown that $\mu$ is a measure, and that we're only concerned with bounded sets: Say $B$ is a bounded closed set and $E\subset B$ is measurable. First, note that $$\mu(E)=\inf\{\mu(V):E\subset V\subset B, V\text{ relatively open}\}.$$ (Here "relatively open" means open relative to $B$; that is, $V=B\cap W$ where $W$ is an open subset of $\Bbb ... 3 He uses the fact that $$\frac1{\mu(E)}\int_E\alpha\,d\mu=\alpha.$$If$\mu(E)=\infty$then$\int_E\alpha\,d\mu$is undefined. Or maybe it's defined for$\alpha\ge0$; in any case, whatever value we give to that integral we still have $$\frac1{\mu(E)}\int_E\alpha\,d\mu=0.$$ The result is still true if$\mu(X)=\infty$. But offhand the only proof I see requires ... 2 The claimed equality $$|A_E(f) - \alpha| = \frac{1}{\mu(E)}\left| \int_E (f - \alpha)\,d\mu\right|$$ depends on the assertion $$\frac{1}{\mu(E)}\int_E \alpha\,d\mu = \frac{1}{\mu(E)}\mu(E)\alpha = \alpha.$$ This is not valid if$\mu(E) = \infty$. And we cannot in general rule this case out if$\mu(X) = \infty$. 3 Edit: this answer addresses a previous version of the question, which since then has been completely modified. The answer to the new question is that the standard Radon-Nikodym theorem applies to measures (i.e., non-negative), not to signed or complex measures. This is wrong. All that can be said is that the integral exists in$[0,\infty]$, not that the ... 2 Claim: Let$m(E)>0$and$f : E \to (0,\infty]$. Then$\int_E f>0$. Proof: Let$E_n = \{ x\in E : f(x) \ge \frac 1n\}$. Then$\cup E_n = E$and so $$\lim_{n\to\infty} m(E_n)= m(E).$$ Since$m(E)>0$, there is$n$so that$m(E_n) >0$. Thus $$\int_E f \ge \int_{E_n} f \ge \int_{E_n} \frac 1n = \frac 1n m(E_n) >0.$$ 1 This does not imply that$f$is integrable. Counter example Consider$f(x,y) = \begin{cases} \frac{xy(y^2 -x^2)}{(x^2 + y^2)} & \text{for } y \in (0,2] \\ 0 & else \end{cases}$and$B_1= [0,1]$and$B_2=[0,2]$. Then $$0 \leq \int_0^1 f(x,y) \, dx = \frac{y}{2( y^2 +1)^2} \leq 1$$ for all$y \geq 0$and for$y < 0$the integral is 0. In ... 0 To me you only demonstrated that$\underline{\mu}(\bigcup S_i) \geq \sum\underline{\mu}(S_i)$. Now you need to show that$\underline{\mu}(\bigcup S_i) \leq \sum\underline{\mu}(S_i)$and for that you need the semifinite property of$\mu$for the case where$\exists A \subset \bigcup S_i, A \in {\cal M}$and$\mu(A) = \infty$exactly like Weltschmerz was ... 0 The composition of measurable functions where we use the same$\sigma$-algebras throughout, is measurable. However, when speaking of Lebesgue measurable functions$\Bbb R\to\Bbb R$, the definition says that the inverse image of a set belonging to the Borel$\sigma$-algebra is a set that belongs to the Lebesgue$\sigma$-algebra. Hence for Lebesgue measurable ... 3 Yes If$\mu(E^c)>0$, then $$\int\varphi\geq\int_{E^c}\varphi=\infty.$$ As mentioned in the book, in the line before$(4)$,$\mu(S^c)=0$. 2 The measure$\frac{dx}{x}$assigns the measure$\int_A \frac{1}{x} dx$to a set$A$(implicitly considered to be a subset of$[0,\infty)$so that you guarantee positivity). Note that in this notation Lebesgue measure is sometimes denoted by just$dx$. 0 This definition is inexact: it should probably be defined on$\text{BOREL}(X)$(by property 2, and the fact that all compact and open sets are measurable). The property non-atomic is not defined (it probably means there are no atoms, where an atom is a Borel (?) set$A$such that$\mu(A) > 0$and for all subsets$B$of$A$either$\mu(B) = \mu(A)$or ... 5 This is not exactly an answer to your question, but is rather a historical digression and is too long for a comment. The exact same question was asked by Axel Harnack in 1885 and he used interval$[0, 1]$in place of whole of$\mathbb{R}$. Harnack convinced himself that his reasoning is correct and the interval$[0, 1]can be covered by a countable number ... 0 Lipschitz functions are almost everywhere differentiable (see the second bullet of the wikipedia article). You ask if they're continuously differentiable almost everywhere. It seems that the easy answer would be $$\mu \left( \left\{ x \in \mathbb R^n \;\middle|\; \tilde f(x) = f(x) \text{ or } D \tilde f(x) = D f(x) \right\} \right) \leq \varepsilon ... 1 Your proof is in the right direction. Here is it in a complete form. 1.7 Proposition - If \varepsilon is an elementary family, the collection \mathcal{A} of finite disjoint unions of members of \varepsilon is an algebra. Proof: We begin by proving two auxiliary results: i.) \varepsilon \subseteq \mathcal{A} Note that if A\in ... 1 Hint: show that \mathcal{B} is generated by \{(-\infty,\alpha]\}_\alpha. If you already know that \mathcal{B} is generated by the open sets, this should be straightforward. You will need the theorem below as well. Theorem: Let (X,\mathcal{M}) and (Y,\mathcal{N}) be given. Assume \mathcal{N} is generated by \mathcal{E} (i.e. ... 2 \mathcal S is not necessarily a \sigma-algebra. For instance let S=\{1,2,3,4\} and let \mathcal S=\wp(\{1,2,3\}). Then \mathcal S satisfies the conditions, but \{1,2,3,4\}\notin\mathcal S. \mathcal S is a so-called \sigma-ring. edit: Another example inspired by the comment of Samuel and emphasizing that there is an essential ... 1 Your proof is in the right direction. Here a copy it and change some points in it to make it a complete proof. Theorem 1.9 - Suppose that (X,M,\mu) is a measure space. Let \mathcal{N} = \{N\in M:\mu(N) = 0\} and \overline{M} = \{E\cup F: E\in M, F\subset N, N\in\mathcal{N}\}. Then \overline{M} is a \sigma-algebra and there is a unique ... 1 Saying the integral is finite for each x cannot be enough... Say (I_n)_{n=1}^\infty is a sequence of disjoint open intervals tending to 0. Say f_n\ge0, f_n is supported in I_n, and f_n(x_n)=3^n for some x_n\in I_n. Take \mu supported on \{f_n\} with \mu(f_n))=2^{-n}. Then the integral, let's call it \phi(x), is finite for every x, ... 1 Note regarding context: The notation m below refers to Lebesgue measure in \Bbb R^d; I'm assuming here that that's already been defined, and we know a few of its basic properties, mainly just that it's finite on bounded measurable sets, is inner regular and satisfies m(rE)=r^dm(E). Now I'm using Lebesgue measure in proving that \mu^*, and hence m, ... 1 See https://www.math.ucdavis.edu/~hunter/measure_theory/measure_notes_ch2.pdf and check if theorem 2.31. (given with proof) on page 21 solves your problem 5 If you require that a Haar measure is also a Radon measure, then yes: a topological group carries a Haar measure if and only if it is locally compact. See 443E in Fremlin's Measure theory Vol. 4 Topological Measure Spaces. As for examples of groups which are not locally compact, try to show that there is no Haar measure on an infinite-dimensional Banach ... 1 Let \{A_j\}_{1}^{\infty} be a sequence of disjoint \mu^*-measurable sets and E\subset X. By countable subadditivity,$$\mu^*\left(E\cap\left(\bigcup_{1}^{\infty}\right)\right) = \mu^*\left(\bigcup_{1}^{\infty}E\cap A_j\right) \leq \sum_{1}^{\infty}\mu^*(E\cap A_j)Let B_n = \bigcup_{1}^{n}A_j. For each n\geq 2, since A_n is \mu^*-measurable ... 0 The following implications are valid if and only if the measure \mu is complete: a.) If f is measurable and f = g \mu-a.e., then g is measurable. b.) If f_n is measurable for n\in \mathbb{N} and f_n\rightarrow f \mu-a.e., then f is measurable. Proof a.): Suppose \mu is complete and E is the exceptional set where ... 1 The proof for \mu(E_2)\leq \mu(E_1) is correct, so you showed it is well defined and obviously it extends \mu. For uniqueness, pick \mu' extending \mu, pick any F, then it is contained in some measure zero set, hence \mu'(F)=0. Pick E\cup F\in\overline M, then \mu'(E\cup F)=\mu'(E\setminus F)+\mu'(F)=\mu'(E\setminus F)\geq \mu'(E\setminus N) ... 3 Suppose there are measurable subsets A,B and next to that some set E with A\subseteq E\subseteq B. If \mu(B-A)=0 (and automatically \mu A=\mu B) then it is quite tempting to say that also E can be labeled to be a set having measure \mu A=\mu B. That is the inspiration for "defining" \mu^*(E)=\mu(A)=\mu(B). But wait a minute... What if ... 0 Suppose B\in M' then there is an E\in M such that B = E\cup F where F\subset N' and N'\in N. Then \begin{align*} X\setminus B &= X\cap B^c\\ &= X\cap (E\cup F)^c\\ &= X\cap ((E^c\cap N'^c\cap F^c)\cup (F^c\setminus N'^c))\\ &= X\cap ((E^c\cap N'^c)\cup (F^c \cap N'))\\ &= X\cap ((E\cup N')^c \cup (N'\setminus F)) \end{align*} ... 3 f \equiv 1 gives an example for which \mu_f is complete, since it coincides with \mu. f \equiv 0 gives an example for which \mu_f is not complete. Since \mu_f(\mathbb{R})=0, every set would have to be measurable, which we know is not the case. 0 I just stumbled across this post after I was trying this problem. Part one of martins incorrect but a small change can fix it. We dont know that there exists a finite open cover, {\cup_{k=1}^n I_k} of E such that if O=\cup_{k=1}^n I_k then m(O\setminus E) < \epsilon since the definintion of lebesgue measure uses countable coverings and finite ... 1 The upper bound on \Bbb{P}(X<0) is \frac{1}{2} for any \mu > 0 and we will show how to construct a probability distribution function with \Bbb{P}(X\leq 0)=\frac12) and how to go from there to a function with \Bbb{P}(X < 0)=\frac12-\epsilon) for arbitrary small positive \epsilon. Let us work to maximize \Bbb{P}(X\leq 0), and call ... 1 If you wish to construct product measure with infinitely many probability spaces as factors, then no topological assumptions are needed. More precisely, if \{(E_t,\mathcal E_t,\Bbb P_t): t\in T\} is a non-empty collection of probability spaces indexed by some set T, then there is a unique probability measure \Bbb P on the product space (\times_{t\in ... 0 Hint: If f=\sum_{i=1}^nc_i\chi_{A_i} where the c_i\in\mathbb R denote constants and the A_i measurable sets then:\int fdP=\sum_{i=1}^nc_iP(A_i)$$(linearity of expectation) So wonder what values are taken by the described function, and on what sets. 1 The problem is not to construct an infinite product of sigma-algebras. This always exists and is defined as you have stated, as the smallest sigma-algebra that makes all the projections measurable. The problem is to find measures on this space. Of course a measure on the infinite product always pushes forward to a measure on the finite products contained in ... 1 I believe your proof solves completely the first part of the question. So I will only adress the second question:$$\mathfrak L^p \otimes\mathfrak L^q \subsetneq \mathfrak L^{p+q}$$The counterexample can be given for p=q=1. Then it can be extended for any indices. Let V be the Vitali set in \mathbb{R}. We get that N := V\times \{0\} \subset ... 1 If S is any set, there is a measure \mu on the set X=\{0,1\}^S such that for each finite F\subseteq S and each a\in\{0,1\}^F, \mu(\{x\in X:x|_F=a\})=2^{-|F|} (this measure can be defined on the \sigma-algebra generated by these sets \{x\in X:x|_F=a\} using the Caratheodory extension theorem, for instance). Intuitively, you can think of this ... 4 This is false in general. For instance, you can take a Jordan curve C in {\mathbb R}^2 of positive 2-dimensional measure (an "Osgood curve") and let A, B be the components of {\mathbb R}^2 -C. 1 Here X=\mathbb N For n=1,2,\dots let f_{n}:\mathbb{N}\to\left[0,\infty\right] and let \mu denote the counting measure on \mathbb{N}. Define f:=\sum_{n=1}^{\infty}f_{n}, in the sense that f:\mathbb{N}\to\left[0,\infty\right] is prescribed by k\mapsto\sum_{n=1}^{\infty}f_{n}\left(k\right). Then:$$\int ... 1 In this case,X = \{1, 2, 3, \ldots\}$and$f_j(i) = a_{i, j}$. If$\mu$is counting measure and $$f(i) = \sum_{j=1}^\infty f_j(i) = \sum_{j=1}^\infty a_{i, j},$$ then by Theorem 1.27 we have $$\sum_{i=1}^\infty \sum_{j=1}^\infty a_{i, j} = \sum_{i=1}^\infty f(i) = \int_X f \, d\mu = \sum_{j=1}^\infty \int_X f_j \, d\mu = \sum_{j=1}^\infty ... 0 Let f(i)=\sum_{j=1}^\infty a_{ij}, with f(i)=0 whenever i is not a positive integer, and \mu(\{i\})=1 for i=1,2,\cdots, and 0 otherwise. Now apply Tonelli's theorem. 0 If we are in (\Omega, F, \mu_c) where \mu_c is the counting measure, the measurable function are of the form f= \{a_i \}_{i\in \mathbb N} and \int_\Omega f d\mu_c = \sum_{i=1}^\infty a_i So that f_j=\{a_{ij}\}_{i,j \in \mathbb N}, thus using the Tonelli theorem in this setting, you get the thesis 1 Looks ok to me, if you justify that$$ |TE|=|\det T|\,|E|. $$The result you are trying to prove is the change of variable formula. Just assuming that f is measurable you get directly from the substitution formula that$$ \int_Ef(y)\,dy=\int_{T^{-1}E} f(Tx)\,|\det T|\,dx, $$where the only verification to be made is that the Jacobian of T is T ... 5 For a counterexample with n=1, take A and B to be disjoint open subsets of [0,1] defined as follows. Start with C_0 = [0,1]. At each stage n \ge 1, C_{n-1} will be a finite union of closed intervals; take two disjoint finite sets of points E_n and F_n in the interior of C_{n-1} such that every point of C_n is within distance 1/n ... 1 Ciao Giuseppe. The answer is negative, as @DavidC.Ullrich has already pointed out. I can provide an explicit counterexample. Consider the function q:\mathbb{Q}\backslash\{0\}\to\mathbb{N} associating to x the unique natural number q=q(x) such that we can write x=\tfrac{p}{q} an irreducible fraction. Then define$$f(x) = \cases{q(x)&if ... 3 The answer is negative. In fact if we define the Baire class$B_\alpha$for countable ordinals$\alpha$by saying$B_0=C(I)$,$B_{\alpha+1}$is the set of pointwise limits of sequences in$B_\alpha$, and$B_\alpha=\bigcup_{\beta<\alpha}B_\beta$for limit ordinals$\alpha$then all the$B_\alpha\$ are distinct. Or so I've read; don't ask me to prove it. ...

1

Hint: $$|f(x)|^q \le \max(|f(x)|^p,1)$$

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