New answers tagged

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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