New answers tagged

1

This should work: you may represent any $r\in(0,1)$ as a continued fraction: $$ r = [0;a_1,a_2,a_3,\ldots] = \frac{1}{a_1+\frac{1}{a_2+\frac{1}{\ldots}}},\qquad a_i\in\mathbb{N}^*$$ then take $A$ as the set of numbers such that $a_1$ is even and $B$ as the set of numbers such that $a_1$ is odd, or something similar to meet the constraint $\mu(A\cap [0,x])=x/...


2

$\def\rr{\mathbb{R}}$Let $f(r) = μ( A \cap [-r,r] )$ for every $r \in \rr_+$. Then prove that $f$ is continuous on $\rr_+$. Note that $f(0) = 0$. Then prove that $f(r) \to μ(A)$ as $r \to \infty$ (say by MCT for sets). Then you're done because for any $b \in [0,μ(A))$ there is some $r \in \rr_+$ such that $f(r) = b$.


0

yeah this is what we call atomless measure for the case of Lebesgue measure the proof is simple you can take A as an interval $A=[c,d]$(because every borel set is a reunion of interval ) $$ B=[c,c+\frac{b}{a}A] $$ Let Now $A'$ a Lebesgue set then there exist a Borel set $A$ such that $\mu(A'-A)=0$ but in every Borel $A$ set with positive measure we have ...


0

If $g$ is a measurable function then $\forall c\in \mathbb{R}, c\ne 0$, we have $c*g$ measurable. Moreover ${x \in X: g(x)> \frac{q}{c}}$ which is measurable.


5

The answer to your question is no, and this is not hard. But first, for the record we should point out that one of your conjectures about this is false: The range of a measure need not be closed. In fact, although the answer to your question is no, there is a measure with range equal to $$\{0,\infty\}\cup(\Bbb Q\cap[1,\infty)).$$ Say $(r_1,r_2,\dots)$ is ...


0

The function $x\to\xi_x$ is not well defined, as you stated it. However, if you make it well defined, for instance saying that $\xi_x$ is the smallest number that satisfy the MVT, then $f'\circ h$ is measurable. This is because $$ (f'\circ h)(x) = \frac{f(x)-f(a)}{x-a} $$ Since the right hand side is a measurable function, $f'\circ h$ is measurable.


2

Your proof is correct. It just need a small adjustment: to make explicit that the representation $\sum_{n}a_n \chi_{E_n}$ of $\phi$ being used satisfies the condition: for all $n$, $a_n>0$. I have also improved the wording in the end of the proof. If $f\in L^+$ and $\int f < \infty$, for every $\epsilon > 0$ there exists $E\in M$ such that $\mu(...


0

Congratulations! Your proof is correct. I have just added some details to make it clearer. Problem 3.2.14 - If $f\in L^{+}$, let $\lambda(E) = \int_{E}f d\mu$ for $E\in M$. Then $\lambda$ is a measure on $M$, and for any $g\in L^{+}$, $\int g d\lambda = \int f g d\mu$.(First suppose that $g$ is simple) Proof - Observe that $\lambda(\emptyset) = \int_{\...


0

note that regular as in Rudin book mean that every borel set is inner and outer regular, it mean : $$ \mu(F)= \sup\{\mu(K) \; K \; \textrm{compact of } F\}\\ \mu(F)=\inf\{\mu(O) \; O \; \textrm{open which contient } F\}\\ $$ it's clear that $\mu(K)\leq \mu(F)$ to proof that it will be the sup we can use the proprety of supremum bounded $$ \forall \epsilon &...


2

Given $E$ of infinite measure and $\epsilon > 0$, let $V$ and $F$ be as in (a). Then $\mu(E-F) < \epsilon$. For each $n$, put $$ F_n = F\cap \left(\bigcup_{i=1}^{n}K_i\right), $$ $F_n$, as a closed subset of a compact set, is itself compact. We have $$ \lim_{n\to \infty} \mu(F_n) = \mu(F) = +\infty, $$ from which we conclude that $\mu$ is regular.


1

Congratulations! Your proof 2 is more direct, and it is correct. If $\{f_n\}\subset L^+$, $f\in L^+$ such that $\{f_n\}$ is dominated by $f$ where $\int f < \infty$ then $$\limsup\int f_n\leq \int \limsup f_n$$ Proof 2 - Consider the sequence $\{f - f_n\}_n\subset L^+$ then applying Fatou's lemma we have \begin{align*} &\int (\liminf(f-f_n))\...


1

You'll want to have a look at "A Signed Measure on Path Space Related to Wiener Measure" by K. Hochberg http://www.jstor.org/stable/2243147 Hocberg constructs a Markovian signed measure (of unbounded variation) on the space of continuous paths, associated with the operator $Lu:={\partial^4u\over\partial x^4}$. The transition density $p$ is the fundamental ...


0

If the $E_n$ are not disjoint, then the claim is not true. To see why it is true, though, if they are disjoint (which is the case): $$\chi_{\cup_{n=1}^{\infty} E_n} (x) = \begin{cases} 1, \text{ $ x \in E_{n_0}$ for some $n_0 \in \Bbb N$} \\ 0, \text{ if for all $n$, $x \notin E_n$} \end{cases}$$ If $x \in E_{n_0}$ for some $n_0$, then $x$ is not in any ...


0

I present a sketch show I made from suggestions @Ian: Suppose that Dirac's delta belongs in $L^2(\mathbb{R})$, i.e., that has $\int\delta(x)f(x)dx\le||\delta||_{L^2}||f||_{L^2}$, for all $f\in L^2(\mathbb{R})$. By definition Dirac's delta, $f(0)=\int\delta(x)f(x)dx\le||f||_{L^2}$ i.e, $f(0)\le||f||_{L^2}$, for all $f\in L^2(\mathbb{R})$. Then we will ...


2

You may try proving and using the following dual to the monotone convergence theorem: Suppose $\{g_n\}\subseteq L^+$, $g_n$ decreases pointwise to $g$ , and $g_1$ is integrable. Then, $\int g=\lim\int g_n$. This is Exercise 2.15 in Folland (1999, p.52).


2

Your proof is essentially OK up to close the end. In the end you can not use Fatou's lemma as you did. I have rewritten your proof, makng minor adjustments and correcting the ending. I also offer you a second proof. This one does not use Monotone Convergence, but actually uses Fatou's lemma. Corollary 2.19 - If $\{f_n\}\subset L^+$, $f\in L^+$, ...


1

Since $E$ is $\sigma$-finite, we can write $E = \cup E_n,$ where $E_1 \subset E_2 \subset E_3 \cdots,$ and $\mu(E_n) <\infty$ for each $n.$ By the result already known, for each $n$ we can choose a compact $K_n \subset E_n$ such that $\mu(E_n\setminus K_n) < 1/n.$ We then have $$\mu(E) = \lim_{n\to \infty} \mu(E_n) = \lim_{n\to \infty} \left (\mu(K_n) ...


0

It's a really vague question: what you want can be achieved in a lot of different ways. This one has just come to mind: $$ SPARSITY([0=a_0, a_1, \ldots, a_{n-1}, a_n=1])= {1\over1+\sum_{k=1}^{n-1}(a_k-k/n)^2} $$ Could that work for you?


0

Basically you are asked to prove the equivallence of to definitions of Lebesgue measurable set. (1) $\implies$(2) Suppose $\mu^*(E)=\mu_*(E)$. Than for any $\varepsilon>0$ there are two sets of open intervals $\{I_n\}_{n\in\mathbb{N}}$ and $\{J_n\}_{n\in\mathbb{N}}$ such that (i) $\quad E\subset I=\bigcup_{n\in\mathbb{N}} I_n$ (ii) $\ \ \ [a,b]\...


1

The connection with the spectral measure $P$ is $$ P(E) = \chi_{E}(A). $$ So, for example, $$ P[a,b] = \chi_{[a,b]}(A), \;\; P(a,b) = \chi_{(a,b)}(A) \\ P[a,b] = P(a,b) + P\{a\}+P\{b\} $$ The spectral measure $P$ is regular in the strong topology, which gives you \begin{align} P[a,b]x & = \lim_{\...


2

Let $E\in\mathfrak{M}$. If $\mu(E)<\infty$ then $E$ is inner regular by $(d)$. Suppose $\mu(E)=\infty$ and $E$ $\sigma$-finite. To show: $E$ is inner regular. My idea Since $E$ $\sigma$-finite, there exists a countable family $\{A_n\}_{n\in\mathbb{N}}\subset \mathfrak{M}$ such that $E=\bigcup_{n\in\mathbb{N}}A_n$ and $\mu(A_n)<\infty$ for all $n\in \...


1

We can endow $\mathcal A$ with a pseudo-metric defining $\rho\left(A,B\right):=\mu\left(A\Delta B\right)$, where $\Delta$ denotes the symmetric difference operator. In this way, $\left(\mathcal A,\rho\right)$ is a complete pseudo-metric space. For the details, see this thread.


2

We have $\int_x^y f = \int_a^y f - \int_a^x f = 0$. Now assume $f(b) > 0$ for some $b$ and that $f$ is continuous at $b$ (otherwise we're still in a set with measure $0$ by Lebesgue's criterion). Then there is $\varepsilon > 0$ with $f(y) > f(b)/2$ for all $y \in (b - 2\varepsilon, b + 2\varepsilon)$. In patricular, since $[b - \varepsilon, b + \...


1

just to find a covering of a Cantor set with open balls so that the desired dimension attains. Yes, the way to go is to find a covering. Not necessarily with open balls: the usual definition of Hausdorff measure allows arbitrary sets. If your definition requires an open ball, observe that every set of diameter $d$ is contained in an open interval of ...


1

Your proof is essentially correct. Just some minor adjustments are required. 2.18 Fatou's Lemma - If $\{f_n\}$ is any sequence in $L^+$, then $$\int \left(\lim_{n\rightarrow \infty}\inf f_n\right) \leq \lim_{n\rightarrow \infty}\inf\int f_n$$ Proof - We know that $$\int \left(\lim_{n\rightarrow \infty}\inf f_n\right) = \int \sup_{k\geq 1}\left(\inf_{n\...


1

@Wolfy your proof is correct. Just some minor adjustments are needed. I Have copied your proof, making such adjustments. Theorem 2.14 (MCT) - If $\{f_n\}$ is a sequence in $L^+$ such that $f_{n}\leq f_{n+1}$ for all $n$, and $f = \lim_{n\rightarrow \infty}f_n (=\sup_n f_n)$, then $\int f = \lim_{n\rightarrow \infty}\int f_n$. Proof - Let $f = \lim_{n\...


0

We assume that $\mu$ is a measure on $X$. Note that is the only standing assumption here; no norms or measures are assumed finite below except where explicitly stated. The question as stated is unclear. The question at the linked post is clear, but slightly silly because the hypotheses are obviously redundant. Here are the facts, or at least some of the ...


2

Note: your proof of 2.17 is incorrect, because you assume that an arbitrary element $f \in L^+$ can be written as $f = \sum_{j=1}^{\infty}a_j \chi_{E_j}$, but in fact this is only true of simple functions. Note that nonnegative simple functions are elements of $L^+$, but a general element of $L^+$ is any measurable function from $X$ to $[0,\infty]$. We wish ...


1

Looks correct to me. You could add that the monotone convergence theorem can be applied because $$\inf_{n\geq k} f_n \leq \inf_{n\geq k+1} f_n$$ Also, your "indexing mistake" is that it should have been $\inf_{n\geq k} f_n \leq f_k$. And it's not "for all $n\geq k$", since $n$ is a variable internal to the $\inf$. It's "for all $k$".


0

Not unless you assume $\mu$ is non-atomic, as has been pointed out. Also of course if $f$ is as you say then $\mu(X)=1$. Yes if $\mu$ is a non-atomic probability measure. (If $\mu$ is just $\sigma$-finite you need to take $f:X\to\Bbb R$.) This is easy to see from the "stronger version" of a theorem of Sierpinski.


0

Let $X=[0,1]$ and $\mu=\delta_0$. Then for any measurable function $f:[0,1]\to[0,1]$ and any Borel set $A$, $\mu(f^{-1}(A))=1$ if $f(0)\in A$ and $\mu(f^{-1}(A))=0$ otherwise. In other words, $\mu\circ f^{-1}=\delta_{f(0)}$, which is not equal to Lebesgue measure for any choice of $f$.


0

Hint: note that two invariant integrable functions are equal almost everywhere if and only if their integrals are equal on every invariant set; why? :)


2

The case $n=2$ generalizes by induction so this is what we'll show. Let $U \subset \mathbb{R}^2$ be open. As $\mathbb{R}^2$ is separable and open balls in $\mathbb{R}^2$ are countable union of open intervals in $\mathbb{R}$ we may write $$U = \cup_{j=1}^\infty ( (a_{j1},b_{j1}) \times (a_{j2}, b_{j2}))$$ for some family of intervals. Here its understood that ...


2

Let $(X,M, \mu)$ be a measure space. Question: Proposition 2.13 - Let $\phi$ and $\psi$ be simple functions in $L^+$. a.) If $c\geq 0$, $\int c\phi = c\int \phi$. b.) $\int(\phi + \psi) = \int \phi + \int \psi$. c.) If $\phi\leq \psi$, then $\int \phi\leq \int \psi$. d.) The map $A\rightarrow \int_{A}d\mu$ is a measure on $M$. ...


2

If I'm understanding correctly, I think you can prove the result by first proving the result when $g$ is an indicator function of sets, then for simple functions $g$ (by linearity), then for positive functions $g$ (by monotonicity), and then finally for integrable functions $g$ (by splitting the positive and negative portions and applying the result ...


1

Let $f_k = f$ for $k$ odd and $f_k = g$ for $k$ even. Then $\max(f(x),g(x)) = \max(f(x),g(x),f(x),g(x),...) = \sup_k f_k(x) = g_1(x)$. Hence by Proposition 2.7 $g_1$ and hence $x \mapsto \max(f(x),g(x))$ is measurable. Similarly, $g_2$ is measurable.


4

An obvious thing to do with the hint is observe that $f > 1$ is the same thing as $\frac{1}{1+f} < \frac{1}{2}$.


1

Problem 2.1.5 - If $X = A\cup B$ where $A,B\in M$, a function $f$ on $X$ is measurable if and only if $f$ is measurable on $A$ and on $B$. Before the proof, let us precise what means " $f$ is measurable on $A$" (and " $f$ is measurable on $B$"). Since $A \in M$, we have that $M_A=\{ C\in M : C\subset A\}$ is a $\sigma$-algebra. It is the "restriction" of $...


1

Let $(X,M)$ be the measurable space. We already know that the proof of Proposition 2.6 promptly adapts to show that if $f,g:X\rightarrow \mathbb{R}$ are measurable then $f+g$ and $fg$ are measurable. So. the key point in Exercise 2 is to take care of the infinity values and extend Proposition 2.6 to the case where $f,g:X\rightarrow \overline{\mathbb{R}}$....


2

Here I assume two things: $\int |f|<\infty$ and $f>-1$ a.e. \begin{aligned} \int_{-\infty}^\infty \frac{1}{1+f} dx & \geq \int_{-\infty}^\infty \frac{1}{1+|f|} dx \\ & = \int_{-\infty}^\infty 1-1+\frac{1}{1+|f|} dx \\ & = \int_{-\infty}^\infty 1-\frac{|f|}{1+|f|} dx \\ & \geq \int_{-\infty}^\infty 1-|f| dx \\ & = \int_{-\infty}^\...


4

If $$ \int _{-\infty }^{\infty}\frac{1}{f+1} d x < \infty$$ Then $$\int _{-\infty }^{\infty}\frac{1/4}{f+1} d x +\int _{-\infty }^{\infty}f d x < \infty$$ But \begin{align*} \int _{-\infty }^{\infty}\frac{1/4}{f+1} d x +\int _{-\infty }^{\infty}f d x&= \int _{-\infty }^{\infty}\frac{1/4}{f+1}+f d x\\ &=\int _{-\infty }^{\infty}\frac{f^2+ f+1/...


8

Suppose that $f$ is nonnegative (you say it's a distribution?). Notice that $1=\frac{1}{1+f}+\frac{f}{1+f}$. Also $0\leq \frac{f}{1+f}\leq f$ which means $\frac{f}{1+f}$ is integrable. Since the l.h.s. is not integrable, it follows that $\frac{1}{1+f}$ is not integrable.


2

Your proof is essentially correct. The proof of the last part, that is (b. $\Rightarrow$ $\mu$ is complete) is very similar to the proof that (a. $\Rightarrow$ $\mu$ is complete). Here is the proof in details. Proposition 2.11 (Exercise 10) - The following implications are valid if and only if the measure $\mu$ is complete: a.) If $f$ is ...


1

I assume that: $$\int_{-N}^{N} \frac{1}{f(x)+1} dx < +\infty,$$ for any $N > 0$ since $f(x) > 0 ~\forall x$. Furthermore, I assume $$\lim_{x \to \pm\infty} f(x) = 0$$. This means that: $$\forall \varepsilon > 0 ~\exists N > 0: |x| >N \Rightarrow f(x) < \varepsilon.$$ Then: $$f(x) < \varepsilon \Rightarrow \\ f(x) + 1 < \...


1

Let $h(x)=f(x)g(x)$. We want to show that $a\in \mathbb R\Rightarrow \left \{ x:h(x)>a \right \}\in \mathscr M$. Note that $F:\mathbb R^{2}\to \mathbb R$ defined by $F(x,y)=xy$ is continuous, so $U=\left \{ (x,y):F(x,y)>a \right \}$ is an open set $\Rightarrow U=\bigcup_nI_n\times J_n$ where $I_n=(a_n,b_n)$and $J_n=(c_n,d_n)$ are non-degenerate ...


0

If $\mu(\Omega)<\infty$ then $L^2(\Omega) \subset L^1(\Omega)$. One proof of this is to use Cauchy-Schwarz. This still works when we replace $2$ with $p>1$, but now the inequality is attributed to Holder instead. If $\mu(\Omega)=\infty$ then either there is an atom of infinite measure or there are sets of arbitrarily large finite measure. (This "or" ...


2

Let $\Omega=\{1,2\}^3$ with the uniform probability law, and let $X_i$ be the projection onto the $i$th coordinate. Then $\mathcal{F}_1=\sigma(X_1,X_2)$ contains the event $$\{X_1=1\}=\{(1,1,1),(1,1,2),(1,2,1),(1,2,2)\}$$ Suppose for contradiction that $\mathcal{F}_2=\sigma(\frac{X_1}{S},\frac{X_2}{S})$ also contains this event. Since $\mathcal{F}_2$ ...


1

@Wolfy , Your proof is correct and, in fact, it is a simple and clear proof. I copy it here, just to add some details, to make it even clearer. If $\{f_j\}$ is a sequence of $\overline{\mathbb{R}}$-valued measurable functions on $(X,M)$, then the functions $$\begin{aligned} g_1(x) = \sup_{j}f_j(x), \ \ \ \ g_3(x) = \lim_{j\rightarrow \infty}\sup f_j(...


0

Note that every set mentioned below is assumed to be measurable. Presumably you know that if $\mu(\Omega)<\infty$ then $L^2\subset L^1$, and meant to ask about the converse. As @zhw comments, letting $X$ be a singleton of infinite measure gives a counterexample. In fact any counterexample must be wacky in an analogous way: Theorem Suppose $\mu$ is a ...


1

Lemma: Let $(X,M)$ be a measurable space and $Y$ be a set. If $f:X\rightarrow Y$ is a function, then $$\Sigma= \{E \subset Y : f^{-1}(E) \in M \}$$ is a $\sigma$-algebra on $Y$. Proof: Since $f^{-1}(\emptyset)= \emptyset \in M$, we have that $\emptyset \in \Sigma$. If $E\in \Sigma$ then $f^{-1}(E) \in M$. So we have $f^{-1}(E^c) = (f^{-1}(E))^c \in ...



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