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Note first that, for every nonnegative $u$, $$u^p = p \int_{0}^u t^{p-1} \mathrm dt = p \int_{0}^{\infty} t^{p-1} \mathbf 1_{\{u>t\}}\mathrm dt.$$ Use this for $u=f(x)$ and integrate the identity with respect to the Lebesgue measure $\lambda^n$ on $\mathbb R^n$. Then the desired result stems from Tonelli theorem and from the fact that, for every ...


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It is still quite trivial but at least... Given the Lebesgue measure $\lambda:[0,1]\to(0,\infty)$. Consider a Vitali set $\mu_*(V)<\mu^*(V)$. Construct the function: $$f:[0,1]\to\mathbb{R}:f(x):=x^2\chi_V(x)$$ This one is not sum of measurable plus step but product.


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Notation $\sigma_X(\mathcal E)$ is the sigma algebra generated by $\mathcal E$ in $X$ $A^c$ is the complement of the subset $A$ in $X$ $\mathcal C\cap A = \{C\cap A| C\in\mathcal C\}$ for a collection of subsets $\mathcal C$ $2^C$ is the collection of all subsets of $C$ To me the crux of the proof is the following Lemma If $A\subseteq X$, $\mathcal B$ ...


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Ok let's go but I will only outline the solution (using theroem 2 of planetmath.org) : Take $\mathcal{K}$ as of processes of the form $\sum_{i=1}^n C_i.1[({s_i},{t_i}]\times A_{s_i}]$ where $0\leq {s_i}<{t_i}$, $s_i$ are increasing and $A_{s_i}\in\mathcal{F}_{s_i}$ (those processes are bounded and predictable, and $\mathcal{K}$ is stable by ...


1

Every Polish space without isolated points admits an atomless strictly positive Borel probability measure. This is a consequence of the fact that such a space has a dense subspace homeomorphic to the irrational reals. The latter is mentioned at the end of exercise 6.2.A in Engelking's General topology.


3

If $e = \{e_n \colon n\in\mathbb{N}\}$ is an orthonormal system (hence $e\subset B$), then $$\rho^2(e_m,e_n) = (e_m-e_n,e_m-e_n) = (e_m,e_m)-2(e_m,e_n)+(e_n,e_n) = 1-2\cdot 0 + 1=2$$ if $m, n$ are distinct natural numbers. Therefore, from $e$ it can't be selected convergent subsequance. Moreover, from this, you can deduce that the unit sphere $S =\{x\in H ...


2

The usual example is $$\mathcal{H} = l^2 = \left\lbrace (a_n)_{n=1}^{\infty} \in \mathbb{R}^{\mathbb{N}} ~ | ~ \sum_{n=1}^{\infty} a_n^2 < \infty \right\rbrace$$ Taking the sequence $(e^{(n)})$, where $e^{(n)}_n = 1$ and $e^{(n)}_i = 0$ for all $i \neq n$, we obtain a sequence that has no convergent subsequence. To see that this holds, note that ...


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I think your estimate is wrong, you should double check it. I'll assume it should be $W^{2,p}$ on the left, and I'll assume $n=3$. Let $V\subset\subset U$. Note then $f\in L^p(V)$ for all $p$. Since $W^{1,2}\subset L^6(V)$ we have $u\in L^6(V)$. This implies, by the equation and the given estimate that $u\in W^{2,p}(V)$ for $p<3/2$. From the equation and ...


2

I learnt about RN derivative from "Real Analysis" by Folland, and would advise you to check it out there (Chapter 3) as it may answer your coming questions. In particular, Theorem 3.5 answers your Q1. It state that If $\nu$ is a finite signed measure and $\mu$ is a positive measure, then $\nu\ll \mu$ iff for any $\varepsilon >0 $ there exists $\delta ...


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Morreys inequality tells us that if $u\in W^{1,p},p>n$, then $u$ is holder continuous, you need to do some kind of covering argument to extend your inequality to the whole of $U$.


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Write $f=u+iv$ with $u,v$ real valued. The reduction of complex case to the real is based on $$ \left|\int_E u\,d\mu \right| = \left|\operatorname{Re}\int_E f\,d\mu \right|\le \left|\int_E f\,d\mu \right| $$ and similarly $$ \left|\int_E v\,d\mu \right| = \left|\operatorname{Im}\int_E f\,d\mu \right|\le \left|\int_E f\,d\mu \right| $$ So, if ...


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Yes. In the definition of premeasure, Folland requires that $\mathcal{A}$ is an algebra (this is important). So $X\in\mathcal{A}$ by definition. The proposition says that the outer measure $\mu^{*}$ (induced by $\mu_0$) agrees with $\mu_0$ on the algebra $\mathcal{A}$. In particular, $\mu^{*}(X)=\mu_0(X)$.


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Gauges are directed: $$\delta\wedge\delta'(\omega):=\delta(\omega)\cap\delta'(\omega)$$ and every gauge admits a partition: $$\Omega=\bigcup_{n=1}^N\delta(a_n)=:\bigcup_{n=1}^NA_n:\quad\mathcal{P}_\delta:=\{A_1',\ldots,A_N'\}\quad(A_n':=A_n\setminus\cup_{m=1}^{n-1}A_m)$$ (Note that compactness was necessary!) That is gauges induce a net of partial sums: ...


3

An "almost everywhere" always refers to the measure under consideration. And if several measures are under consideration, it should be specified to which one refers, usually by prefixing "$\nu$-almost everywhere". [If all measures under consideration have exactly the same null sets, that can be omitted without causing ambiguity, but it's not necessarily ...


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Let $U\subset\mathbb{R}^{2}$ be open and let $\mathcal A$ denote the $\sigma$-algebra generated by the sets $A\left(\theta,\eta,r,R\right)$. Then: $$U-\left\{ \left(0,0\right)\right\} =\bigcup\left\{ A\left(\theta,\eta,r,R\right)\mid\theta,\eta\in[0,2\pi)\cap \mathbb Q\wedge r,R\in\mathbb{Q}_{+}\wedge A\left(\theta,\eta,r,R\right)\subset U\right\} $$ It ...


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The pointwise limit of your indicator functions is the zero-function since given any $\alpha \in \mathbb R$ there is some $N$ with $|\alpha|\notin [n,n+1]$ for all $N \geq n$.


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The Borel $\sigma$-algebra ${\cal B}$ is the smallest $\sigma$-algebra containing the open sets. Suppose ${\cal S}\subset {\cal B}$ is a countable collection of sets such that for any open set $U$ and for any $x \in U$, there is some $A \in {\cal S}$ such that $x \in A \subset U$. Then we have $\sigma({\cal S}) = {\cal B}$. To see this, note that $U = ...


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Take $$B_M = \left\{x\in[0,1] \colon |f(x)|<M\right\}.$$ If $f$ is not infinity almost everywher, then $\mu(B_M)>0$ for some $M$.


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Let $B_n = \{x \in [0,1] \mid |f(x)| \leq n\}$. Then $[0,1] = \bigcup_n B_n$ (why?) with $B_n \subset B_{n+1}$. Now use "continuity of the measure from below" to conclude $\mu(B_n) \to \mu([0,1]) = 1$ to conclude that for every $1>\varepsilon > 0$, there is some $B_n$ with $\mu(B_n) > 1-\varepsilon > 0$. Hence, you can even come arbitraryly ...


2

One of the characteristics of an outer measure is: $$A\subset B\Rightarrow \mu^*(A)\leq\mu^*(B)$$ If $\mu^*$ coincides with $m$ on $C$ then it cannot have that property since: $$\left\{ a\right\} \subset\left\{ a,b\right\} \wedge m\left(\left\{ a\right\} \right)=2>1=m\left(\left\{ a,b\right\} \right)$$


0

Wikipedia answers both of your questions. Specifically, let $X$ be a topological vector space and $M \subset X$ be a subspace. Then $X/M$ in its usual topology is Hausdorff if and only if $M$ is closed. This is essentially because a topological vector space is Hausdorff if and only if $\{ 0 \}$ is closed, and $\{ 0 \}$ is closed in $X/M$ if and only if $M$ ...


2

Let $h_{n}\downarrow0$ and define $$ f_{n}:\left[0,s\right]\to\mathbb{R},t\mapsto\frac{1}{h_{n}}\int_{t}^{t+h_{n}}\left|f\left(u\right)-f\left(t\right)\right|\,{\rm d}u. $$ By the Lebesgue differentiation theorem, $f_{n}\left(t\right)\xrightarrow[n\to\infty]{}0$ for almost all $t\in\left[0,s\right]$. Furthermore, $$ 0\leq ...


2

Let $(A_n)_n$ be as in your question. Then $$\infty = \mu (X) = \mu \left( \bigcup_n A_n \right) = \lim_n \mu (A_1 \cup \dots \cup A_n)$$ so, by the definition of limit, for all $C>0$ there exists $n$ such that $\mu (A_1 \cup \dots \cup A_n) > C$. Finally, note that $\mu (A_1 \cup \dots \cup A_n) < \infty$ since the $A_n$'s have all finite measure. ...


1

As David Mitra suggests in the comments, Apply Littlewood's first principle to find a finite collection of open intervals whose union $U$ satisfies $\mu(U\Delta E)<\varepsilon$. Take $F$ to be $(U\Delta E)^C$ and define the step function to be $1$ on $U$ and $0$ off $U$.


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As pointed out in the comments and answers, there is a disctinction between isomorpshisms of measurable spaces and measure spaces. The former one is defined in the answer by Rafflesia adnoldii, and is very much similar to homeomorpshim in spirit: we just compare two structures on sets being $\sigma$-algebras (rather than topologies). There is a famous Borel ...


2

The theory of outer measures was basically developed by Carathéodory. A reference would be C. Carathéodory, Vorlesungen über reelle Funktionen, 1st ed, Berlin: Leipzig 1918, 2nd ed, New York: Chelsea 1948. Vitali found the first example of a non-measurable set in 1905.


2

If there is a countable dense subset $\{x_n \mid n \in \Bbb{N}\}$, we can define $\mu =\sum_n 2^{-n} \delta_{x_n}$, where $\delta_{x_n}$ is the dirac measure (point mass) at $x_n$. As every open set contains some $x_n$, this does what you want.


0

Hint: it's a simple function, also for simple functions integrals are sums.


0

Shiyu, If m* denotes the Lebesgue outer measure and if the measurability is defined as you did, what you obtain is the Lebesgue measure in $\mathbb{R}$ (a complete measure), which differs from a Borel measure. There are more non-Borel sets and a non-Borel set may still be Lebesgue measurable by a cardinality argument. As others have pointed out, finding a ...


1

Any measurable solution is continuous. Any nonmeasurable solution is very "wild": in particular its graph is dense in $\mathbb R^2$, so it has no one-sided limits anywhere.


2

Let $d=\dfrac{b-a}{n}$ and $a_1 \in [a,a+d]$. Using the stacking argument in the linked question, the measure of the $\{a_1\}$ for which the requirement is true is at least $m([a,a+d]) -\left( m([a,b]) - m(A) \right)$. Using the hint, this is greater than $\dfrac{b-a}{n} - \dfrac{b-a}{2n}=\dfrac{b-a}{2n} \gt 0$. A set with positive measure has at least ...


1

Two measurable spaces $(X,\mathcal {A})$ and $(Y,\mathcal{B})$ are called isomorphic if there exists a bijection $f:X\to Y$ such that $f$ and $f^{-1}$ are measurable (such $f$ is called an isomorphism). This is from the EOM article Measurable space. Same definition is found, e.g., in Real Analysis by Royden. There is a weaker notion of isomorphism ...


0

If I understood you correctly, from the link you've provided it follows that the Riesz measure is just a regular measure, so in particular it is $\sigma$-finite. For the latter reason, $\mu$ has countably many atoms, let them be a separate measure $\rho$. Define $\lambda = \mu-\rho$: clearly it does not have atoms, so you only need to show that $\lambda $ is ...


2

Leta $A_n =\left\{x\in X :f(x)>\frac{1}{n} \right\} .$ If for some $k$ we have $\mu (A_k ) >0$ then $$\int_X fd\mu \geqslant \int_{A_k} fd\mu \geqslant \frac{1}{k} \mu (A_k ) >0 .$$ Therefore $\mu (A_k ) =0 $ for all $k\in\mathbb{N}$ and hence $$\mu (\{x\in X :f(x)>0\} ) =\lim_{n\to \infty } \mu (A_n )=0 .$$


1

I don't know how Lebesgue measure was defined on ${\mathbb T}^d$ in your case. I shall adopt a pedestrian version in the following. It will be sufficient to consider the case $d=1$. Any subset $X\subset{\mathbb T}$ has a representant $\hat X$ in the fundamental domain $[0,1[\ $, and one puts $$\lambda(X):=\mu(\hat X)\ ,$$ where $\mu$ is the usual Lebesgue ...


3

Yes. Let $A_n = \{x : |f(x)| \ge \frac{1}{n}\}$. Note that if $f$ is integrable we must have $\mu(A_n) < \infty$ (Markov's inequality). The sequence $|f| 1_{A_n}$ converges to $|f|$ pointwise and is dominated by the integrable function $|f|$, so we have $\int |f| 1_{A_n} \,d\mu\to \int |f|\,d\mu$. Subtracting, $\int |f| (1 - 1_{A_n}) \,d\mu= \int_{X ...


1

This is true. Just apply monotone convergence or dominated convergence to $|f|$ times the indicator function on $[-n,n]$.


0

Let $f$ be the function defined from $\mathbb{R}^2 - \mathbb{R}_+ \times \{0\}$ to $\mathbb{R}^2$, defined by $$ f(x) = (\|x\|,\omega(x)).$$ Then $f$ is a continuous function; the only thing that really needs to be checked is that $\omega$ is continuous on the set $\mathbb{R}^2 - \mathbb{R}_+ \times \{0\}$ and can be proved by giving an explicit formula ...


1

Take $B$ to be the complement of the linear space generated by your $n$. Clearly, this has full measure. Consider $f$ to be the characteristic function for $B$. This is an invariant function for $T^k_A$, but not necessarily invariant for just $T_A$. Now, what techinique have you learned that makes functions into invariant functions? Play with it a bit.


2

Here's a sketch (the details are easy): Suppose $f$ isn't constant almost everywhere and get two disjoint sets $A, B$ of positive measure such that $f[A] \cap f[B] = \phi$. Next observe that the set $G$ of integer linear combinations of $1, \sqrt{2}$ is a dense additive subgroup of $\mathbb{R}$. It follows that $A + G$ and $B + G$ are conull subsets of ...


0

Ok, I think it is clear to me now... Consider the finite Borel measure: $$\mu(E):=\int_Ee^{-x^2}\mathrm{d}x$$ Restrict it to the ring of bounded Borel sets: $$E\in\mathcal{B}(\mathbb{R}):\quad|E|<\infty$$ Then the alternative construction does not agree with the original one: $$\mu_C'(\mathbb{R})=\infty\neq\sqrt{\pi}=\mu_C(\mathbb{R})$$ Moreover, the ...


1

Your proof by separating $e^{-iu}$ into its real and imaginary parts is fine, well done. But how would you solve this without separating into real and imaginary parts? Pretty much in the same way, we only need to consider the difference quotient $$\frac{e^{-i(t+h)x} - e^{-itx}}{h}$$ and need to show that its modulus is bounded by $C\cdot\lvert ...


1

If we use this with $f_n:=nf$ where $f$ is some integrable function, then the set $\{\delta_n,n\in\mathbb N\}$ has a lower bound, but it is zero. The problem is that this sequence is not Cauchy. However, if we fix $\varepsilon$ then there is an $N_0$ such that $\int_X |f_n-f_{N_0}|\mathrm d\mu\lt\varepsilon$ for each $n\geqslant N_0$. The $\delta$ which ...


1

A tentative for a proof. 1) Suppose in addition to the hypothesis that $f$ is a bounded function (a.e). Then $\displaystyle F(x)=\int_0^x f(t)dt$ is a continuous function on $\mathbb{R}$. If $a\in E=\mathbb{Z}+ \mathbb{Z}\sqrt{2}$, then we get easily as $f(t+a)=f(t)$ that for all $x$: $$F(x+a)=F(x)-F(-a)$$ Putting $x=0$ gives $F(a)=-F(-a)$. Hence ...


1

Hints: By definition, there exists a sequence $(E_n)_n \subseteq \mathcal{E}$ such that $\varrho(E_n) \downarrow \gamma$. Consider $$E_0 := \bigcap_{n \in \mathbb{N}} E_n.$$ Let $A$ be a measurable set such that $\mu(A)=0$. In order to prove absolute continuity, we have to show $\varrho(A \cap E_0)=0$. Show that $\mu(A \cup E_0^c)=0$ and therefore $(A \cup ...


2

The prototypical example in $\mathbb{R}$ is $1/|x|^\alpha$. Varying $\alpha$ puts this function in various $L^p(\mathbb{R})$-s. You can adapt this example of $\mathbb{R}^n$; you should integrate in polar coordinates to see which $\alpha$-s are needed.


0

Ok, so let me introduce the idea of a period and the fundamental period. A function $f$ has period $t$ if for all $x$ in the domain it is true that $f(x+t) = f(x)$. A function is called periodic if it has (at least one) period. Take any periodic function $f$ and define its fundamental period $T=T(f)$ as $T=\inf \{t>0:$ t is a period of $f\}$. Take the ...


0

Yes, $\mu((C+\alpha)\cap C)=0$ for all irrational $\alpha$ and for most rational $\alpha$. This can be inferred from the description of $(C+\alpha)\cap C$ on page 57 of [DH95]. I offer a translation of their terminology ("(1,1)-string sequence", etc). Assume $\alpha$ is not a ternary rational (i.e., does not admit a finite ternary expansion). In the ternary ...


2

It's trivial that $\mu^*(A\cap E)\le\mu^*(B\cap E)$, so let's check the other inequality: $$\mu^*(A\cap E)=\mu^*(A)-\mu^*(A\setminus E)\ge\mu^*(A)-\mu^*(B\setminus E)=\mu^*(B)-\mu^*(B\setminus E)=\mu^*(B\cap E)$$


0

I'm not entirly sure, but lets give it a try.. one inequallity is obvious from monotonic: $\mu(A\cap E)\le \mu(B\cap E)$ for the other one, consider the sub-additivity: $\mu(B\cap E)\le\mu((B\cap E)\cap A)+\mu((B\cap E)\cap A^c)=\mu(A\cap E)+\mu((B-A)\cap E)$ So all's left is to show $\mu(E\cap(B-A))=0$. I think that it should be pretty easy to complete ...



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