New answers tagged

1

...but the union is our box before is not correct. In the two dimensional case, let $B=[0,1]\times[0,1]$ $B_1=[0,1]\times\{0\}$ $B_2=\{0\}\times[0,1]$ Then $B$ is a square, of which $B_1,B_2$ are two sides, and $B\neq B_1\cup B_2$. Higher dimensions are similar.


1

$B_1 \cup B_2 \neq B$. For example $(a_1, \dots, a_b) \notin B_1$ and $(a_1, \dots, a_b) \notin B_2$ while $(a_1, \dots, a_b) \in B$.


4

Consider any $A\subset[0,1/2)$. Let $\bar{A}=[0,1/2)\setminus A$. (I think the question is strange with the element $1/2$, but we are not too much interested in this one point so I ignore) Let $X=A\cup(1-\bar{A})$. Then, every condition you mentioned is satisfied. However, to make your expectation true, we should be able to say that every subset of $[0,1/2)...


1

For $n \ge 2$ take $$U =\{(x_1, \dots x_n) \in \mathbb R^n \mid x_1 >0 \textrm{ and } x_2^2 + \dots x_n^2 < e^{-x_1}\}$$ $U$ is open as it is the intersection of two open subsets. Namely the half open plane $\{(x_1, \dots x_n) \in \mathbb R^n \mid x_1>0\}$ and the inverse image of the open subset of the reals $\{x \in \mathbb R \mid x>0\}$ ...


2

There clearly is no such example if $n=1$. For $n=2$ a simple example would be $\{(x,y):|y|<1/(x^2+1)\}$. You could give an analogous example for $n>2$. Now it turns out that "I need easy examples of unbounded open and connected subsets of $\mathbb{R}^n$ with finite Lebesgue measure." is not what the OP actually wanted; he says "It is interesting to ...


1

The importance of absolute continuity comes from the Radon-Nikodym theorem. It states if $\mu \ll \nu$ then $\mu$ has a density function with respect $\nu$. A lot of the time it's much easier to prove things if all the distributions we work with have a density function, i.e. are dominated by some common measure. It's also possible phrase absolute continuity ...


0

I owe the idea of this answer to the earlier answer by Andreas Blass. I will post this answer because it took me some effort to formalize the argument made by him. Lemma. Let $(X,\mathcal M, \mu)$ and $(Y,\mathcal N,\nu )$ be complete $\sigma$-finite measure spaces, and let $(X\times Y, \mathcal L, \lambda )$ be the completion of $(X\times Y, \mathcal M\...


0

We exploit the Zeno's paradox and the Heine Borel theorem. Suppose that $\{(a_i,b_i]\}_{i=1}^\infty\subset\mathcal S$ is disjoint and its union equals $(a,b]\in\mathcal S$. We clearly have $\sum_{i\leq N}(b_i-a_i) \leq b-a$ for all $N$, so $\sum_{i}(b_i-a_i) \leq b-a$. To prove the reverse inequality, let $\varepsilon>0$. Define $$a_i'=a_i-2^{-(i+1)}\...


0

The answer to the title question is that no, a Borel measurable subset of $\mathbb{R}^2$ need not contain a measurable rectangle of positive measure. See the first answer to the following: Measurable rectangles inside a non-null set.


0

I am pretty sure this is a homework question, so I will just leave an outline and you will need to fill in the gaps. Assuming your test function $\phi$ is locally Lipschitz (which will be the case if it is continuously differentiable), then $\|\phi(x)-\phi(x_0)\|<C(\|x-x_0\|)$ for all $x\in B(x_0,1/n)$ with $n$ sufficiently large. Now combine the ...


5

You correctly demonstrated with a counterexample that $$ \left\{ {x \in \left[ {0,1} \right]|f\left( x \right) \ge {{\left\| f \right\|}_\infty }} \right\} $$ does not necessarily have measure zero. Correct would be that $$ E = \left\{ {x \in \left[ {0,1} \right]|f\left( x \right) \color{red}{>} {{\left\| f \right\|}_\infty }} \right\} $$ has measure ...


5

Well, yes. Just take any set $B$ of Lebesgue measure $m$ (e.g. $B=[0,m]$) and consider $$A := B \cup (\mathbb{Q} \cap [0,1]).$$ The set has Lebesgue measure $m$ and it is dense in $[0,1]$. It's a bit more tricky to construct an open dense set $A$ with small Lebesgue measure $m$. Here, one approach is to consider an enumeration $(q_n)_{n \in \mathbb{N}}$ ...


7

The answer is yes. For $m\in [0,1]$ consider the set $A:=[0,m]\cup(\mathbb{Q}\cap [0,1])$. This is clearly dense and has measure $m$.


2

Sure. Take $$ [0,m]\cup(\mathbb{Q}\cap [0,1])=[0,m]\overset{\cdot}{\cup}(\mathbb{Q}\cap(m,1]). $$


2

Your solution seems fine. For a bit of further simplification, we may do as follows. Write $I_n$ for the integral. Then \begin{align*} &\left| I_n - \int_{1}^{\infty} \frac{\cos(n\pi x)}{x^{2}} \, \mathrm{d}x \right| \\ &\leq \int_{0}^{1} \left| \frac{x^{n-2}\cos(n\pi x)}{1+x^n}\right| \, \mathrm{d}x + \int_{1}^{\infty} \left| \frac{x^{n-2}}{1+x^n} -...


1

Hint: If $x \geq 1$ then $$ \frac{x^{n-2}}{1+x^n} \leq \frac{1}{x^2}$$ and if $0\leq x \leq 1$ then $$ \frac{x^{n-2}}{1+x^n} \leq x^{n-2} \leq 1 $$


1

Hint: apply Fatou's lemma to the sequence of functions $f_k = f 1_{E_k}$.


3

Hints: Since the intervals $[n,n+1)$ are disjoint, it holds that $$|f| = \sum_{n =1}^{\infty} |a_n| \chi_{[n,n+1)}.$$ Conclude that $$\int |f| = \sum_{n=1}^{\infty} |a_n|$$ (e.g. using monotone convergence). This gives you a sufficient and necessary condition for integrability of $f$. To show that convergence of $\sum_{n=1}^{\infty} a_n$ is not sufficient, ...


1

Too long for a comment. A density of a set can be defined by means of any net in $\mathcal P_f(S)$ (see Definition 2.1 of the paper). But I guess that from the amenability point of view a purpose of Følner nets is to define a left-shift invariant measure on the family of subsets of $S$, see, for instance, paragraph after Definition 1.6, Theorems 4.5 and 4....


1

Which is the definition of outer measure which you are using, in what space is it defined and which is the function that induces the outer measure? The most common example which I think you are using is the outer measure in $\mathbb{R}$, which is defined by \begin{equation}\tag{1} m^*(A)=\text{inf }\bigg\{\sum_{j=1}^{\infty} \lambda(I_j): A\subseteq \...


0

When thinking about integration on exotic spaces, as a general rule, it's "easy" to deal with exotic ranges and "harder" to deal with exotic domains. In particular, integrating maps from $\mathbb{R}$ to $X$ is not really any more difficult than integrating a function from $\mathbb{R}$ to $\mathbb{R}$ (the tricky part in either case is defining the Lebesgue ...


1

The reason why you can do it with intervals is because the definition of this outer measure is: (assuming that you already know how it is defined for intervals) $m^*(A)=\inf\{\sum_{n=1}^\infty m^*(I_n): A\subseteq \cup_{n=1}^\infty I_n, \ I_n$ are intervals$\}$ So obviously if $\cup A_n\subseteq \cup I_m^n$ when $(I_m^n)_{n=1}^\infty$ is a sequence of ...


0

Claim: if $X$ is a compact metric space and $\mu$ is a finite measure on $\mathcal{B}(X)$, the Borel sets of $X$, then $L^p(X, \mathcal{B}(X), \mu)$ is separable for each $1\leq p<\infty$. We can prove it as follows: For each $n\in \mathbb{N}$, consider a finite collection of balls of radii $1/n$ that cover $X$, and let $\mathcal{P}_n$ be the partition ...


3

This is related to the "duality characterization" of weak $L^2$- space. For a given $\lambda>0$ (which will be chosen later), let $\displaystyle E = \left(E\cap \{|f|>\lambda\}\right) \cup \left( E\cap \{|f| \le \lambda\}\right) = E' \cup E'' $. Then we have \begin{align*} \left|\int_E f\right| \le& \int_{E'} |f| + \int_{E''}|f| \\ \le& \int_\...


3

The theorem is proved by applying the second mean value theorem for integrals (given that $\Phi$ is monotone and $f$ is simply integrable). For $c_2 > c_1 > a$, there exists $\xi \in (c_1,c_2)$ such that $$\tag{*}\left|\int_{c_1}^{c_2}\Phi(x) f(x) \, dx\right| = \left|\Phi(c_1)\int_{c_1}^{\xi}f(x) \, dx + \Phi(c_2)\int_{\xi}^{c_2}f(x) \, dx \right| ...


0

Just filling the details of Michael's answer: If $A\in\mathcal{L}'$, then consider the family $\mathcal{C}$ of every sequence of products of open intervals $\{R_i\}_{i=1}^{\infty}$ such that $A\subset\bigcup_{i=1}^{\infty}R_i$. Now, by definition of the Lebesgue outer measure, $\lambda^*(A):=\inf\left\{\sum_{i=1}^{\infty}\mathrm{vol}(R_i):\{R_i\}_{i=1}^{\...


1

Let $U$ be open. The collection of all closed cubes which are contained in $U$ is a Vitali cover of $U$. Hence there is a sequence of disjoint closed cubes $B_i$ each contained in $U$ such that $m(U\setminus \cup_i B_i)=0$. Now $\mu (U) \geq \mu (\cup_i B_i)= \sum \mu (B_i) \geq \sum m (B_i)=m(\cup_i B_i) =m(U)$.


0

Cover $ A $ with a countable union of intervals $\bigcup_iI_i \supseteq A$ such that $\mu (A) + \epsilon \sum_i \mu (I_i)> \sum_i \mu (I_i)$ $\sum_i \mu (A \bigcap I_i) \ge \mu (A) > (1-\epsilon) \sum_i \mu (I_i)$ This implies there exists $ I_i $ such that $\mu (A \bigcap I_i) >(1-\epsilon) \mu (I_i)$ However this does not imply Lebesgue density ...


1

You can use the canonical identification $\mathbb C = \mathbb R^2$, which happens to preserve measure, together with the fact that the imaginary axis is the zero set of the nonconstant real analytic function $f(x,y)=x$.


0

$A$ has a measurable superset $B$ whose Lebesgue measure equals the outer measure of $A$. In other words, $B-A$ has inner measure $0$. Then, for any interval $I$, $(B\cap I)-(A\cap I)$ also has inner measure $0$, and so $\mu(A\cap I)=\mu(B\cap I)$. Apply the Lebesgue density theorem to $B$ and get the result you want for $A$.


1

Take $E=[0,1]$ and, for each $x\in\mathbb R$, take $y=x-\lfloor x\rfloor$. Then $x-y\in\mathbb Z\subset\mathbb Q$.


2

Hint: given any interval $I_0$, find a nested sequence of intervals $I_0 \supset I_1 \supset I_2$ such that, for every $j\ge0$, there exists $n(j)$ such that $$ I_j \subset \bigg[ q_{n(j)}-\frac1{{n(j)}^2},q_{n(j)}+\frac1{{n(j)}^2} \bigg]. $$ The density of the rational numbers in $\Bbb R$ should help you construct such intervals. Do you see why we have $f(x)...


2

Enumerate the rationals $q_0, \ldots, q_n, \ldots$. Then consider the balls $B_{2^{-n-3}}(q_n)$. Then $\lambda(\bigcup_{n \in \mathbb{N}} B_{2^{-n-3}}(q_n)) \leq \sum_{n \in \mathbb{N}} \lambda(B_{2^{-n-3}}(q_n))) = \sum_{n \in \mathbb{N}} 2^{-n-3} = \frac{1}{4}$. $M = \bigcup_{n \in \mathbb{N}} B_{2^{-n-3}}(q_n)$ is one of the sets you are searching.


0

$E(\delta;k)=\cup_F \cap_{n \in F} (E\cap {2^{-n}} E)$ where the union is taken over all finite subsets of $\mathbb Z$ with cardinality $k$.


0

No.Think about $f(x)=\frac{1}{\sqrt{x}}\ge 0$ on $(0,1)$. It's integrable but not bounded a.e.


1

I disagree with your assessment that the integral over a set of zero measure should be nonzero or undefined; this is not intuitive to me. Here is why. For simplicity, consider $[0,1]$. Let $A \subseteq [0,1]$ be a measurable set, and let $f \equiv 1$. Then: 1) As we agreed above, intuitively, the lebesgue measure of A is the probability $P(x \in A)$ where ...


1

What a coincidence! I just have finished this exercise today! You need to refer to section 5 in Chapter 6 to find the definition of "measure-preserving transformation". In this exercise, our measure space is $(X,\mathcal M,\mu)=(S^{d-1},\mathcal M,\sigma)$. Given any set $E\subset S^{d-1}$ we let $\widetilde{E}=\{x\in\mathbb R^d: x/|x|\in E,0<|x|<1\}$....


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