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The map $\sum \frac {2a_n} {3^{n}} \to \sum \frac {a_n} {2^{n}}$ ($a_n \in \{0,1\}$ for all $n$) is a continuous map from the Cantor set $C$ onto $[0,1]$. This map extends to a continuous function on $\mathbb R$ since its vales at the end points of the intervals removed in the construction of $C$ are equal). This extended function is called the Cantor ...

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The problem you pose is the weak version of a classical result in harmonic analysis, which does not require the continuity nor the compactness of the support of the datum $f$: in order to prove this stronger result, first note that \begin{align} \int\limits_{\Bbb R} f\big(l(t)\big)\mathrm{d}t=0\:\text{ for every}&\text{ straight line } l\subsetneq\Bbb ... 3 If \mu is counting measure on a finite set then uniform absolute continuity trivially holds for any sequence (f_n): take \delta <1. Obviously, (f_n) need not be L^{1} bounded. 3 As a general rule, anything that you can define explicitly in terms of a measurable function or sequence of functions is going to turn out to be measurable; it takes a little trick to show that not every set is measurable. That's too broad to be provable, but this particular example is trivial. Say f=u+iv where u and v are real-valued. Then \{f(x)\... 3 The support is usually defined as the closure of that set, but I'll answer your specific question. Note that your set is preciselyf^{-1} ((-\infty,0)\cup (0,\infty))=f^{-1}((-\infty,0))\cup f^{-1}((0,\infty)).$$Now, just recall the definition of a Lebesgue measurable function. 3 No, except in the trivial sense that any set is the union of "intervals" that are single points [x,x]. If your intervals are open intervals, any union of intervals is an open set and the intersection of a sequence of these is a G_\delta set. If your intervals are closed intervals (but not singletons), the union of their interiors is an open set, which ... 3 You can easily check the following proposition. Let E\subset \mathbb R^N be a measurabe set. If f and g are two functions defined on E satisfying f=g for a.e.x\in E, then f is measurable if and only if g is measurable. In this problem, we can thus assume f\neq0 on I. Since \frac1x is a continuous function on \mathbb R- \{0\}, \... 2 The second statement is wrong. Consider the characteristic function of a set which is not Lebesgue measurable. The construction of a non-measurable set is classical. For instance, see https://www.math.purdue.edu/~zhang24/NonMeasurableSet.pdf 2 No, it does not. x \mapsto \mu(A\cap [x - 1; x + 1]) is continuous on \mathbb{R} as \forall \epsilon > 0 and \forall x \in \mathbb{R} we have |\mu(A\cap [x - 1; x + 1]) - \mu(A\cap [x + \epsilon - 1; x + \epsilon + 1])| \leq 2\epsilon. Thus, by intermediate value theorem if A \notin \{\mathbb{R}, \emptyset\}, then \exists x \in \mathbb{R}, ... 2 |\varphi|(x)=|\varphi(x)|=|\sum_{k=1}^na_k\chi_{E_k}(x)|. Suppose x\in E_j, 1\le j\le n. Then |\varphi|(x)=|a_j|=|a_j|\chi_{E_j}(x)=\sum_{k=1}^n|a_k|\chi_{E_k}(x). Since this holds for any arbitrary x , we have |\varphi |=\sum_{k=1}^n|a_k|\chi_{E_k}. 2 The end of your proof is incorrect. The problem is that you have proven that for each x and \varepsilon there is a k with a < f_k(x) + \varepsilon. But that does not imply that there is a k with a < f_k(x). For example, consider the case where a = 0 and f_k \equiv 0 for all k. Then a < f_k(x) + \varepsilon for every \varepsilon &... 2 Hint: E turns out to be a G_\delta set. Here is a complete answer. For every k\in \mathbb N, there exists open set E_k with A\subset E_k, such that$$m(E_k)\leq m^*(A)+\frac{1}{k}.$$Now let$$E=\bigcap_{k=1}^\infty E_k,$$then E is a G_\delta set, which is measure and contains A. Since$$m^*(A)\leq m(E)\leq m(E_k)\leq m^*(A)+\frac 1k,$$... 2 The idea is to make an assumption of Corollary 3.3 not satisfied. Let \{x_n\} be an enumaration of [0,1]\cap\mathbb{Q}. Consider$$ E=\cup_{n=1}^{\infty} (x_n-\frac 1{4^n}, x_n+\frac1{4^n}).$$This is an open set with Lebesgue measure at most \sum 2/4^n = 2/3. Also, E is dense in [0,1]. Then m(O_n)\geq 1 for each n. 2 In [0,1] let C be a Cantor like set of positive measure and E=C^{c}. Then m(O_n) \to m(\overline {E})\neq m(E) because \overline {E} is the complement of the interior of C and interior of C is empty. 2 The pointwise limit f surely exists as a measurable function from E to [0,+\infty] as pointwise we have an increasing real sequence that converges to its supremum. You only have to argue that A=\{x: f(x)=+\infty\} has measure 0. If you've done that MCT (e.g. as given here, will do the rest. So suppose it has not, and deduce a contradiction with ... 2 Suppose there exists p\in\mathbb{R}^2 with f(p)\neq 0, then WLOG (by replacing f by -f if necessary) we might assume f(p)>0. But then by continuity of f there exist \varepsilon>0 such that f(x)>f(p)/2 for all x with \|x-p\|\leq\varepsilon, which then gives that the integral of f over every line contained in B(p,\varepsilon) ... 2 You may find this useful for real analysis, and this for functional analysis. Moreover, you can find more video courses here about some other topics in mathematics. If Chinese is OK to you, this course from Shanghai Jiao Tong University is acessible. 1 (Assuming that f(x)=n>0 for all real numbers x.) For m\in\Bbb N, let g_m := n \cdot \chi_{[-m,m]}. Then all g_m are bounded, simple, supported on a set with finite measure, and 0\le g_m\le f. So, by definition of the supremum,$$\int_{\Bbb R} f \geq \int_{\Bbb R} g_m = n \cdot 2m$$for all natural numbers m. Since the right-hand side gets ... 1 Hint: By using translation invariance of Lebesgue's measure and changing to polar coordinates you get$$\int_{B(x,r)}\left(1+\left|y\right|\right)^{\alpha}\,dy= n V_n\int_0^r\left(1+s\right)^\alpha s^{n-1}\,ds,$$where V_n denotes the volume of the n-dimensional unit ball. 1 If you want to prove that something is a vector space, you have to check the axioms https://en.wikipedia.org/wiki/Vector_space#Definition Note here that the elements of your space are functions f. Thus for example if you want to check if the addition of f_1 and f_2 are in \mathcal{L} for f_1,f_2\in\mathcal{L}, you have to check the condition$$\...

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This is a direct result from the definition. We call $f$ a step function if it is a finite sum $$f=\sum_{k=1}^N a_k\chi_{R_k},$$ where each $R_k$ is a rectangle, and the $a_k$ are constants. What you cited is Theorem 4.5. If you feel not comfortable with the choice of $E_n$, we can consider in the setting of $\mathbb R^1$, in which case $f$ a step function ...

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Hint: We have that $[-n, 0)$ is measurable, then $(-\infty , 0) = \bigcup_{n \in \mathbb{N}}[-n, 0)$ is measurable. Hence, $(\infty, 0)^{c} = [0, \infty)$ is measurable To prove that $E_{\alpha}$, $0 < \alpha < 1$, observe that $[0, 1] = \bigcup_{n \in \mathbb{N}}[\frac{1}{2^{n+1}}, \frac{1}{2^{n}}]$. Hence, exist $m \in \mathbb{N}$ such that $\... 1 Not true. It is well known that the set of all rational numbers cannot be written in this form. See Show that$\mathbb Q$is not a$G_{\delta}$-set. and the comments below the question. 1 Yes, there is. Let$E^*$be the subset of$E$where$f'$exist.$\mu(E^*)=\mu(E)$since$f'$exist almost everywhere. Pick any$x_0\in E^*$which is not isolated, so we have a sequence$\{x_n\}_n\subset E^*$that converges to$x_0$.$f'(x_0)$exists. Therefore, $$f'(x_0)=\lim_{n\to\infty}\frac{f(x_n)-f(x_0)}{x_n-x_0}=\lim_{n\to\infty}\frac{0-0}{x_n-x_0}=... 1 No, (1) (and hence (2) as well) cannot happen. Remember that the measurable sets form a \sigma-algebra: in particular, the union of countably many measurable sets is measurable. Suppose E,f are as in (1). We have$$E=\bigcup_{\alpha\in\mathbb{R}}\{x\in E: f(x)>\alpha\}.$$But every real number is bigger than some rational, so this is the same as$$\... 1 Since the measure space is finite convergence in$L^p$implies convergence in$L^1$. So for a given$\epsilon$choose$n_0\in\mathbb{N}$such that$||f_n-f||_1<\frac{\epsilon}{2}$for all$n>n_0$. Now, since$f\in L^1$we know that there is$0<\delta$such that for all$E\subseteq X$with$\mu(E)<\delta$we have$\int_E |f|<\frac{\epsilon}{2}$.... 1 Note that$E=\bigcup\limits_{n=1}^{\infty}E \cap [-n,n]$. Hence, by the continuity of measure,$m(E)=\lim\limits_{n \rightarrow \infty}m(E \cap [-n,n])$. This implies that given an$\epsilon>0$we can find a$n_{\epsilon}\in\mathbb{N}$such that$m(E)-m(E \cap [-n_{\epsilon}, n_{\epsilon}])<\epsilon$. By the additivity of measure the LHS equals$m(E\...

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First of all, your $g^{(-n)}$ is defined only up to a polynomial of degree $n-1$, so it's not quite correct to talk about the $n$-th integral. But, to make a definite choice, one may take definite integrals, say $$g^{(0)}(x):=g(x),\quad g^{(-n-1)}(x):=\int_{0}^{x}g^{(-n)}(y)\,dy,$$ and see what happens. We have in this case $\color{gray}{\text{[induction on$...

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No: let $a_n$ be an enumeration of the rationals in $[0,1]$, and take $A_n=(0,1)\backslash \{a_0, \ldots, a_n\}$. Then $A$ is the set of irrationals in $[0,1]$ thus has empty interior yet full measure.

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The problem is not trivial. Let $\pi_X: X\times Y\rightarrow X$ the projection onto $X$, that is $\pi:(x,y)\mapsto x$. Your question is whether $\pi(E)$, which is $\bigcup_yE_y$ , belongs to $\mathcal{M}$ for any $E\in\mathcal{M}\otimes\mathcal{N}$.The answer is no in general. This can be studied under what is called Analytic sets and universal measurability....

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