# Tag Info

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 ...

3

$\def\rr{\mathbb{R}}$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 $f(r) = b$.

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 ...

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 $[b - \varepsilon, b + \... 2 Let$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 \...

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.

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 $\mu(... 2 The answer is that we cannot find such subsets$A$and$B$. Of course,$A$and$B$need to be Lebesgue measurable so that$\mu (A)$and$\mu(B)$are well-defined. Let us defined$\nu (C) = \mu (A \cap C)$for all Lebesgue measurable subsets$C$. Then:$\nu$is a measure on Lebesgue-measurable sets, as it satisfies all axioms;$\nu(C) = \mu(C)/2$for all ... 2 You may try proving and using the following dual to the monotone convergence theorem: Suppose$\{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^+$, ... 2 Yes, we can choose such a$K$, or just notice that $$\lim_{K\to +\infty}K^{1-p}A=0$$ and that the bound of$\mathbb E\left(\left|X\right|;\left|X\right|\gt K\right)$is uniform with respect to$X\in\mathcal C$. The follows from the inclusion $$\left\{\left|X\right|\gt K+k^{-1}\right\}\subset \left\{\left|X-X_n\right|\gt k^{-1}\right\}.$$ To see this, it ... 2$\mathcal{C}$is UI$\Rightarrow(i)$Let$\epsilon>0$. Then, since$\mathcal{C}$UI there exists$K>0$s.t. $$E[|X|] = E[|X|1_{|X|>K}+|X|1_{|X|\le K}]\le \epsilon + K$$ for all$X\in\mathcal{C}$. Thus$(i)$.$\mathcal{C}$is UI$\Rightarrow(ii)$You want to find for a given$\epsilon>0$a$\delta>0$such ... 2 Since$\|f - f_n \|_p \to 0$, we can extract a subsequence$f_{n_j}$so that$\| f - f_{n_j} \|_p \le \frac{1}{2^j}$. Put $$g = \lvert f \rvert + \sum^\infty_{j=1} \lvert f - f_{n_j} \rvert.$$ Then$g \in L^p$and by the triangle inequality we have $$\lvert f_{n_j} \rvert \le g \,\,\, (\text{almost everywhere}).$$ From this subsequence, you can extract a ... 2 Suppose, for the moment, that the family$\{f_n\}_n \cup \{f\} \subseteq L^1$is uniformly integrable, i.e. that for any$\epsilon>0$there exists$\delta>0$such that $$\sup_{n \in \mathbb{N}} \int_A |f_n| \, d\mu + \int_A |f| \, d\mu < \epsilon \tag{1}$$ for any Borel set$A$with$\mu(A)\leq\delta$(here, and in what follows,$\mu$denotes a ... 2 I believe your approach is correct. I wrote something up before realizing you had already provided a proof (more-or-less what you have, just by proving the contrapositive): Let$E_{n,\epsilon}=\{x\colon|f_{n}(x)-f(x)|\geq\epsilon\}$. Suppose$f_{n}$does not converge to$f$in measure so that there exists an$\epsilon>0$and$\delta>0$with$\mu(E_{n,\...

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 $$\mu(E) = \lim_{n\to \infty} \mu(E_n) = \lim_{n\to \infty} \left (\mu(K_n) ... 1 Usually, if \mu  is a measure and f  is a function, then f\mu  is the measure given by$$ (f\mu)(A) =\int_A f \, d\mu. $$This is indeed a measure as long as f \in L^1 (\mu) or if f : X \to [0,\infty] . 1 Your proof of item a is correct. I just copy it here for sake of completeness. Your proof of item needs a little adjustment, mainly for clearness. Theorem 2.26 - If f\in L^1(\mu) and \epsilon > 0, then a.) there is an integrable simple function \phi = \sum_{1}^{n}a_j\chi_{E_j} such that \int |f - \phi|d\mu < \epsilon. b.) If \mu... 1 @Wolfy , From Theorem 2.15 we know$$\int \sum_{1}^{\infty}|f_j| = \sum_{1}^{\infty}\int |f_j| $$So, of course, we also know$$\int \sum_{1}^{\infty}|f_j| \leq \sum_{1}^{\infty}\int |f_j| $$but we don't need this inequality since we know the equality holds. Here is the proof with some adjusments / clarifications Theorem 2.25 - Suppose that \{f_j\}... 1 @Wolfy, It is true: since we can deduce Fatou's lemma from the Monotone Convergence Theorem, and the Dominate Convergence Theorem is a consequence of Fatou's lemma, we can prove the Dominate Convergence Theorem from the Monotone Convergence Theorem. However, such path would make us to prove again (inside the proof of the Dominate Convergence Theorem) ... 1 Proposition 2.22 - If f\in L^1, then |\int f|\leq \int |f| Proof: 1. Note that if a,b \in \mathbb{R} and a, b>0 then |a-b|\leq a+b. So, if f if a real-valued function then$$\left|\int f\right| = \left|\int f^+ - f^-\right|=\left|\int f^+ - \int f^-\right|\leq \int f^+ + \int f^- = \int (f^+ + f^-)= \int |f|$$2. If f is complex valued,... 1 I have made the required adjustments /corrections in your proof. Proposition 2.21 - The set of integrable real-valued functions on X is a real vector space, and the integral is a linear functional on it. Proof - To prove that the set of integrable real-valued functions on X is a real vector space, it is enough to prove that for all a,b\in \... 1 Use the Co-area formula with g=|\nabla u|^{-1} \chi_{\Omega_t} to get$$ \int_{\Omega_t} dA=\int_t^\infty \left(\int_{\partial \Omega_x} \frac{ds}{|\nabla u|}\right)dx.$$Differentiating this with respect to t gives the formula you're looking for. 1 Theorem: Let (X,\mathcal{M}) and (Y,\mathcal{N}) be given. Assume \mathcal{N} is generated by \mathcal{E} (i.e. \mathcal{N}=\sigma(\mathcal{E})). If f\colon X\rightarrow Y satisfies$$ E\text{ is in }\mathcal{E}\implies f^{-1}(E)\text{ is in }\mathcal{M}, $$f is measurable. Proof: Define the pullback$$ f_{*}(\mathcal{M})=\left\{ E\...

1

The usual definition of $\sigma$-algebra states that it contains the empty set. An algebra of sets is closed under intersection and complement, so if it contains $A$ it contains $A \cap A^c = \emptyset$.

1

Both results are actually equivalent. You can prove one from the other. Regarding the first result: Let $\mathcal{C}$ be a class of subsets of $\Omega$ under finite intersections and containing $\Omega$. Let $\mathcal{B}$ be the smallest class containing $\mathcal{C}$ which is closed under increasing limits and by difference. Then $\mathcal{B} = \... 1 Congratulations! Your proof 2 is more direct, and it is correct. If$\{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$\{f - f_n\}_n\subset L^+then applying Fatou's lemma we have \begin{align*} &\int (\liminf(f-f_n))\... 1 The connection with the spectral measureP$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$Pis regular in the strong topology, which gives you \begin{align} P[a,b]x & = \lim_{\... 1 Sayf\in C_c(X)$. Standard construction: For$n=1,2\dots$and$j\in\Bbb Z$define $$E_{n,j}=\{x:j/n<f(x)\le(j+1)/n\}$$and set $$\phi_n=\sum_{j\in\Bbb Z}\frac jn\chi_{E_{n,j}}.$$Then$\phi_n$is$\sigma(\mathcal E)$-measurable and$\phi_n\to f$(uniformly). 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.

Only top voted, non community-wiki answers of a minimum length are eligible