I'm wondering if there exists a Lebesgue compact set (with respect to the usual topology on $[0,1]$) in  $[0,1]\setminus \mathbb{Q}$ whose Lebesgue measure is positive. In fact we've just seen the Egoroff theorem in class, and I thought of the functions $$ f_n(x)= \left\{     \begin{array}{ll}         \frac{1}{n} &amp; \mbox{if } x\in[0,1]\setminus \mathbb Q\\         1 &amp; \mbox{otherwise.}     \end{array} \right. $$ Obviously $f_n\to f\; \mu$-a.e., but since $\mathbb Q$ is dense, I'm wondering how we can find a compact set $F\subset [0,1]$ for a given $\delta&gt;0$ such that $\mu([0,1]\setminus F)&lt;\delta$ and $\sup_{x\in F} |f_n(x)-f(x)|\to 0 $ as $n\to 0$. Could you please explain why Egoroff holds in this case?
122585
Does there exist $K \subseteq \mathbb{R} \backslash \mathbb{Q}$ such that $K$ is compact, and has Lebesgue measure greater than $0$? As I have been trying to think of examples, I suspect that any subset of $\mathbb{R} \backslash \mathbb{Q}$ that is closed can be at most countable, since the closure of an uncountable subset of irrationals should contain some rationals. And, the Lebesgue measure of a countable set is $0$. If there are any examples of such a set, I would be very interested to know how it is constructed.
110983
Given a $\sigma$-finite measure $\mu$ on a set $X$ is it possible to formulate a topology on the space of functions $f:X \rightarrow \mathbb{R}$ that gives convergence $\mu$-almost everywhere?  I can't seem to find any way to write this and am suspecting that no such topology exists! Is this true?  If so, is there some generalisation of a topological space where one can make sense of convergence without having open sets?  Any comments, references or tips would be greatly appreciated.
63270
Just having trouble with this problem. First, it says to prove that if $E$ is Lebesgue Measurable, and $\epsilon&gt;0$ is arbitrary, then there is an open $O$ such that $E \subset O$ and $m(O\setminus E)&lt;\epsilon$. Now for this part, since $E$ is Lebesgue measurable, I was able to take the definition of being Lebesgue outer measurable (the infimum of open coverings of $E$) and easily construct $O$ from that, and it works out.  But now I want to show that there is an $F$ closed with $F \subset E$ and $M(E\setminus F)&lt;\epsilon$. The problem is, I can't figure out how to construct this $F$! The idea seems obvious intuitively, but I feel like I am given no tools for constructing a closed subset of an arbitrary set that satisfies these conditions. Am I missing something obvious here?  Thanks!!
121200
(Not mentioned in title but $\epsilon$ is a number greater than $0$ and $E$ a Lebesgue measurable subset of $\mathbb{R}^n$.) I know a question equivalent to this one has been asked (). But it was not answered. Also my thinking is a bit different. So I would love to get some help on this problem. So far I've got:  $$\frac{\lambda(E \cap I)}{\lambda(I)} &gt; 1- \epsilon \iff \frac{\lambda(I)-\lambda(E\cap I)}{\lambda(I)} &lt; \epsilon$$  And also, a theorem states that, for all $\epsilon$ there exists a closed set $F\subset E$ such that $\lambda(E\setminus F) &lt; \epsilon$ so it would sufice to prove  that there exists an interval $I$ such that $$\frac{\lambda(I)-\lambda(E\cap I)}{\lambda(I)}&lt;\lambda(E\setminus F).$$ I'm not sure if this helps me much but any help is appreciated. What would you do?
121971
I was wondering if someone could please give me an example of a sequence of decreasing sets where the first set has infinite Lebesgue measure; i.e., $\{B_{n}\}_{n=1}^{\infty}$ such that $m(B_{1}) = \infty$ but $m(\cap_{n=1}^{\infty} B_{n}) \neq \lim_{n \to \infty}m(B_{n})$?  thank you.
100102
Let $(X, \mathcal{B}, \mu)$ be a measure space and $f$ a measurable function on $X$ and suppose that $\forall E \in \mathcal{B}$ we have that $\int_E f = 0$.  Then I want to show that $f = 0$ almost everywhere (a.e.).   Suppose for sake of contradiction that $f \ne 0$ a.e. Then $\not\exists E \in \mathcal{B}$ s.t. $\mu(E) = 0$ and $f(x) = 0$, $\forall x \in X - E$  Then $\{x : f(x) \ne 0\} = A$ is not measure zero so that either $\mu(A) &gt; 0$ or $A \notin \mathcal{B}$. Now if $\mu(A) &gt; 0$ then it is easy to see that $\int_A f \ne 0$ so that we have a contradiction of our original hypothesis that $\forall E \in \mathcal{B}, \int_E f = 0$. But if on the other hand $A \notin \mathcal{B}$, I cannot no longer appeal to $\int_A f \ne 0$ since $\int_A f$ is non-sense.  So I'm having trouble with this part of the argument.
29696
I'm trying to solve this previous qual problem from my univeristy: Let $L_n$ be the continuous linear functions on $L^\infty(\mathbb{R})$ given by $$L_n(\phi)=\frac{1}{n!}\int_0^\infty x^ne^{-x}\phi(x)\,dx.$$  Prove that $L_n$ has no subsequence that converges in the weak* topology of $L^\infty(\mathbb{R})^\ast$. So far, I managed to see that $\frac{1}{n!}x^ne^{-x}$ converges uniformly on compact sets to 0, hence the weak* limit of any subsequence of $L_n$ must take compactly supported $L^\infty$ functions to 0. I've also tried to construct a test function directly for any subsequence but so far I have yet to come up with such a construction. Can anyone pleas help me? Some hints would be very much appreciated! Thanks!
23547
Possible Duplicate:          Is there a measurable set $E \subset [0,1]$ such that for any $0 &lt; a &lt; b&lt;1$, the Lebesgue measure $$m(E \cap [a,b])= \frac{b-a}{2}?$$  I am stumped and have no idea.
104753
If there is a closed subset $B$ of $\mathbb{R}^d$ and $z\in \mathbb{R}^d$ then show there is a point $b_0\in B$ so that we have $\delta(z,B) = \|z-b_0\|.$  I'm not sure how to begin this exercise and what theorems are available to help me solve it
35734
Let $X=\mathbb{R}$ and let $\mathcal{S}=\{A\subseteq \mathbb{R}:$ either $A$ is countable or $\mathbb{R}\setminus A$ is countable }. Let $f:\mathbb{R}\rightarrow [0,\infty)$ be any function, and define $\mu_{F}:\mathcal{S}\rightarrow[0,\infty)$ by $\mu_{f}(A)=$sup $\{\Sigma_{x\in F}f(x):F\subseteq A$ is finite$\}$ Show that $\mu_{f}$ is a measure on $(\mathbb{R},\mathcal{S})$. Now I know that a measure on $(X,\mathcal{S})$ is a function $\mu:\mathcal{S}\rightarrow[0,\infty]$ such that $\mu(\bigcup_{k=1}^{\infty}E_{k})=\Sigma_{k=1}^{\infty}\mu(E_{k})$, but I am unsure of how to incorporate the disjoint sets into the definition for $\mu_{f}(A)$. I know that if $E_{1}\cup E_{2}\cup...=A$ where each $E_{i}$ is disjoint then we have \begin{equation*} \begin{split} \mu_{f}(E_{1}\cup E_{2}\cup...)=\mu_{f}(A)&amp;=\text{sup}\{\Sigma_{x\in F}f(x):F\subseteq A \space  \text{is finite}\}\\ &amp;=\text{sup}\{f(x_{1})+...f(x_{n}):\{x_{1},...,x_{n}\}\in F\} \end{split} \end{equation*} and somehow the last line should equal $\mu_{f}(E_{1})+\mu_{f}(E_{2})+...$ . Any help would be appreciated. Thank you
73914
In my self-study, I came across the following two interesting, related results:  Let $E$ be Lebesgue measurable, with $\mu(E)&gt;0$ (here $\mu$ denotes the Lebesgue measure). Then:   for any $0&lt;\rho&lt;1$, there exists an open interval $I$ such that $\mu(E \cap I)&gt; \rho \cdot \mu(I)$. In turn we should be able to use (1) with $\rho &gt; 3/4$ to prove that  the set $E-E = \{x-y : x, y \in E\}$ contains an (open) interval centered at $0$ (in particular, if $\rho &gt; 3/4$, the text I am using suggests that $(-\frac{1}{2} \mu(I), \frac{1}{2}\mu(I)) \subseteq E-E$).   I would like to see if anyone visiting today would be up for proving (1) or (2) (inclusive-or). I have tried to work out (1) a few times, but none of my attempts have been satisfactory.
3865
Is $[0,1] \cap \Bbb Q$ a compact subset of $\Bbb Q$?  I get the feeling it isn't compact, but I can't figure out a way to prove this.  I understand that,  in $\Bbb Q$, any open set $U = \Bbb Q \cap O_i$ such that $O_i \subseteq R$.  But I can't figure out a way to construct a finite set that doesn't contain the intersection set.
47979
Let me start with a precise statement of the question.  For a subset $A\subseteq \mathbb{N}$ and a series of positive real numbers $\sum_{n=0}^\infty a_n$, I'll use the notation $\sum_A a_n$ as a shorthand for $\sum_{n\in A} a_n$.  Is there a bijection $f:\mathbb{N}\rightarrow \mathbb{N}$ with the property that for every $A\subseteq \mathbb{N}$, $\sum_A \frac{1}{n}$ converges iff $\sum_{f(A)} \frac{1}{\sqrt{n}}$ converges?  Background: Fix a series of positive terms $\sum_{n=0}^\infty a_n$.  Given a subset $A\subseteq \mathbb{N}$, call it $a_n$-small if $\sum_A a_n$ converges.  The following proposition is easy to prove:  Proposition:  The set $\{A\subseteq \mathbb{N}: A\text{ is } a_n\text{-small}\}\cup \{\mathbb{N}\}$ is a topology of closed sets on $\mathbb{N}$.  I will use the notation $(\mathbb{N},a_n)$ to refer to this topology.  Then one can ask how topological properties of $(\mathbb{N},a_n)$ are related to series properties of $\sum a_n$.  For example, I can show:  Proposition:  The following are equivalent.  $\sum a_n$ converges.  $(\mathbb{N},a_n)$ is discrete.  $(\mathbb{N},a_n)$ is disconnected.  $(\mathbb{N},a_n)$ is Hausdorff    And there are other nice things.  For example, $(\mathbb{N},a_n)$ is compact iff $(\mathbb{N},a_n)$ is cofinite iff $\liminf a_n &gt; 0$. With this language, my question can be reformulated as...  Are $\left(\mathbb{N}, \frac{1}{n}\right)$ and $\left(\mathbb{N}, \frac{1}{\sqrt{n}}\right)$ homeomorphic?  I have made very little progress on this.  Of course, the identity function $i:(\mathbb{N}, \frac{1}{\sqrt{n}})\rightarrow (\mathbb{N}, \frac{1}{n})$ is a continuous bijection, but the inverse map is not continuous.  Also, if there is such a bijection, there there is such a bijection with agrees with $i$ on any preassigned finite set. Edit  I thought I add in a slightly suprising (to me, at least) example when things work out to be homeomorphic. Begin with a convergent series $\sum c_n$. and divergent series $\sum a_n$ with $\lim a_n = 0$.  Create a new series $b_n$ using all the terms of $c_n$ and $a_n$ (in whatever order you wish).  Then $(\mathbb{N}, a_n)$ and $(\mathbb{N}, b_n)$ are homeomorphic.  The idea is that since $\lim a_n = 0$, there is a convergent infinite subseries $\sum_A a_n$.  Then we use a bijection which with $A$ and $A\cup \{\text{indices of }c_n\}$ to &quot;squeeze&quot; the $c_n$ in without changing the topology.  Of course, I am glossing over many details, but I can include them if desired.
15171
This question is exercise 4.10 of Richard F. Bass's Real Analysis for Graduate Students, 2nd edition.  Let $\epsilon \in (0,1)$, let $m$ be Lebesgue measure, and suppose $A$ is Borel Measurable subset of $\mathbb R$. Prove that if $$ m(A\cap I)\leq (1-\epsilon)m(I)$$  for every interval $I$, then $m(A)=0$.  Collection of my thoughts  1. Try to prove by contradiction, suppose $m(A)=a&gt;0$ Borel Measurable Set satisfies the condition given. A property of any Borel Measurable Set on $\mathbb R$ is $A$ contains a closed set $F$ and is contained in open set $O$, and the measure of $F$ and $O$ can be arbitrarily close to $a$. 2. Obviously $A$ contains no interval, but it does not lead to contradiction, because we can find an example of $m(A)&gt;0$ that contains no interval(fat Cantor Set(remove middle 1/4 each time) has measure $1/2$ and contains no interval).  2'. For fat Cantor Set $C$ and given $\epsilon&gt;0$ how could we construct an interval I satisfies $m(C \cap I)&gt;(1-\epsilon)m(I) ?$   Maybe try to write Borel measurable set as countable union/intersection of intervals.   I'm stucked here and will appreciate any help, some examples/special case will be great too.
63245
Let $\gamma : [0,1]\rightarrow\mathbb R^2$ be a continuous curve whose image has positive Lebesgue measure. Must $\gamma$ have self-intersections? Intuitively this seems like it should be true, but I could not find a proof. An equivalent question is whether there is a positive measure subset of $\mathbb R^2$ which is homeomorphic to $[0,1]$.   Note that it is shown here  that $\gamma$ must in fact have lots of intersections if the image of $\gamma$ is a rectangle. But, here I am not making any topological assumptions on the image of $\gamma$. The answer to the linked question does not work in this generality.
116890
Given a positive real number construct a non measurable set of that outer measure   I've been asked this question and am not able to construct it.  I am well aware of the construction of a Vitali set, and I know that their outer measure are not known explicitly.  I can construct a set dense in any interval by taking the union of the standard non measurable set with rationals of that interval but that doesn't serve my purpose because measure of countable set is zero and doesn't do anything good to the required result.
25173
Let $f:R^n \rightarrow R^m$ be any function. Will the graph of $f$ always have Lebesgue measure zero?  $(1)$ I could prove that this is true if $f$ is continuous.  $(2)$ I suspect it is true if $f$ is measurable, but I'm not sure. (My idea was to use Fubini's Theorem to integrate the indicator function of the graph, but I don't know if I'm using the Theorem properly).  If $(2)$ is incorrect, what would be a counterexample where the graph of $f$ has positive measure?  If $(2)$ is correct, can we prove the existence of a non-measurable function whose graph has positive outer measure?