Let $\epsilon \in (0, 1)$, let $m$ be Lebesgue measure, and suppose $A$ is a Borel measurable subset of $\mathbb{R}$. If$$m(A \cap I) \le (1 - \epsilon)m(I)$$for every interval $I$, then does it follow that $m(A) = 0$?
122265
Let $A\subset\mathbb{R}$ a measurable where $0&lt;m(A)&lt;\infty$. Show that exists for each $0&lt;\alpha&lt;1$ an interval $I$ such that  $$ \frac{m(A\cap I)}{m(I)}&gt;\alpha. $$  MY ATTEMPT: Following a hint a give that:  Let $\varepsilon&gt;0$, exists $G$, a open set such that $m(A)\leq m(G)&lt;m(A)+\varepsilon$. As $G$ is open, we can write as the disjoint sum of open intervals $\dot{\bigcup}_n I_n = G$.So, $$ m(A)\leq m(G)=m\left(\sum \dot{\bigcup_n} I_n \right)\leq \sum m\left(I_n \right)&lt;m(A)+\varepsilon $$ Suppose that $\varepsilon=(\alpha^{-1}-1)m(A)$:  $$ m(A)+\varepsilon=m(A)+m(A)(\alpha^{-1}-1)=m(A)(1+\alpha^{-1}-1)=m(A)\alpha^{-1} $$ and $$ \alpha\sum m(I_n)&lt;m(A)=\sum m(A\cap I_n) $$ $$ \Rightarrow \alpha&lt;\frac{\sum m(A\cap I_n)}{\sum m(I_n)}  $$  But I have a insecurity with that:   (1) I can write that: $m(A)=\sum m(A\cap I_n)$?   (2) How to get the result with this: $$ \alpha&lt;\frac{\sum m(A\cap I_n)}{\sum m(I_n)}  $$
77791
I want to prove that a function $f:\mathbb{R}^n\rightarrow \overline{\mathbb{R}}$ that is continuous everywhere except for a set $E$ of Lebesgue measure zero is a Lebesgue measurable function. We know that $f$ is not continuous, so there are some open sets ${U_k}$ in $\overline{\mathbb{R}}$ such that $f^{-1}(U_k)$ is not open in $\mathbb{R}^n$. But $f^{-1}(U_k)\subseteq E$, and so $f^{-1}(U_k)$ is still measurable because it has measure zero. I don't know if this works, because I'm not sure if $E$ contains actually the preimage of those open sets $U_k$.
27690
Is there a measurable set $A$ such that $m(A \cap B)= \frac12 m(B)$ for every open set $B$?  Edit: (t.b.) See also  for further answers.
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.
42425
Let $f$ be a real-valued Lebesgue measurable function on $\mathbb{R}$. Prove that there exist Borel measurable functions $g$ and $h$ such that $g(x)=h(x)$ almost everywhere and $g(x)\le f(x) \le h(x)$ for every $x \in \mathbb R$. I know that $f$ is measurable since there exists a sequence of simple function that converges to $f$. I have no further idea how to tackle this problem.
21076
The solution of problem 6 (page 3)  uses a measurable set $E\subset\mathbb{R}$ such that $0&lt;\mu(E\cap I)&lt;\mu(E)$ for all interval $I\subset\mathbb{R}$. Could someone give me an example of a set that satisfies this condition?  ($\mu$ is the Lebesgue measure)  Thanks.
122149
Let $n$ be a positive integer.  $\;  $  Let $f : \mathbb{R}^n \to \mathbb{R}^n$ be everywhere .  Let $S$ be a subset of $\mathbb{R}^n$ with  zero.  Does it follow that $\{f(s) : s\in S\}$ has Lebesgue measure zero?  (I know it would if $D(f)$ is continuous.)
46567
Let $(X, \mathbb{X}, \mu)$ be a finite measure space and let $1 \leq p&lt; \infty$. Let $\phi$ be continuous on $\mathbb{R}$ to $\mathbb{R}$ and satisfy the condition: $(*)$ there esists $K&gt;0$ such that $|\phi (t)|&lt;K|t|$ for $t \geq K$. Show that $\phi \circ f$ belongs to $L_p$ for each $f \in L_p$ . Conversely , if $\phi$ does not satisfy $(*)$, then there is a function $f$ in $L_p$ on a finite measure space such that $\phi \circ f$ does not belong to $L_p$.  Any suggestion? I tried to use $|\phi (t)|&lt;K|t|$ exchanging $t$ by $f(t)$ and I calculated $||\phi \circ f||_p$, but without success.
77245
I am a graduate student at Iowa State University attempting to return after a five-year hiatus and take the Real/Complex Analysis qualifier on January 8 for potential reinstatement. Since the professors here are on hiatus, I hope you guys don't mind if I bombard the board with a few past questions from our quals here, like this one from Spring 2013:      Let $m$ denote Lebesgue measure on $\mathbb R$ and $M$ the $\sigma$-algebra of Lebesgue measurable subsets. Given $E \in M$ with $m(E) &gt; 0$ and $\alpha \in (0, 1)$, prove that there is an interval $I$ such that $m(E \cap I) &gt; \alpha \, m(I)$.   I've attempted to force the issue using the definition of outer measure but it doesn't seem to work. I've also attempted using the definition of Lebesgue measurability but that doesn't seem to work either - so I'm at a loss!   Thanks in advance!   -Darrin Rasberry
30733
Suppose that $f\colon[a, b] \to \mathbb{R}$ is a function of bounded variation. Show that $f$ is Borel measurable.  I was wondering if I could get a hint.
49184
This is from problem $8$, Chapter II of Rudin's Real and Complex Analysis.  The problem asks for a Borel set $M$ on $R$, such that for any interval $I$, $M \cap I$ has measure greater than $0$ and less than $m(I)$.  I was thinking of taking the Cantor approach: taking $R$ to be the union of $[a,b]$ with $a$ and $b$ rationals, and for each $[a,b]$ we construct Cantor sets inside it. During theconstruction of each Cantor set, in order to have positive measure on it, we need to take off smaller and smaller intervals from it, namely the proportion goes to $0$. As a result, these Cantor sets are extremely "dense" on their ends. If for an interval $I$ it intersects with the Cantor set on $[a,b]$ while $b-a&gt;&gt;m(I)$, we shall expect the measure of intersection to be rather close to $m(I)$ and then we lose control on these cases.  Is there any way to fix this or shall I consider other approaches?  Thank you
80138
Suppose $E \subset \mathbb{R}$ is Lebesgue measurable. Define $$ E^2 = \{x^2 : x \in E\}. $$ Is $E^2$ Lebesgue measurable as well?  I believe the answer is yes, but I am struggling to prove it. I tried forming interval covers of $E$ for which the sum of the lengths of the intervals is very close to $m(E)$, then considering the square of these intervals. However, if $E$ is unbounded, there is no guarantee these intervals will remain "small".
53080
(1)  If $E$ is a subset of $\mathbb R$ with finite outer measure, i.e. $m^{*}(E) &lt;\infty$; and   (2) $E$ is not Lebesgue measurable, i.e. there exists $F$ such that  $m^{*} (F) &lt; m^{*}(EF) + m^{*}(FE^{c}).$  [Claim] There exists an open set $O\supset E$ with finite outer measure, such  that   $$m^{*}(O E^{c})  &gt; m^{*} (O)  - m^{*} (E).$$  My  questions are the following.  (1) Is the above claim correct? I've seen this in Royden's book, but have a bit concern. Note that a set $E$ is Leb measurable if  it satisfies caratheodory condition: $m^{*} (F) = m^{*}(EF) + m^{*}(FE^{c})$ for  all subset $F$. If the claim is yes, it seems that measurability condition can  be reduced from all subset $F$ to Borel set $F$.  (2) If the claim is correct, a proof is needed.
79526
Suppose $A=\{S\;|\;S \subset \mathbb R^n, S\text{ is Lebesgue measurable}\}$. What is the cardinality of $A$? Is it the same as the cardinality of all of the real numbers?
100435
Let $f\in C^2\{[0,1],\mathbb{R})$ and $m\{x\in [0,1]:f(x)=0\}&gt;0$. Prove that $$m\{x\in [0,1]:f'(x)=0\}&gt;0,$$ where $m$ denotes Lebesgue measure.  I don't have any clue to solve this exercise..
22070
$\mu^*$ is an outer measure on $X$ and $\{A_j\}_{1}^\infty$ is a sequence of disjoint $\mu^*$-measurable sets. Prove that $\mu^*(E\cap(\bigcup_1^\infty A_j))=\sum_1^\infty \mu^*(E\cap A_j)$ for any $E\subset X$.   This is what I have shown so far:   $\mu^*(E \cap (\bigcup_1^\infty A_j))=\mu^*(\bigcup_1^\infty (E \cap A_j))\leq \sum_1^\infty \mu^*(E\cap A_j) ;$ since $\mu^*$ is an outer measure.   If I can show the reverse inequality I will win. Can anyone help me?
25170
In a space with measure $1$, $||f||_p$ is an increasing function with respect of $p$. To show that $\lim_{p \rightarrow \infty} ||f||_p=||f||_{\infty}$ we have to show that  $||f||_{\infty}$ is the supremum, right??   To show that, we assume that $||f||_{\infty}-\epsilon$ is the supremum.  From the essential supremum we have that $m(\{|f|&gt;||f||_{\infty}-\epsilon\})=0$.  So, we have to show that $m(\{|f|&gt;||f||_{\infty}-\epsilon\})&gt;0$.   Let  $A=\{|f|&gt;||f||_{\infty}-\epsilon\}$.   We have that $\int_A |f|^p \leq \int |f|^p \leq ||f||_{\infty}^p$.   $\int_A |f|^p &gt;\int_A (||f||_{\infty}-\epsilon)^p=(||f||_{\infty}-\epsilon)^p m(A)$   So, $m(A)^{1/p} (||f||_{\infty}-\epsilon)&lt;||f||_p \leq ||f||_{\infty}$  How could we continue to show that $m(A)&gt;0$??   EDIT:   Is it as followed??  We have that $0&lt;||f||_{\infty}-\epsilon&lt;||f||_{\infty}$ for some $\epsilon&gt;0$.  $||f||_{\infty}$ is the essential supremum. So, from the definition we have that $m\left ( \{|f(x)&gt;||f||_{\infty}-\epsilon\} \right )&gt;0$.   Let  $A=\{|f|&gt;||f||_{\infty}-\epsilon\}$.   We have that $\int_A |f|^p \leq \int |f|^p \leq ||f||_{\infty}^p$.   $\int_A |f|^p &gt;\int_A (||f||_{\infty}-\epsilon)^p=(||f||_{\infty}-\epsilon)^p m(A)$   So, $m(A)^{1/p} (||f||_{\infty}-\epsilon)&lt;||f||_p \leq ||f||_{\infty}$  Taking the limit $p \rightarrow +\infty$ we have the following:  $$\lim_{p \rightarrow +\infty}m(A)^{1/p} (||f||_{\infty}-\epsilon)&lt;\lim_{p \rightarrow +\infty} ||f||_p \overset{ m(A)&gt;0 \Rightarrow \lim_{p \rightarrow +\infty}m(A)^{1/p}=1}{\Longrightarrow} \\ ||f||_{\infty}-\epsilon&lt;\lim_{p \rightarrow +\infty} ||f||_p$$   Is this correct?? How do we conclude that $\lim_{p \rightarrow +\infty} ||f||_p=||f||_{\infty}$ ??