$\quad A\subset\mathbb{R}$ is measurable and $m(A)&gt;0$. then $m\left(\mathbb{R}-\bigcup_{q\in\mathbb{Q}}q\cdot A\right)=0$.  How to prove it ?  $ q \cdot A $ is a dilation or contraction of each element in $ A$ for example if $ A=(1 ,3],  2A= 2(1 ,3] = (2,6].$ The proof  is trivial if $ A $ is an interval or countable union of intervals but not so for arbitrary sets.  This question is not a duplicate of  because taking logarithm $ log_{10} (qA) =log_{10}A+log_{10}q$ , $ log_{10} q $ is not always a rational number
121294
Let $A\subset \mathbb{R}$ be of positive Lebesgue measure, i.e. $\mu(A)&gt;0$. Is it then true that $\mu(\mathbb{R}\setminus (A+\mathbb{Q})) = 0$?   I am quite sure that if $\mu(A)&gt;0$, then $A-A$ contains a rational, as well as an irrational number, but I'm not sure if this actually helps.
20983
Suppose $\mu$ is a Radon measure on $X$ such that $\mu(\left\lbrace x \right\rbrace)=0$ for all $x \in X$, and $A \in \mathbb{B}_X$ satisfies $0 &lt; \mu(A) &lt; \infty$. Then for any $\alpha$ such that $0 &lt; \alpha &lt; \mu(A)$ there is a Borel set $B \subset A$ such that $\mu(B) = \alpha$   I'm not even sure how to start this problem. It seems like a nice result, but any hints on how to start?
115389
Suppose that $(X,\mathcal{E},\mu)$ is a non-atomic finite measure space (i.e. for every $E \in \mathcal{E}$ with $\mu(E)&gt;0$ there exists $F \subset E$ measurable such that $0&lt;\mu(F) &lt;\mu(E)$.)  a) Prove that for every $ \varepsilon &gt;0$ there is a finite partition of $X$ in measurable subsets $X_1,..,X_n$ such that $\mu(X_i)\leq \varepsilon$.  b) Prove that for every $\alpha \in [0,\mu(X)]$ there exists $E \in \mathcal{E}$ with $\mu(E)=\alpha$.    I guess that a) is given to prove b) more easily. I have the following idea of solution for a) (inspired from , which proves b)  )  Denote  $$ \Gamma = \{ (X_1,..,X_n) : n &gt;&gt; \mu(X)/\varepsilon, X_i \text{ are disjoint }, \mu(X_i)\leq \varepsilon \}$$ ordered by componentwise inclusion. Totally ordered parts $(Y_\alpha)$ of $\Gamma$ have an upper bound element $(\bigcup Y_\alpha^i)_{i=1}^n$ which is still in $\Gamma$.   By Zorn's lemma $\Gamma$ has maximal elements. If a maximal $(X_1..X_n)$ element is not a partition, then we can replace it with something better, of the form $(X_1,..,X_n\cup A)$ where $A \subset X \setminus (X_1 \cup..\cup X_n)$ and $\mu(A)&gt;0$.     My questions are:  1) Is my solution of a) correct? I feel that the part with the upper bound element may not work, since there might be noncountable unions.  2) What is a simpler solution of a)?
92071
I have to show that $E=\{(x,\alpha)\mid  0\leq \alpha&lt;|f(x)|\}$ is measurable if $f$ is measurable.   My attempt :  If $f=\boldsymbol 1_F$ where $f$ is measurable, then $$E=\big((\mathbb R\cap F)\times [0,1[\big)\cup \big((\mathbb R\cap F^c)\times \{0\}\big)$$ which is measurable since it's intersection, union and product of measurable set. If $f$ is simple (remark that $f$ simple $\iff$ $|f|$ simple) i.e. $$|f|=\sum_{i=1}^n a_i\boldsymbol 1_{F_i}$$ where $a_i&gt;0$ and $F_i$ measurable, we obtain in the same way $$E=\bigcup_{i=1}^n\big((\mathbb R\cap F_i)\times [0,a_i[\big)\cup \bigcap_{i=1}^n \big((\mathbb R \cap F_i^c)\times \{0\}\big)$$ which is also measurable since it's finite union, intersection and product of measurable set. Finally, if $|f|\geq 0$ we have a sequence of simple function $\{\varphi_k\}_{k\in\mathbb N}$ s.t. $$\varphi_k(x)\nearrow |f(x)|\quad\text{ for all }x.$$  Let $$E_n=\{(x,\alpha)\mid 0\leq \alpha\leq \varphi_n(x)\}.$$ Therefore $E_n\subset E_{n+1}$ and thus $E_n\nearrow E$ i.e. $$E=\bigcup_{n=1}^\infty  E_n,$$ and since $E_n$ is measurable, the set $E$ is measurable.  Is it correct ?
96126
Fix $b&gt;1$.   Problem 6(a): Let $m$, $n$, $p$, $q$ be integers such that $n&gt;0$, $q&gt;0$, and $r = m/n = p/q$. Then I've managed to prove that $$b^{m/n} = b^{p/q}.$$ So we can reasonably define $b^r$ as $$b^r \colon = \sqrt[n]{b^m}.$$   From this definition, we can Problem 6(b) prove that $$ b^{r+s} = b^r \cdot b^s, $$ where $r$ and $s$ are any rational numbers.   Problem 6(c): Now for a rational number $r$, let the set $B(r)$ be defined as follows:  $$B(r) \colon= \{ \, b^t \colon \, t\in \mathbb{Q}, \, t \leq r \, \}. $$  Then it is clear that $$b^r = \sup B(r).$$ So for every real $x$, we can define $b^x$ as follows: $$b^x \colon= \sup B(x) = \sup \{ \, b^t \, \colon \, t \in \mathbb{Q}, \, t \leq x \, \}. $$   Problem 6(d): Using this definition, we are required to prove that, for every pair of real $x$ and $y$, the equation $$b^{x+y} = b^x \cdot b^y$$ holds.   My work:   If $r$, $s \in \mathbb{Q}$ such that $r \leq x$ and $s \leq y$, then $r+s \in \mathbb{Q}$ also and $r+s \leq x+y$ so that we can write  $$ B(x) \cdot B(y)  = \{ \, b^r \cdot b^s \, \colon \, r \in \mathbb{Q}, s\in \mathbb{Q}, \, r \leq x, s\leq y \, \} \subseteq \{ \, b^t \, \colon \, t \in \mathbb{Q}, \, t \leq x+y \, \} = B(x+y),$$ and hence  $$b^x \cdot b^y = \sup B(x) \cdot \sup B(y) = \sup \{ \, b^r \, \colon \, r \in \mathbb{Q}, \, r \leq x \, \} \cdot \sup \{ \, b^s \, \colon \, s \in \mathbb{Q}, \, s \leq y \, \} = sup \{ \, b^{r+s} \, \colon \, r \in \mathbb{Q}, \, s\in \mathbb{Q}, \,  r \leq x, \, s \leq y \, \} \leq \sup \{ \, b^t \, \colon \, t \in \mathbb{Q}, \, t \leq x+y \, \} = \sup B(x+y) = b^{x+y}.$$  Here we have used the following definition: Given two non-empty sets $U$ and $V$, say of real numbers, we define the set $U \cdot V$ as follows:  $$U \cdot V \colon= \{ \, u \cdot v \, \colon \, u \in U, \, v \in V \, \}.$$   And we have also used the fact that if $W$ and $Z$ are two non-empty bounded above subsets of the set of positive real numbers such that $W \subseteq Z$, then we must have  $$ \sup W \leq \sup Z.$$   So far, we have shown that $$b^x \cdot b^y \leq b^{x+y}.$$  Now how to prove the reverse inequality using the machinery developed above?  That is, how to prove that $$b^{x+y} \leq b^x \cdot b^y?$$
107439
$f:(X,\mathcal{M}\to (\overline{\mathbb{R}},\mathcal{B}(\mathbb{\overline{R}}))$ then TFAE  a)$f$ is measurable  b)$f^{-1}([a,\infty])$ is measurable $\forall a\in\mathbb{R}$  c)$f^{-1}([-\infty, a))$is measurable $\forall a\in\mathbb{R}$  Just need some hints.   By def $f$ is measurable if $f^{-1}(A)\in\mathcal{M}$ (i.e measurable ), $\forall A\in\mathcal{B}(\mathbb{R})$  why is $a\Rightarrow b$ proven in the following way  $f^{-1}([a,\infty])=\displaystyle\bigcap_{n=1}^{\infty}f^{-1}(a-\frac{1}{n},\infty])$  and for example $b\Rightarrow c$  $f^{-1}([-\infty, a))=X\setminus f^{-1}([a,\infty])$  I think we are supposed to use the following theorem here: Let $\mathcal{E}$ be the class of subsets of $Y$ assume $\mathcal{N}=\sigma({\mathcal{E}})$. Then $f:X\to Y$ is measurable iff $f^{-1}(A)\in\mathcal{M}$ for all $A\in\mathcal{E}$  so if $f$ is measurable now then by the above theorem $f^{-1}([a,\infty])$ is measurable as $[a,\infty]$ generates $\mathcal{B}(\mathbb{R})$ so what is the point of writing the inverse image as the intersection of the inverse images ?
93477
I am trying to prove that Given $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and a basis for $\mathbb{R}$, then $|B|=2^{\aleph_0}$.  Proof: Suppse that $|B|&lt;2^{\aleph_0}$, then $\mathbb R = span\{ b_i ; b_i \in B \}$ So $ \mathbb R = \{ \Sigma_{i=1}^{\infty}\alpha_i b_i ; \alpha_i \in \mathbb Q, b_i \in B, i \in \mathbb N \} = \bigcup_{n=1}^{\infty}\{\Sigma_{i=1}^n \alpha_ib_i\}$. Which is a contradiction since a countable union of countable union of countable sets is countable and $\mathbb R$ is not.  What do you think?  Thank you, Shir
67122
We define the Lebesgue Outer Measure of an interval $[a,b]$ by   $$\lambda([a,b])= \inf \left\{\sum_{j=1}^\infty |I_j|: [a,b]\subset \bigcup_{j=1}^{\infty} I_j\right\}$$  where $\{I_j\}$ is a sequence of open intervals that cover $[a,b]$, and we define the length of an open interval $|(a,b)|= b-a$. I want to show $\lambda([a,b])= b-a$  The proof of the book that I am using starts with the following:  Let $\epsilon &gt;0$. Because   $$[a,b] \subset \left(a- \frac{\epsilon}{4}, b+ \frac{\epsilon}{4}\right) \cup \bigcup_{n=2}^\infty \left(- \frac{\epsilon}{2 \cdot 2^n},\frac{\epsilon}{2 \cdot 2^n}\right)$$, we obtain $\lambda([a,b]) &lt; b-a +\epsilon$ which implies $\lambda([a,b])\le b-a$.  I can understand $[a,b] \subset (a- \frac{\epsilon}{4}, b+ \frac{\epsilon}{4})$, but I do not get the part $\cup \bigcup_{n=2}^{\infty} (- \frac{\epsilon}{2 \cdot 2^n},\frac{\epsilon}{2 \cdot 2^n})$ Since $(a- \frac{\epsilon}{4}, b+ \frac{\epsilon}{4})$ already covers $[a,b]$, and I do not think that $\bigcup_{n=2}^{\infty} (- \frac{\epsilon}{2 \cdot 2^n},\frac{\epsilon}{2 \cdot 2^n})$ will necessarily cover an arbitrary closed interval $[a,b]$, so I do not really know why you need that union of arbitrary small intervals. (My guessing is that you add the countable union so that it coincides with the definition of Lebesgue outer measure)  In addition, I do not know how can we arrive at the inequality $\lambda([a,b]) &lt; b-a +\epsilon$. I do know that $b-a + \epsilon$ is essentially the length of $(a- \frac{\epsilon}{4}, b+ \frac{\epsilon}{4}) \cup \bigcup_{n=2}^{\infty} (- \frac{\epsilon}{2 \cdot 2^n},\frac{\epsilon}{2 \cdot 2^n})$ since I already computed it out. Taking the infimum of the length, we get $b-a$ and this is the outer measure (I think this statement should be wrong though)  Sidenote: I have not officially encountered Lebesgue Measure yet. I encountered this outer measure thing when the book is trying to show function continuous a.e is Riemann-integrable.  I really hope someone can shed some light since I have been puzzled by this the entire day.
118109
This is an exercise in real analysis:     Let $E\subset{\Bbb R}^d$ be Lebesgue measurable. Show that    $$ m(E)=\sup\{m(K):K\subset E, K \text{compact}\}. $$   When $E$ is bounded, this can be done by the following proposition:       $E\subset{\Bbb R}^d$ is Lebesgue measurable if and only if for every $\varepsilon&gt;0$, one can find a closed set $F$ contained in $E$ with $m^*(E\setminus F)\leq\varepsilon$.    How can I deal with the case that $E$ is unbounded?
111317
Below is an attempt at a proof of the following problem. Any feedback would be greatly appreciated. Thx!  Let $E$ be a set and $O_n = \{x: d(x, E) &lt; \frac{1}{n}\}$. Show  If $E$ is compact then $m(E) = lim_{n \rightarrow \infty} m(O_n)$.  This is false for $E$ closed and unbounded or $E$ open and bounded.   Note that all sets are real and measurable refers to Lebesgue measurable.  Suppose $E$ is compact. Then $m(E) &lt; \infty$. Also, since each $O_n$ is an open set of $\mathbb{R}^d$, $m(O_n) &lt; \infty$. If either of the following is true, we have that $lim_{n \rightarrow \infty}m(O_n)=m(E)$: $O_n \nearrow E$ $O_n \searrow E$ Now  $\bigcap_{n=1}^{\infty} O_n = \{ x:d(x,E) = 0\}$. That is, $x \in \bigcap_{n=1}^{\infty}O_n \iff \exists n&gt; \frac{1}{\epsilon},\ n \in \mathbb{N} \iff x\in \{x \mid d(x,E) =0\}$. Then we have that $\bigcap_{n=1}^{\infty}O_n = E$ and $O_n \searrow E$. Thus $ \ m(E) = lim_{n \rightarrow \infty} m(O_n)$. Suppose $E$ is closed and unbounded. Let $E = \{(x,y): y = 2 \}$ and $O_n = \{(x,y): d(x,E) &lt; \frac{1}{n}\}$. Since $E$ is a line in $\mathbb{R}^2$, $m(E) = 0$. Also, since the measure of a rectangle in $\mathbb{R}^d$ is its volume, $m(O_n) = \left| O_n \right| = \infty$, since the rectangle $O_n$ is unbounded. It is therefore apparent that $m(E) \not = lim_{n \rightarrow \infty} m(O_n)$. Suppose $E$ is open and bounded. Let $E=(0,1)$ and $O_n = \{x \mid d(x,E) &lt; \frac{1}{n}\}$. $\mathbb{R}-E$ is closed and $O_n \in \mathbb{R}-E$. Since $\mathbb{R}-E$ contains all its limit points $lim_{n \rightarrow \infty}O_n = O \in \mathbb{R}-E$. Thus $lim_{n \rightarrow \infty} m(O_n) \neq m(E)$.
76078
Suppose $(\mathbb{R},\Sigma(m),m)$ is our measure space, where $m$ is Lebesgue measure.  Also, suppose $f : \mathbb{R} \to [-\infty, \infty]$ is a Lebesgue measurable function.  The problem:     Prove that $f$ is equal almost everywhere to a Borel measurable function.   My attempt:  We only have to prove this assertion for a non-negative Lebesgue measurable function $f$ because the result will follow for all Lebesgue measurable functions.  Furthermore, since any Lebesgue measurable function can be approximated by a sequence of non-negative, monotonically increasing, Lebesgue measurable simple functions $s_{n}$, we only need to prove the claim for an arbitrary Lebesgue measurable simple function.  So, let $s: \mathbb{R} \to [-\infty, \infty] $ be a Lebesgue measurable simple function.  We can write $s$ canonically as $$ s(x) = \sum \limits_{i = 1}^{n} \alpha_{i} \chi_{A_{i}}(x)$$ where $\alpha_{i} \in [-\infty, \infty]$, $\bigcup \limits_{i = 1}^{n} A_{i} = \mathbb{R}$, and $A_{i} \cap A_{j} = \emptyset$ if $i \neq j$.  For each $i$, since $A_{i}$ is Lebesgue measurable, we can find a Borel set $B_{i}$ such that $A_{i} \subseteq B_{i}$ and $m(B_{i} \setminus A_{i}) = 0$.  Clearly, this implies that $\bigcup \limits_{i = 1}^{n} B_{i} = \mathbb{R}$.  Now we just need the $B_{i}$'s to be pairwise disjoint, with each $B_{i}$ still retaining $A_{i}$.  To make them pairwise disjoint, I constructed the following sets:  $\tilde{B_{i}} = [B_{i} \setminus (\bigcup \limits_{j \neq i} B_{j})] \cup A_{i}$.  This construction gives us that the $\tilde{B_{i}}$'s are pairwise disjoint (I think....) and $A_{i} \subseteq \tilde{B_{i}}$.  But I don't know that $\tilde{B_{i}}$ is necessarily still a Borel set. :( :(  Am I approaching this problem all wrong?
72339
I am reading through Folland's Real Analysis: Modern Techniques and Their Applications, and they have the following proposition and proof:     Proposition: If $\{f_{j}\}$ is a sequence of $\bar{\mathbb{R}}$- valued measurable functions on $(X, \mathcal{M})$, then the functions:      $$g_{1}(x)=\sup_{j} f_{j}(x)$$ $$g_{2}(x)=\inf_{j} f_{j}(x)$$ are   measurable.       Proof:      We have       $$g_{1}^{-1}((a,\infty])= \bigcup_{j=1}^{\infty}  f_{j}^{-1}((a,\infty])$$ $$g_{2}^{-1}([-\infty,a))=  \bigcup_{j=1}^{\infty} f_{j}^{-1}([-\infty,a))$$      Where the result follows from the fact that $g_{1}^{-1}((a,\infty])\in  \mathcal{M}$ for all $a \in \mathbb{R} \iff g$ is measurable, and   $g_{2}^{-1}([-\infty,a))\in \mathcal{M}$ for all $a \in \mathbb{R}  \iff g$ is measurable.   I understand the last part of the proof, but how do we know $g_{1}^{-1}((a,\infty])= \bigcup_{j=1}^{\infty}f_{j}^{-1}((a,\infty])$ and $g_{2}^{-1}([-\infty,a))=\bigcup_{j=1}^{\infty} f_{j}^{-1}([-\infty,a))$? I would appreciate any help with this.
26172
As a problem during introduction to measure theory, I am asked to prove the following.  Let $A \subset [0,1]$ such that $A$ is measurable with $\mu(A) = 0$ ($\mu$ being the Lebesgue measure). Prove that $A^c =  [0,1] \setminus A$ is dense in $[0,1]$.  This makes intuitive sense to me $-$ since $\mu(A) = 0$, using that $\mu(X) = \mu(X \cap A) + \mu(X \cap A^c)$ for every $X \subseteq \mathbb{R}$ we get $\mu([0,1]) = \mu(A) + \mu(A^c) = \{\mu(A)=0\} = \mu(A^c)$. Thus, $A^c$ has the same measure as $[0,1]$. I am aware that this is not the definition of $A^c$ being dense in $[0,1]$ though, and I do not know how to formalize this proof.  Any advice would be appreciated.
105043
Suppose we have a premeasure $\mu$ on a space $X$ such that $\mu(X) &lt; \infty$. Prove that $E \subset X$ is Caratheodory measurable iff $ \mu^*(E)+ \mu^*(E^C) = \mu(X)$.  Going in the forward direction is fairly easy. Assuming that E is Caratheodory measurable, we can just substitute X into $\mu^*(A) = \mu^*(A \cap E) + \mu^*(A \cap E^C) $, and then we note that the outer measure and the premeasure of X themselves would have the same value.   The other direction is more difficult however. My primary idea of how to solve this part is to show that $\mu^*(A)$ and $\mu^*(A \cap E) + \mu^*(A \cap E^C) $ are both greater than and less than each other to show equality. However, I am not completely sure how to proceed with this. Can anybody provide any pointers as to how to prove the equality between these two expressions?
47378
Could someone please check if I'm saying something that make sense? I am trying to check the following:     If $f$ is a measurable function, then $\text{supp}(f)$ is measurable set?   Is it simply because, if we consider $E_1 = (0,\infty)$ and $E_2 = (-\infty,0)$, since they are open, they are measurable. So their union $\mathbb{R}-\{0\}$ is measurable as well. Since $f$ is measurable, then the preimage $f^{-1}(\mathbb{R}-\{0\})$ is measurable. But the preimage is just the support?
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
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