Let A be a Lebesgue measurable set. Let f: $\mathbb{R} \rightarrow \mathbb{R}$ be a  function of the class $C^1$;  Is this true that f(A) is lebesgue measurable?  I know that this is true when f is injective.  But I don't know that  when f is not injective.  Could you teach me it is true or not?  Sorry for my poor English.
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.)
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 ?
10873
I don't know how can I imply Fatou's lemma for any measurable sets $A_k$, that is, $$\lambda(\liminf A_k)\le\liminf\lambda(A_k).$$ How can I prove it? And is there any example in $\mathbb R$ of sequence of measurable sets $A_k$ such that $A_k\subset[0,1]$, $\lim\lambda(A_k)=1$, but $\liminf A_k=\varnothing$ ? Thanks for your help!
13273
Let $f:\mathbb R \rightarrow \mathbb R$, and for every $x,y\in \mathbb R$ we have $f(x+y)=f(x)+f(y)$.  Show that $f$ measurable $\Leftrightarrow f$ continuous.
64074
I have an intuition problem regarding a certain claim.  I've learned that since $\mathbb{Q}$ is measurable than  $$\lambda(\mathbb{Q})= \inf\left\{\sum_{n \ge 1} \lambda([a_n,b_n)) : \mathbb{Q} \subset \bigcup_{n \ge 1} [a_n,b_n)\right\}$$  This is because Lebesgue measure is the Caratheodory extension over the semiring whose elements are $[a,b)$.  But it seems so false, since as we know $\lambda(\mathbb{Q}) = 0$, But I can't even imagine a countable family of sets such that $\mathbb{Q} \subset \bigcup_{n \ge 1} [a_n,b_n)$ and  $\sum_{n \ge 1} \lambda([a_n,b_n)) &lt; \infty$.  How can the infimum of the above set be $0$?
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?
54748
Let $E$ be measurable and define $f:E\rightarrow\mathbb{R}$ such that $\{x\in E : f(x)&gt;c\}$ is measurable for all $c\in\mathbb{Q}$, is $f$ measurable?  There are a number of equivalent definitions for the measurability of a function and the most obvious one would be to show that $\{x\in E : f(x)&gt;c\}$ is measurable for all $c\in\mathbb{R}$.  Thus my strategy has been to consider an arbitrary irrational $y$ and use the density of the rationals in the reals to show that there exists some open set $O$ such that $m^*(O-\{x\in E : f(x)&gt;y\})&lt; \varepsilon$.  I would do this by choosing some rational $q&lt;y$, close enough to $y$, such that the set of values in $E$ which get sent under $f$ to values strictly greater than $q$ and less than or equal to $y$ is small enough.  But that doesn't in general seem possible since I could keep sending intervals of some fixed positive length to smaller and smaller regions just beneath $y$.  However conversely, such a function would not be enough for a counter example, since finding such a function only rules out this strategy, but doesn't necessarily entail non-measurability.  Is there a better definition of the measurability of a set I should use? or does the statement actually not imply measurability?  Thanks.
92802
My exercise says: Let $f:\mathbb{R} \rightarrow \mathbb{R}$ a continuous function e suppose that exists $k$ such that:  $$|f(x)-f(y)|\geq k|x-y|$$  Then $f$ is bijective and  its inverse is continuous.  Well, there's a Theorem , Invariance of domain, that says   "If $U$ is an open subset of $\mathbb {R^n}$ and $f : U \rightarrow\mathbb{R}$ is an injective continuous map, then $V = f(U)$ is open and $f$ is a homeomorphism between $U$ and $V$".  But I'm not knowing how to proceed...need a clue...Thanks for attention!!!
40132
Let $f : \mathbb{R} \to [ 0, \infty)$ be a measurable function. If $\int_{- \infty}^{\infty} f(x) dx =1$. Then I want show that $ \int_{- \infty}^{\infty}  \frac{1}{1+ f(x)} dx = \infty.$  Any help will be appreciated.  Thank you in advance.
36112
I am trying to show that for $A$, $B \; \subset \mathbb{R}$ $A \cup B$ is Lebesgue measurable such that \begin{align} \infty &gt; \mu(A \cup B) = \mu^{*}(A) + \mu^{*}(B), \end{align} Then $A$ and $B$ are Lebesgue measurable.  Thus far, I have tried reversing the directions in the proof of Caratheodory.  I seem to want to do the following:  If $A \cup B$ is Lebesgue measurable, then, for any $E \subset \mathbb{R}$, \begin{align} \mu^{*}(E) \geq \mu^{*}(E \cap (A \cup B)) + \mu^{*}(E \cap (A \cup B)^{c}). \end{align}  I want to use this and work backwards from here adding and subtracting sets to get  \begin{align} \mu^{*}(E) \geq \mu^{*}(E \cap A) + \mu^{*}(E \cap A^{c}). \end{align} yet I am having trouble.
116357
If $X$ is measurable space, $f : X\to [-\infty,+\infty]$, function such that $f^{-1}([-\infty,r))$ measurable for all rational numbers $r$,prove than $f$ is measurable  I know that $f$ is measurable iff set $\{x\in X : f(x)&lt;a\}$ is measurable for all $a\in \mathbb{R}$ but for $a\in\mathbb{Q}$ I don't know how to proceed.
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
92462
Let $f:(a,b)\rightarrow\mathbb{R}$.  The statement to prove is that if $\forall x \in (a,b)$ Lebesgue integral $\int_{(a,x)}fd\lambda=0$, then $f(x)=0$ $\lambda$-almost everywhere.  So if it wouldn't be true then we would have $\forall x \in (a,b):\int_{(a,x)}f_+d\lambda=\int_{(a,x)}f_-d\lambda\neq0$, so every interval contains positive and negative values of the function. It is  such an $f$ so my guess would be that this function is not Lebesgue measurable and we can't take this integral.  Thanks in advance!
66911
Given an example of a continuous function $\phi$ and a lebesgue measurable function $F$. both defined on $[0,1]$, such that $F\circ\phi$ is not lebesgue measurable.
7448
I have to prove the following:  Let $f: \mathbb{R^2}\to \mathbb{R}$ such that $f_x:y\to f(x,y)$ is Borel measurable for all $x\in\mathbb{R}$ and that $f^y:x\to f(x,y)$ is continuous for all $y\in\mathbb{R}$. Prove that $f$ is Borel measurable.  What I have tried to do is to find a sequence of functions $f_n(x,y)$ s.t for a fixed $y$ $f_n(.,y)$ is a linear approximation of $f(.,y)$..
4361
Let $(X , \mathscr{M}, \mu)$ be a measure space. Prove that if $f$ is measurable, then $|f|$ is measurable.    A function $f$ is measurable if $f^{-1}((a , \infty)) = \{x : f(x) &gt; a\} \in \mathscr{M}$ for every $a \in \mathbb{R}$.   Ok so this is what I have so far:  Consider two cases: $a &lt; 0$ and $a \geq 0$. If $a &lt; 0$, then we have $$|f|^{-1}((a , \infty)) = \{x : |f(x)| &gt; a \} = X \in \mathscr{M}$$ since $\mathscr{M}$ is a $\sigma$-algebra. I am confused on how to prove the case when $a \geq 0$. If $a \geq 0$, we have $$|f|^{-1}((a , \infty)) = \{x : |f(x)| &gt; a \} = ?$$ I am assuming we will have to use that fact that $f$ is measurable.  Any help will be appreciated!
45278
Consider the space $\mathcal{K}(\mathbb{R}^{n})$ of compact subsets of $\mathbb{R}^{n}$ endowed with the Hausdorff metric $\rho$, and let $\lambda$ denote the $n$-dimensional Lebesgue measure on $\mathbb{R}^{n}$. Now, I understand that $\lambda$ is not continuous with respect to $\rho$, i.e. that $[ \lim_{k \to \infty} \rho(K, K_{k}) = 0 ] \not \Rightarrow [ \lambda(K) = \lim_{k \to \infty} \lambda(K_{k}) ]$. But my question is: Suppose $\lambda(K_{i}) = \lambda(K_{j}) = C$ for every $i, j \in \mathbb{N}$, i.e. the sequence $k \mapsto \lambda(K_{k})$ is constant; does this imply that $\lambda(\lim_{k \to \infty} K_{k}) = C$? If not, I'm interested to see a counterexample.  Thanks.