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7 votes
Accepted

Is there a function $f:\mathbb{R}\rightarrow\mathbb{R}$ that maps every subset of cardinality $|\mathbb{R}|$ to the whole real line?

Suppose such a function $f$ existed. Since $f$ is surjective, the pre-image of $\mathbb{R}\setminus\{0\},$ $X$, has cardinality $\vert \mathbb{R}\vert,$ and so we must have $f(X)=\mathbb{R}.$ But $f(X)...
Adam Rubinson's user avatar
4 votes
Accepted

If $\int_0^1 f(x) \, dx = I$, then $m\left\{x: f(x) > \frac{I}{2}\right\} \geq \frac{I}{2}$

Just decompose the integral into two parts: \begin{align*} I &= \int_0^1 f(x)dx = \int_{\{x: f(x) > I/2\}}f(x)dx + \int_{\{x: f(x) \leq I/2\}}f(x)dx \\ & \leq m(\{x: f(x) > I/2\}) + \...
Zhanxiong's user avatar
  • 14.3k
3 votes
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Let a real measurable function $f$ map every open set to the whole real line. Is there always a restriction to a set of measure zero doing the same?

I think the following is a counterexample. Enumerate the intervals with rational endpoints as $I_1,I_2,I_3,\dots$. Constract pairwise disjoint nowhere dense closed sets $A_n\subset I_n$ with Lebesgue ...
user14111's user avatar
  • 1,491
3 votes
Accepted

A non-measurable set with density $\frac{1}{2}$

The answer is no. Following the notation in the question, I will use $\mu$ to denote the Lebesgue outer measure; I will use $|\cdot|$ to denote the Lebesgue measure. We will use the following standard ...
Willie Wong's user avatar
  • 73.6k
3 votes
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What is the measure of the set of values $S=\{0.x_1x_2... \in (0,1)$ (in binary) for which $\sum_{n \geq 1}\frac{(-1)^{x_n}}{n}$ converges$\}$?

The OP’s main question has already been resolved in the comments. The following is a long comment addressing OP’s follow-up question in comments, namely, we show that under AC, we can construct a ...
David Gao's user avatar
  • 7,545
2 votes
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Confused about the Radon-Nikodym theorem applied to the counting measure

The Radon-Nikodym Theorem requires the measure to be $\sigma$-finite. What you found is an example that shows that the hypothesis is crucial, as the counting measure is not $\sigma$-finite on any ...
Martin Argerami's user avatar
2 votes

For every measurable $E$ with $0<m(E)< \infty$, there is $[a,b]$ such that $m(E \cap [a,b]) > \frac{1}{2} m(E)$

In this question, we have that, if $0 < m(E) < \infty$, for any $\epsilon > 0$, there exists $M > 0$ such that $$ m(E \setminus [-M, M]) < \epsilon $$ Now, choose $\epsilon = \...
Thành Nguyễn's user avatar
2 votes

Complemented subspace of set o functions whose integral is zero

Choose any $f_{0}\in L^{1}([0,1])$ such that $\int_{[0,1]}f_{0} \, {\rm d}\mu = 1$. Such an example of a function is the constant function which maps each element of $[0,1]$ to $1$. Set $Z := \text{...
Dean Miller's user avatar
  • 2,006
2 votes
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Ordinal Event Ordering $P(E)\geq P(E')\iff Q(E)\geq Q(E')$, so we can conclude something like $P$ and $Q$ are the same measure?

We immediately get $$P(E) = P(E') \iff Q(E) = Q(E')$$ Let $p=P(E)$, then $P(E)=P([0,p))$ so $Q(E)=Q([0,p))=f(p)$ Which tells us $Q$ is just some increasing function applied to $P$. $P([0,x))=P([x,2x))$...
Zoe Allen's user avatar
  • 4,798
2 votes

$L^\infty(\Omega)$ is dense in $L^{p,\infty}(\Omega)$ if $\Omega$ is compact

Actually, the closure of $L^\infty$ in $L^{p,\infty}$ is $$ L^{p,\infty}_0:=\left\{ f:\lim_{t\to\infty}t^p\mu\{x:\lvert f(x)\rvert>t\}=0 \right\}. $$ Indeed, if $f$ belongs to the closure of $L^\...
Davide Giraudo's user avatar
1 vote

$L^\infty(\Omega)$ is dense in $L^{p,\infty}(\Omega)$ if $\Omega$ is compact

This density is not true. Let $\Omega=(0,+1)$. Let $p\in (0,\infty)$ and define $$ f(x) = x^{-\frac1p}. $$ Then $$ \mu(\{x:\ |f(x)| > t\} ) = t^{-p} $$ for $t\ge1$, so that $\|f\|_{L^{p,\infty}} = ...
daw's user avatar
  • 49.9k
1 vote

$\mu[X\geq x]\geq \mu[Y\geq x] \iff \nu[X\geq x]\geq \nu[Y\geq x] $ implies that the two measures $\mu,\nu$ are the same measure?

Assume that $\mu, \nu$ satisfy $X \: \mu\text{-FOSD} \: Y \Longleftrightarrow X \: \nu\text{-FOSD} \: Y$ for all X, Y. For $A, B \subset S$, such that $\mu(A) < \mu(B)$ denote $$X := 2\chi_{\bar A}...
SamedyMaj's user avatar
1 vote
Accepted

Rudin’s RCA Theorem $7.26$ part $1$.

I'm writing down this answer both because I think you understood the gist of it and to make this question leave the Unanswered queue. For $x \in E_0 \cap X \subset X$, $T$ is differentiable at $x$ by $...
Bruno B's user avatar
  • 5,702
1 vote

Is a Borel measurable function always Lebesgue measurable?

This is not true. Let $A \subset \mathbb{R}$ be a fat Cantor set. Then $A$ is homeomorphic to the usual Cantor set $B \subset \mathbb{R}$ and let $g: A \to B$ be such a homeomorphism. Define, $$f: \...
David Gao's user avatar
  • 7,545
1 vote
Accepted

Given $f \in L^p_{\text{loc}}(\Omega) \setminus L^\infty(\Omega)$, does it follow that $A \cap S(f,K) \neq \emptyset$ for all $K > 0$?

This answer just replicates my comment: For any two sets $A,B$ we can write $B$ as a disjoint union $$ B = (B\cap A) \cup (B \cap A^c)$$ If we know the sets $A,B$ are Lebesgue measurable we know $A^c$,...
Michael's user avatar
  • 24.8k
1 vote
Accepted

Continuity on the torus

Take the Fourier transform: for every $n\in \mathbb Z^d$, since the Fourier transforms changes convolution into product, $$ \mathcal{F}\left( \mathcal F^{-1} \eta_r * v_0 \right)(n) = \eta_r(n) \...
Thomas Lehéricy's user avatar
1 vote

For every measurable $E$ with $0<m(E)< \infty$, there is $[a,b]$ such that $m(E \cap [a,b]) > \frac{1}{2} m(E)$

Hint: Consider a function $\phi: \mathbb{R} \to [0, \infty]$ defined by $\phi(x) = m(E \cap B(0,x))$ where $B(0,x)$ is the ball centred at $0$ with radius $x$. First show that $\phi$ is continuous, ...
approximate-identity's user avatar
1 vote
Accepted

On surface of $C = \{ x^2+y^2=1,0<z<1 \}$ there is subset A, $A_t=A \cap \{z <t \}$. Prove $\int_0^1 \lambda_2(A_t) dt= \iint_A (1-z)d \lambda_2$

Let $a_{t}={d\lambda_2(A_t)}/{dt}$ represent the 1-dimensional volume of the part of $A$ at height $t$ (defined for almost every $t\in[0,1]$). Then the left hand side is equal to $$\int_{t=0}^1\int_{z=...
Lieven's user avatar
  • 1,943

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