Questions tagged [lebesgue-measure]

For questions about the Lebesgue measure, a measure defined on the Borel or Lebesgue subsets of the real line or $\mathbb R^d$ for some integer $d$. Use it with (tag: measure-theory) tag and (if necessary) with (tag:lebesgue-integral).

Lebesgue measure is the classical notion of length and area to more complicated sets, and its assigns a measure to subsets of $n$-dimensional Euclidean space. Some examples of Lebesgue any closed interval, any cartesian product of intervals, any Borel set, and any countable set of real numbers (which has Lebesgue measure zero).

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Lebesgue measurable but not Borel measurable

I'm trying to find a set which is Lebesgue measurable but not Borel measurable. So I was thinking of taking a Lebesgue set of measure zero and intersecting it with something so that the result is not Borel measurable. Is this a good approach? Can…
JT_NL
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What is the difference between outer measure and Lebesgue measure?

What is the difference between outer measure and Lebesgue measure? We know that there are sets which are not Lebesgue measurable, whereas we know that outer measure is defined for any subset of $\mathbb{R}$.
lavy
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What's the relationship between a measure space and a metric space?

Definition of Measurable Space: An ordered pair $(\Omega, \mathcal{F})$ is a measurable space if $\mathcal{F}$ is a $\sigma$-algebra on $\Omega$. Definition of Measure: Let $(\Omega, F)$ be a measurable space, $μ$ is an non-negative function…
Bear and bunny
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What does it mean for a set to have Lebesgue measure zero?

I am studying examples of sets with Lebesgue measure zero (e.g. the Cantor Set) but wanted an intuitive description of what this means rather than a formal definition. Thank you.
Enigma123
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Are most matrices diagonalizable?

More precisely, does the set of non-diagonalizable (over $\mathbb C$) matrices have Lebesgue measure zero in $\mathbb R^{n\times n}$ or $\mathbb C^{n\times n}$? Intuitively, I would think yes, since in order for a matrix to be non-diagonalizable…
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What is wrong in this proof: That $\mathbb{R}$ has measure zero

Consider $\mathbb{Q}$ which is countable, we may enumerate $\mathbb{Q}=\{q_1, q_2, \dots\}$. For each rational number $q_k$, cover it by an open interval $I_k$ centered at $q_k$ with radius $\epsilon/2^k$. The total length of the intervals is a…
yoyostein
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Apparent inconsistency of Lebesgue measure

Studying the Lebesgue measure on the line I've found the following argument which concludes that $m(\mathbb{R}) < +\infty$ (where $m$ denotes the Lebesgue measure on $\mathbb{R}$). Obviously it must be flawed, but I haven't been able to find the…
Jonatan B. Bastos
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Lebesgue density strictly between 0 and 1

I am having trouble with the following problem: Let $A\subseteq \mathbb{R}$ be measurable, with $\mu(A)>0$ and $\mu(\mathbb{R}\backslash A)>0$. Then how do I show that there exists $x\in \mathbb{R}$ such that $$\lim_{\varepsilon\to 0}…
digiboy1
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Intuitive, possibly graphical explanation of why rationals have zero Lebesgue measure

I know that rationals, being a countable set, have zero Lebesgue measure. I think one way to prove it is to find an open set containing rationals that has measure less than $\epsilon$ for every $\epsilon >0$ fixed. You can do it by taking the…
tommy1996q
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Vitali set of outer-measure exactly $1$.

I know that for any $\varepsilon\in (0,1]$ we can find a non-measurable subset (w.r.t Lebesgue measure) of $[0,1]$ so that its outer-measure equals exactly $\varepsilon$. It is done basicly with the traditional Vitali construction inside the…
T. Eskin
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Prove that every Lebesgue measurable function is equal almost everywhere to a Borel measurable function

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…
layman
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Is there a set $A \subset [0,1]$ such that $\int_{A \times A^\text{c}} \frac{\mathrm{d} x \, \mathrm{d} y}{\lvert x - y\vert}=\infty$?

The above question came up when I was trying to find a counterexample related to this problem. Clearly, the integral of $(x,y) \mapsto \lvert x-y \rvert^{-1}$ over $[0,1]^2$ is divergent. When integrating over a subset of the form $A \times…
ComplexYetTrivial
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When does $\lim_{n\to\infty}f(x+\frac{1}{n})=f(x)$ a.e. fail

We know that if $f\in L^1(\mathbb{R})$, then $\|f(\cdot+1/n)-f(\cdot)\|_{L^1}\to 0$ as $n\to \infty$, which implies that there exists a subsequence $f_{n_k}=f(x+\frac{1}{n_k})$ such that $f_{n_k}\to f$ a.e. My problem is that Is there any $f\in…
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Construct a Borel set on R such that it intersect every open interval with non-zero non-"full" measure

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…
VADupleix
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Show that if the integral of function with compact support on straight line is zero, then $f$ is zero almost everywhere

I want to prove that that given $f:R^2 \rightarrow R$ which is continuous with compact support s.t the integral of $f$ for every straight line $l$ is zero ($\int f(l(t))\mathrm{d}t=0$) then $f$ is almost everywhere $0.$ Well I know how to proof it…
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