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).

7855 questions
82
votes
4 answers

Differences between the Borel measure and Lebesgue measure

I'm having difficult time in understanding the difference between the Borel measure and Lebesgue measure. Which are the exact differences? Can anyone explain this using an example?
60
votes
2 answers

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…
user1120
49
votes
3 answers

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
  • 611
49
votes
2 answers

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 $(\Omega, \mathcal{F})$ be a measurable space, $μ$ is an non-negative…
43
votes
2 answers

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…
42
votes
3 answers

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
  • 611
  • 1
  • 7
  • 12
37
votes
5 answers

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
  • 20,428
31
votes
2 answers

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}…
31
votes
4 answers

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…
30
votes
3 answers

Does taking countably-many "greedy" closed-ball bites from an open subset of "ice cream" in Euclidean space always leave a set of measure zero?

Let $W$ be an open subset of $\mathbb{R}^n$ called an Ice Cream. There is a greedy person who eats it and bites off closed balls from $W$. Since he is a greedy person, he takes as much as he can, so he chooses a ball with radius $r_0 = \sup\{r: B(r)…
30
votes
1 answer

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
  • 8,531
28
votes
2 answers

Completion of borel sigma algebra with respect to Lebesgue measure

There are two ways of extending the Borel $\sigma$-algebra on $\mathbb{R}^n$, $\mathcal{B}(\mathbb{R}^n)$, with respect to Lebesgue measure $\lambda$. The completion $\mathcal{L}(\mathbb{R}^n)$ of $\mathcal{B}(\mathbb{R}^n)$ with respect to…
Blunka
  • 965
27
votes
4 answers

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…
27
votes
4 answers

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…
26
votes
1 answer

Find $f$ such that $f \star f(x) = \frac{1}{1-x}$.

I'm looking for a measurable function $f$ defined on $]0,1[$ such that : $$f \star f(x) = \int_{0}^1 f(x-y) f(y) \ \mathrm{d}y = \frac{1}{1-x}$$ for (almost) any $x \in ]0,1[$. Is it possible to find or construct such a function f ? Eventually, we…
1
2 3
99 100