Linked Questions

10 votes
2 answers

set in $\mathbb{R}$ which is not a Borel-set [duplicate]

Possible Duplicate: Lebesgue measurable but not Borel measurable Constructing a subset not in $\mathcal{B}(\mathbb{R})$ explicitly if i start from the topology of $\mathbb{R}$, i.e. all open ...
StefanH's user avatar
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8 votes
2 answers

Is there a null set that is not a Borel set? [duplicate]

In my module notes, if $A$ is a Borel and $m(A)=0$, then it is not necessarily true that any subset $B$ of $A$ (with $m(B=0)$) is Borel. So I am wondering if there is a null set that is not a Borel ...
王向皓's user avatar
0 votes
0 answers

How to construct a Lebesgue measurable set which is not a Borel set? [duplicate]

How to construct a set $E \subseteq \mathbb{R}^n$ satisfying the following two conditions: (i) $E$ is Lebesgue measurable; (ii) $E$ is not a Borel set. (Here a Borel set is a member of Borel $\...
ScienceAge's user avatar
42 votes
5 answers

Is there a $\sigma$-algebra on $\mathbb{R}$ strictly between the Borel and Lebesgue algebras?

So, after proving that $\mathfrak{B}(\mathbb{R})\subset \mathfrak{L}(\mathbb{R})$, I asked myself, and now asking you, is there a set $\mathfrak{S}(\mathbb{R})$, which satisfies: $$\mathfrak{B}(\...
Salech Alhasov's user avatar
56 votes
1 answer

Lebesgue measurable set that is not a Borel measurable set

exact duplicate of Lebesgue measurable but not Borel measurable BUT! can you please translate Miguel's answer and expand it with a formal proof? I'm totally stuck... In short: Is there a Lebesgue ...
example's user avatar
  • 2,085
15 votes
2 answers

between Borel $\sigma$ algebra and Lebesgue $\sigma$ algebra, are there any other $\sigma$ algebra?

Is there any $\sigma$-algebra that is strictly between the Borel $\sigma$-algebra and the Lebesgue $\sigma$-algebra? How about not in between the two, but in general, are there any other $\sigma$ ...
Qiang Li's user avatar
  • 4,157
14 votes
1 answer

What sets are Lebesgue measurable?

I cannot detect the fallacy in the set of the following statements in my inconsistent notes: A sigma algebra is a set of the sets in the generating set closed under the set operations countable union,...
Dávid Natingga's user avatar
2 votes
1 answer

Lebesgue and Borel Measurable

If a real-valued function on $R$ is measurable with respect to the $\sigma$-algebra of Lebesgue measurable sets, is it necessarily measurable with respect to the Borel measurable space ($R$, $B(R)$...
Jake Casey's user avatar
  • 1,003
0 votes
2 answers

Borel subsets of the unit square

Let $I^2:=[0,1]^2\subseteq \mathbb{R}^2$ be the closed unit square in the plane. Open and closed subsets of $I^2$ are Borel measurable for trivial reasons. Also, every set obtained from open and ...
user17240's user avatar
  • 389
0 votes
0 answers

Measurable non-analytic set

I know the construction by Lusin of a measurable set that is non-Borel. But that set turns out to be analytic. Are there some examples of non-analytic sets that are measurable? Maybe the set that one ...
NPHA's user avatar
  • 339
0 votes
1 answer

Characterisation of null sets (sets of measure zero) as divergences of sequences of step functions.

In A. J. Weir’s Lebesgue Integration and Measure (CUP 1973) the author proves that, given an increasing sequence of step functions $\phi_n$ for which the sequence $\smallint \phi_n$ converges, the ...
Steve Powell's user avatar
0 votes
2 answers

If $M$ is $F$-measurable, then is it also $F'$-measurable with $F'\subset F$?

$F$ and $F'$ are $\sigma$-algebras, and $M$ is a function from $(\Omega,F)$ to $(\mathbb{R},B(\mathbb{R}))$ If this statement is true, how to reason or understand it in a simple way?
Cancan's user avatar
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