Skip to main content

Questions tagged [signed-measures]

A signed measure is a countably additive set function on a sigma-algebra and taking values in the extended reals, but not permitted to assign negative infinity to a set.

1 vote
0 answers
58 views

Prove: $F:M(X,\mathscr{A},\mathbb{R})\to\mathbb{R}$ defined by $F(\mu)=\int fd\mu$ is a linear functional. [closed]

I need to prove the following result: Remark 4.30$\quad$ Let $M(X,\mathscr{A},\mathbb{R})$ be the collection of all finite signed measures on $(X,\mathscr{A})$. Let $B(X,\mathscr{A},\mathbb{R})$ be ...
Beerus's user avatar
  • 2,493
2 votes
1 answer
71 views

Signed measure and bounded total variation on algebras

Let $X$ be a compact Hausdorff second countable topological space. Let $\mathcal{B}$ a countable basis of the topology of $X$, which we can assume to be closed under finite unions and intersections, ...
Ennio's user avatar
  • 23
1 vote
0 answers
51 views

Why does $\mu\mapsto\int fd\mu$ defines a linear functional on $M(X,\mathscr{A},\mathbb{R})$?

I am self-studying signed measure.I got stuck on the following remark: Remark$\quad$ Let $M(X,\mathscr{A},\mathbb{R})$ be the collection of all finite signed measures on $(X,\mathscr{A})$. Then the ...
Beerus's user avatar
  • 2,493
2 votes
1 answer
52 views

Question About Signed Measures

I am self-studying signed measure, and I come across the following construction: Let $\mu$ be a signed measure on the measurable space $(X,\mathscr{A})$, and let $A$ be a subset of $X$ that belongs ...
Beerus's user avatar
  • 2,493
2 votes
0 answers
31 views

Prove: Define $\nu$ on $\mathscr{A}$ by $\nu(A) = \int_Afd\mu$. Then $\nu$ is a signed measure on $(X,\mathscr{A})$.

I am self-studying measure theory and I come across the following question: Prove: Let $(X,\mathscr{A},\mu)$ be a measure space, let $f$ belong to $\mathscr{L}^1(X,\mathscr{A},\mu,\mathbb{R})$, and ...
Beerus's user avatar
  • 2,493
0 votes
0 answers
23 views

Relation between total variation measure and total variation of a function

It is well known that if $f$ defined on a compact interval $[0,T]$ is a continuous function with finite variation, then $f$ induces a signed measure $\mu$ on $[0,T]$. Let $|\mu|$ be the total ...
George's user avatar
  • 105
0 votes
1 answer
50 views

Sum of two signed measures is well defined iff both do not take the same sign of $\pm \infty$

Let $(X,\mathcal{A})$ be a measurable space and $\mu$ and $\nu$ be a signed measures on $(X,\mathcal{A})$. We say that $\lambda =\mu+\nu$ is well-defined as a function $\lambda \: \mathcal{A} \to \...
Squirrel-Power's user avatar
0 votes
0 answers
58 views

Uniqueness of extension of signed measures

By Caratheordory's extension theorem, if the measure space is $\sigma-$ finite and we have two measures that are equal on an algebra generating the $\sigma-$ algbera, then they are equal. Will the ...
vkk's user avatar
  • 31
0 votes
0 answers
27 views

Sigma additivity of signed measure

Suppose we have $(X, M, v)$ as a measure space. $v$ is a signed measure, then it also satisfies sigma additivity, i.e. If {$E_j$} is a sequence of disjoint sets in $M$ , then $v(\cup_1^\infty E_j)=\...
Andrew_Ren's user avatar
5 votes
2 answers
167 views

Why are complex measures not allowed to attain $\infty$ while signed measures are?

I saw another question similar to this one but I'm not satisfied by answers. Here I changed the question to clearify the point I am interested in. I study Measure Theory for Real & Complex ...
Alileo's user avatar
  • 121
1 vote
1 answer
88 views

Characterization of absolute continuity for finite additive functions

Let $\nu:\mathbb{X}\longrightarrow \mathbb{R}$ and additive function, where $(X,\mathbb{X},\mu)$ is a measure space. That is, $\nu(\emptyset)=0$, and for any finite disjoint family of measurable sets $...
isaac098's user avatar
0 votes
1 answer
49 views

Clarification on the definition of signed measure: Is the rearrangement issue assumed to be impossible by the definition?

Let $(\Omega,M,\mu)$ be a $\underline{\text{signed}}$ measure space. Is it true that the definition of signed measure implicitly implies that Either for any sequence of disjoint measurable sets with ...
Asigan's user avatar
  • 1,932
1 vote
2 answers
107 views

Folland Theorem 3.22 Proof Explanation

At the first step of Theorem 3.22 of Folland, he mentioned that $d\nu = d\lambda + f\,dm$ implies $d|\nu| = d|\lambda| + |f| dm$: I am aware of the similar questions has been asked before on this ...
Partial T's user avatar
  • 583
0 votes
0 answers
51 views

Won't signed measures upset Riemann?

I can't seem to wrap my head around signed measures. If $\mu$ is a signed measure on a measurable space $(X, \Sigma)$, then there'll be sets of both positive and negative measure. Let $E_1, E_2, \...
Atom's user avatar
  • 4,119
1 vote
0 answers
36 views

boundedness of signed measures

Let us consider signed charges, these are finitely additive signed measures (I suppose this would also work with sigma additive signed measures). We work on a measure space $(\Omega, \mathcal{A})$, ...
guest1's user avatar
  • 365

15 30 50 per page
1
2 3 4 5
9