Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [singular-measures]

Two measures are said to be singular w.r.t. each other if they are supported on disjoint sets.

Often "singular measure" means a measure that is singular with respect to Lebesgue measure (or Hausdorff measure).

Lebesgue's decomposition theorem, given two measures $\mu$, $\nu$, decomposes $\nu$ in two parts, one absolutely continuous with respect to $\mu$, and the other singular to it.

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How is a singular continuous measure defined?

On a measurable space, how is a measure being singular continuous relative to another defined? I searched on the internet and in some books to no avail and it mostly appears in a special case - the Lebesgue measure space $\mathbb{R}$. Do you know…
Tim
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Is the set of non-differentiable points for a singular continuous function nowhere dense?

A function $f:[0,1]\rightarrow\mathbb{R}$ is called singular continuous, if it is nonconstant, nondecreasing, continuous and $f^\prime(t)=0$ whereever the derivative exists. Let $f$ be a singular continuous function and $T$ the set where $f$ is not…
anon
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Mutually singular measures with the same support

Let $X$ be a compact metric space and let $\mu$ be a measure on $(X,\mathcal{B})$, where $\mathcal{B}$ is the Borel $\sigma$-algebra of subsets of $X$. We define the support of $\mu$ as the smallest closed set of full $\mu$ measure, i.e.,…
Cantor
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An example of a a convolution of singular distribution and a Gaussian distribution that has a 'simple' pdf

I am looking for a nontrivial example of a singular distribution that when convolved with a Gaussian distribution has a pdf of a 'simple' form. I let 'simple' be something that you interpret yourself. Singular distributions are an import class…
Boby
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Singular memoryless distribution

Context: A probability distribution is said to be memoryless if its range $R$ is stable under addition, and for all $a,b\in\mathbb R$, $$\mathbb P(X>a+b|X>b)=\mathbb P(X>a)$$ It is well known geometric and exponential distributions are memoryless.…
Kolakoski54
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Relative entropy between singular measures

Usually, to define relative entropy between two probability measures, one assumes absolute continuity. Is it possible to extend the usual definition in the non absolutely continuous case?
mellow
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How we compute expectation of a singular random variable?

In probability (or measure) courses, we often see the Cantor distribution that is singular with respect to the Lebesgue measure. Its CDF is increasing but whenever its differentiable, the corresponding PDF is zero. But afterwards, we never use…
Sergio Parreiras
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A sequence of singular measures converging weakly* to a continuous measure

Can anyone provide a sequence of singular (w.r.t. Lebesgue measure) measures $\in\mathcal{M}([0,1])=C[0,1]^*$ converging $weakly^*$ to an absolutely continuous (w.r.t. Lebesgue measure) measure?
checkmath
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Do there exist two singular measures whose convolution is absolutely continuous?

Let $\mu, \nu$ be finite complex measures with compact supports on the real line, and assume that they are singular with respect to the Lebesgue measure. Can their convolution $\mu\ast\nu$ have a nonzero absolutely continuous component?
limanac
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Mutually Singular measures

c.f. Rudin's Real and Complex Analysis (Third Edition 1987) Chapter 6 Q9 Suppose that $\{g_n\}$ is a sequence of positive continuous functions on $I=[0,1]$, $\mu$ is a positive Borel measure on $I$, $m$ is the standard Lebesgue measure, and that (i)…
Cecilia
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Singular measures - approximate characteristic function

One can decompose a $\sigma$-finite measure $\mu$ on $\mathbb{R}$ in three parts: $\mu_{ac}$: absolutely continuous $\mu_{sc}$: singular continuous $\mu_{pp}$: pure point A common example for a singular continuous probability measure is Cantor's…
user13655
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Singular measures on Real line

Can some one please give me an example of continuous Singular measure on Reals which is not absolutely continuous to cantor-type sets. Thank you.
chandu1729
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Reverse type $1-1$ inequality

If $\mu \in \mathcal{M}(\mathbb{T})$ is nontrivial and singular with respect to lebesgue measure, then $$|\{\theta \in \mathbb{T} : M\mu(\theta) >\lambda\}| \ge \frac C\lambda \|\mu\|$$ where $| \cdot | $ denotes Lebesgue measure, $|\lambda| >…
David Bowman
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Properties of mutual singular measures

Let $\mu$ be a positive measure and $\nu_1, \nu_2$ arbitrary measures, all defined on the same measurable space $(S,\Sigma)$. We say that two arbitrary measure $\mu, \nu$ are mutually singular (notation $\nu \perp \mu$) if there exist disjoint sets…
iJup
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Singular measure with positive Lebesgue measure support

I was wondering whether there exists a Borel measure $\mu$ on $\mathbb{R}$ such that $ \text{Leb}\big(\text{supp}(\mu)\big)>0$ while $\mu$ is singular with respect to the Lebesgue measure? I have a feeling that this should not be true by basic…
Keen-ameteur
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