Are Measures m and μ Mutually Singular Given Convergence Conditions?

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In summary: This sequence satisfies (i) and (ii) since $g_n(x)\to 0$ for almost all $x\in[0,1]$ and $\displaystyle{\int_0^1} g_n(x)\, dx=1$ for all $n\in\mathbb{N}$. However, for any $x\in [0,1]$ and $f\in C(I)$, we have$$\lim_{n\to\infty}\int_0^x fg_n(t)\, dt= \lim_{n\to\infty}\int_0^x f(t) \dfrac{\dfrac{e^{
  • #1
MarketPantry
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Suppose that [tex]\{g_n\}_{n=1}^\infty[/tex] is a sequence of positive continuous functions on [tex]I=[0,1][/tex], [tex]\mu[/tex] is a positive Borel measure on [tex]I[/tex], [tex]m[/tex] is the standard Lebesgue measure, and


(i) [tex]\lim_{n\rightarrow\infty} g_n =0 \qquad a.e. [m][/tex];
(ii) [tex]\int_{I}{g_n}\;dm = 1 \qquad[/tex] for all [tex]n\in\mathbb{N}[/tex];
(iii) [tex]\lim_{n\rightarrow\infty}\int_{I} fg_n\;dm = \int_{I} f\;d\mu \qquad[/tex] for all [tex]f \in C(I).[/tex]

Does it follow that the measures [tex]m[/tex] and [tex]\mu[/tex] are mutually singular?

I kind of guess that it is a no and have been trying hard to come up with a counterexample. Some of my thoughts:
1 . If [tex]\mu[/tex] is singular relative to [tex]m[/tex], then it has to be concentrated on a set which does not contain any open interval.
2. A typical sequence that satifies both (i) and (ii) above is the familiar "triangular sequence", i.e. the sequence of functions whose graphs are triangles with increasing heights, each having area 1. But I don't know how to deal with (iii).

Please help me. Thank you in advance.
 
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  • #2
I would try
$$
g_n(x)= \dfrac{\dfrac{e^{x^n}-1}{e-1} }{ \displaystyle{\int_0^1} \dfrac{e^{x^n}-1}{e-1}\, dx}
$$
 

FAQ: Are Measures m and μ Mutually Singular Given Convergence Conditions?

What are mutually singular measures?

Mutually singular measures are a concept in measure theory where two measures are considered to be mutually singular if they do not share any common subsets, or if their intersection is a null set. This means that the two measures are completely independent of each other and cannot be combined or compared.

How are mutually singular measures different from absolutely continuous measures?

Mutually singular measures are the opposite of absolutely continuous measures. While mutually singular measures have no common subsets, absolutely continuous measures have no disjoint subsets. In other words, absolutely continuous measures can be combined and compared, while mutually singular measures cannot.

Can mutually singular measures be equivalent?

No, mutually singular measures cannot be equivalent. Equivalent measures are measures that have the same sets with measure zero. Since mutually singular measures have no common subsets, they cannot have any common sets with measure zero and therefore cannot be equivalent.

How are mutually singular measures used in real-world applications?

Mutually singular measures have various applications in probability theory, statistics, and mathematical analysis. One example is in the study of probability distributions, where mutually singular measures can help to distinguish between different types of distributions and their characteristics.

Are mutually singular measures always finite measures?

No, mutually singular measures can be both finite and infinite. A measure is considered finite if it assigns a finite value to the entire space, while an infinite measure assigns an infinite value to the entire space. Both types of measures can be mutually singular, as long as they do not share any common subsets.

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