Are Measures m and μ Mutually Singular Given Convergence Conditions?

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SUMMARY

The discussion centers on the conditions under which the measures m (Lebesgue measure) and μ (a positive Borel measure) are mutually singular given a sequence of positive continuous functions {g_n} on the interval I=[0,1]. The conditions specified include the pointwise convergence of g_n to 0 almost everywhere with respect to m, the integral of g_n being equal to 1 for all n, and the convergence of the integral of fg_n to the integral of f with respect to μ for all continuous functions f. The consensus leans towards the conclusion that m and μ are not mutually singular, as evidenced by the proposed triangular sequence and the need for further exploration of condition (iii).

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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.
 
Last edited:
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I would try
$$
g_n(x)= \dfrac{\dfrac{e^{x^n}-1}{e-1} }{ \displaystyle{\int_0^1} \dfrac{e^{x^n}-1}{e-1}\, dx}
$$
 

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