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Hello,

Here is my original question: let $\mu$ be a singular measure with respect to the Lebesgue's measure on $R$. Is it true that $\int \psi \mu(d x)=0$ for any test function $\psi\in C_c^\infty(R)$ (smooth functions with compact support)?

Following the suggestion by André Henriques, I will put the answer in the answer box. :-)

Thanks again for everyone! :-)

Anand

p.s. this is related to my previous post.

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    $\begingroup$ After reading your other post, I think that I now understand what you mean by "singular". You're probably thinking of an increasing functions on $\mathbb R$, and you want the derivative of that function to be a measure that is singular w.r.t Lebesgue measure. $\endgroup$
    – André Henriques
    Commented Jul 12, 2011 at 21:28
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    $\begingroup$ In view of your answer to André Henriques's comment, I suppose the $f$ and $dx$ in the integrand in your question actually refer to the singular measure that acts as the "derivative" of some increasing, continuous function $g$. In that case, the answer to your question is no, the integral need not vanish. Suppose, for example, that $g$ is constant outside some interval $[a,b]$ (so the measure $f\,dx$ concentrates on $[a,b]$) while $\psi$ is identically 1 on $[a,b]$. Then the integral will be $g(b)-g(a)$. (Apologies if I misunderstood the question.) $\endgroup$
    – Andreas Blass
    Commented Jul 12, 2011 at 22:43
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    $\begingroup$ Anand, you should edit your question and replace it by the comment you gave (second in the thread right now). As it stands, what you wrote about the function $f$ is not true. The Cantor function is a singular continuous function, and clearly if $\psi = 1$ on $[1/3,2/3]$ and is non-negative elsewhere, $\int f\psi dx > 1/6$. $\endgroup$
    – Willie Wong
    Commented Jul 13, 2011 at 11:20
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    $\begingroup$ The simplest example of a singular measure is the so-called "delta function". The derivative of the unit step function. But of course $\int_R \psi(x)\delta(dx) = \psi(0)$. $\endgroup$
    – Gerald Edgar
    Commented Jul 14, 2011 at 12:42
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    $\begingroup$ @Anand: I think that the correct protocol is to type the answer to your question in the answer box, and then click on the check mark to accept your own answer. (Moderators correct me if I'm wrong) $\endgroup$
    – André Henriques
    Commented Jul 14, 2011 at 19:30

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After reading comments by Andreas and Wong, I understand that the above statement is wrong. Professor Gerald Edgar gives the simplest example (at least for me :-) ): delta function is a singular measure, but for any test function $\psi$ not vanishing at zero $\psi(0)\ne 0$, we have that $\int \psi \delta(d x)=\psi(0)\ne 0$.

Thanks everyone for your patience and helps! :-)

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