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.

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.

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)?

After reading comments by Andreas and Wong, I understand that the above statement is wrong. Gerald Edgar gives a simplest example: 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! :-)

Anand

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

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.

Hello,

The question might be too naive. I am just not very confident about it. Let $\mu$ be a singular measure with respect to the Lebesgue's measure on $R$. We have $\int \psi \mu(d x)=0$ for any test function $\psi\in\mathcal{D}(R)$ (smooth functions with compact support). So the singular measures are vanishing distributions. Am I right?

After reading comments by Andreas and Wong, the above statement is wrong. :-)

Thank you very much for any hints!

Anand

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

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)?

After reading comments by Andreas and Wong, I understand that the above statement is wrong. Gerald Edgar gives a simplest example: 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! :-)

Anand

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