Show $\hat f(\xi)=\hat{\nu}(\xi)$ where $\nu$ is probalility measure

Question

Let $f :\mathbb{R}\to\mathbb{R}$ be such that $\int_{\mathbb{R}} f \ d(x)=1$.

now define a measure $\nu$ on $\mathbb{R}$ such that $\nu(A)=\int_{A}f \ d(x)$ for any subset $A$ of $\mathbb{R}$.

now i am defining fourier transform of $\nu$ as $\hat{\nu}(\xi)= \int_{\mathbb{R}} e^{-i\xi x} d{\nu}$.

Now theorem says $$\hat f(\xi)=\hat{\nu}(\xi)$$ but i do not undestand this. $\hat f(\xi)=\int_{\mathbb{R}}f(x)e^{-i{\xi}x} d(x)$

how will i change the integration from measure $\nu$ to lebesgue measure.

i am very confused. any hint .

Answer 1

If $f$ is an absolutely Lebesgue integrable function on $\mathbb{R}$, and if $m$ is Lebesgue measure on $\mathbb{R}$, then $$ \nu(E) = \int_{E}fdm $$ defines a complex or finite signed measure on $\mathbb{R}$. Then $$ \int_{\mathbb{R}}gd\nu = \int_{\mathbb{R}}g\frac{d\nu}{dm}dm=\int_{\mathbb{R}}gfdm. $$ So the Fourier transform of $\nu$ becomes $$ \int_{\mathbb{R}}e^{-ist}d\nu(t)=\int_{\mathbb{R}}e^{-ist}\frac{d\nu}{dm}dm =\int_{\mathbb{R}}e^{-ist}fdm. $$