My professor mentioned briefly during the class, that if $\nu$ is defined as above, then $\nu$ is a regular measure. The first criterion - every compact set has a finite $\nu$ measure is easy to show, but I have been struggling with the second piece.
My ultimate goal is to show: $\nu(E) = \inf \{\nu(U); E\subset U, U \text{open}\}$.
My thought is that since $m$ is regular
Folland theorem 1.18 implies for all $m$-measurable set $E$, $m(E) = \inf \{m(U); E\subset U, U \text{open}\}$
and the way that $\nu$ is defined, I can use
Folland corollary 3.6 implies if $f\in L^1(m)$, for every $\epsilon>0$, there is $\delta >0$ such that $|\int_E fd\nu|<\epsilon$ whenever $m(E)<\delta$.
to somehow connect $\nu(E)$ and $\inf \{m(U); E\subset U, U \text{open}\}$. However, when I try to write this, I had a hard time connecting the quantifiers and making things work. Thank you for the help!
Update: so I read over the paragraph on Folland that proves the statement given that $f\in L^+\cap L^1_{loc}$ and understood the argument. May I ask if the result holds for $f\in L^+\cap L^1_{loc}(m)$, can I prove it holds for $f\in L^1(m)$ using linearity?
$d\nu = f dm$ implies $$\nu(E) =\int_Efdm$$
Then $\nu$ is a signed measure (need not be non negative)
If $f\ge 0$ then $\nu$ is a measure.
Then rest of the proof follows from the properties of Lebesgue measure and Lebesgue integral.