Let $dx$ denote the Lebesgue measure on $[0,1]$. Are there any measures $\mu$ on $[0,1]$ which are a.c. with respect to $dx$ that have support $[0,1]$, but are not equivalent to the Lebesgue measure, i.e. they have different null-sets? If not, then this implies that support is an invariant for measures which are a.c. w.r.t. to the Lebesgue measure.
Since all measures which are a.c. w.r.t $dx$ have the form $d\mu = \int_0^1 fdx$ for some $f\in L^1[0,1]$, I feel like the answer should be no. It seems that the only way to get the measure to not be equivalent is if $f = 0$ on some subset of $[0,1]$, but then the support of $\mu$ is not all of $[0,1]$.
It is well known that there is a measurable set $E$ such that $0<m(E\cap I)<m(I)$ for every open interval $I \subset (0,1)$. [$m$ being Lebesgue measure]. This has been shown many times on MSE. Restriction of $dx$ to $E$ is a counter-example. Note that $E^{c}$ has measure $0$ under this restriction but not under Lebesgue measure.