Here's a "terminological" answer with no actual content. See also this question, whose answers subsume both Michael Greinecker's and mine.
A measure $\tilde{\nu}$ is absolutely continuous with respect to another measure $\nu$ if it doesn't assign positive measure to any set which $\nu$ says has measure zero.
The Radon-Nikodym theorem says that if $\tilde{\nu}$ is absolutely continuous with respect to $\nu$, and both have finite total measure, then $\tilde{\nu}$ can be expressed in terms of $\nu$ in the way you describe: $\tilde{\nu}(A) = \int_A \lambda\;d\nu$ for some measurable function $\lambda$. On the other hand, if $\tilde{\nu}$ isn't absolutely continuous with respect to $\nu$, it's clear that $\tilde{\nu}$ can't be expressed in this way.
So, the measures which can't be expressed in terms in the way you describe are precisely the ones which are not absolutely continuous with respect to whatever measure $dx$ means.
Michael Greinecker's answer gives an excellent example of a non-atomic measure on $[0, 1]$ which is not absolutely continuous with respect to the Lebesgue measure.