@yada, did you ever end up figuring this out for the case where $X = C_c (\mathbb{R})$?

Why would this not contradict Rademacher's theorem? (especially in the case of say $C^1$ boundary)

if you know of a reference in the literature to "natural l.c. topology on the space of measures which is complete and such that the associated convergent sequences are the ones in your question" and add it to your answer, I will accept it.

@DieterKadelka actually the Wikipedia page en.wikipedia.org/wiki/… contains a proof that the sequentially open sets form a topology.

Thank you for spotting this serious issue, and for your long comment! It is certainly food for thought.

@DieterKadelka this is discussed in Chapter 1 of Buttazzo's Semicontinuity, Relaxation and Integral Represenentation in the Calculus of Variations. Unfortunately the reference therein for the fact that the sequentially closed sets always form a topology is to a paper by Dolcher which is in Italian.

@900edges, there is a notion called either "displacement convexity" or "geodesic convexity" which I believe is what you are looking for.