There are two problems for inf D. Its supremum might not be 1, and it might fail the left continuity. I don't think good conditions can be given other than checking those conditions directly.

However, if you define the distribution function as being right continuous instead of left continuous, then automatically inf D is right continuous as well.

That is so because a right continuous nondecreasing function is upper semicontinuous, and the infimum of an arbitrary family of upper semicontinuous functions is still upper semicontinuous.

With your definition and the condition at 0, any supremum of distributions functions would be a distribution function (by a similar argument involving lower semicontinuity), but left continuity and taking infima do not work well together.

Thank you for answer. Lets do some changes on the problem: 

How about: D(t)= Sups

Since each Fi is nondecreasing, their infimum is so as well, so D(t) is just  infi Fi(t).

Do you really need to use the left-continuous version of the distribution function?

Actually, it is a part of a problem in probabilistic normed spaces (in this spaces distribution functions are define left-continuous). In classical normed spaces we can easily define sup norm and sup metric. For example If X is some set and M is a metric space, then, the set of all bounded functions f  from X to M  can be turned into a metric space by defining d(f,g) = sup{x \in X} d(f(x),g(x)) for any two bounded functions f and g.

I am trying define inf probabilistic metrics in probabilistic metric spaces. 

Perhaps a naive comment, but if you take F_i to be the distribution function of a Dirac centered at i \in \N, then F_i(s) = 0 for all s

OK, I see now that my latter remark was wrong. Yes, indeed it is left-continuous. I guess you would have to use that as a definition and checo that everything works fine.

Dear Pedro, 

I didn't understand your last comment completely. If it is possible please explain it.

Thanks

To my mind the infimum not always is a distribution function, even if exists. Let, for example, F(x) be any cdf (between others also continuous) with F(x) infinity) F^n(x) = 0, for each x. Such a limit (as constant function) is not a cdf.

I think a necessary condition for the inf {( )} to be a cdf is existence of a cdf being any lower bound of the considered family. Such a lower bound for a family which posses the infimum as a cdf. always exists (being the same thing as the (existing) infimum). Probably, this is also the sufficient condition for the question's positive answer. The order I considered is defined coordinatewise.      

A book that might help is

http://www.amazon.com/Modern-Summation-Variables-Probability-Statistics/dp/9067642703

In this book several metrics on the space of distributions (or equivalently distribution functions) are defined and their properties analyzed. Maybe it helps with your problem.

The book is certainly worth to read. However, I think that for analysing the notion of infimum an order is more important than metrics. The latter rather more for analysing the convergence which is related to the main problem of infimum as well.