Indeed, if is any set with , then represents a point of , and defines the value of our function at this point. If is another set representing the same point, then . By absolute continuity, as well, and so . Thus our function doesn’t depend on the representative we use.
As for continuity at a point , given an , we want to find a so that if then . We calculate
Since is finite, we know that for every there is a so that if then . Using this , our assertion of continuity follows.
Now, if is a sequence of finite signed measures on that are all absolutely continuous with respect to , and if the limit exists and is finite for each , then the sequence is uniformly absolutely continuous with respect to . That is,
For any we can define the set
Since each is continuous as a function on , each of these is a closed set. Since the sequence always converges to a finite limit, it must be Cauchy for each , and so the union of all the is all of . Thus the countable union of these closed subsets has an interior point. But since is a complete metric space, it is a Baire space as well. And thus one of the must have an interior point as well.
Thus there is some , some radius , and some set so that the ball is contained in . Let be a positive number with , and so that whenever and This will suffice (by definition) for all up to . We will show that it works for higher as well. Note that if , then
so and are both inside . And so we calculate
The first term is less than by the definition of . The second and third terms are less than or equal to because and are in . Since the same works for all , the absolute continuity is uniform.