In Progress... Ben
$ g(x) $ is monotone: Take $ x<y $. Then $ g(x) = \sum{f_n(x)} \le \sum{f_n(y)} = g(y) $.
Since $ f_n(x) \in AC[0,1] $ and $ f_n(0)=0, f(x) = \int_0^x{f_n'(t)dt} $
| $ g(x) $ | $ = \sum{\int_0^x{f_n'(t)dt}} $ |
| $ = \sum{\int_0^x{f_n'(t)dt}} $ |
