As @jonrsharpe says, one can use lookaround to reduce the number of characters splitted away, for instance to a single one. For instance if you don't mind losing space characters, you could use something like:

>>> re.split('\s(?=[A-Z])',content)
['Is this the number 3?', 'The text goes on...']

You can split using spaces with the next character an uppercase. But the T is not consumed, only the space.

Alternative approach: alternating split/capture item

You can however use another approach. In case you split, you eat content, but you can use the same regex to generate a list of matches. These matches is the data that was placed in between. By merging the matches in between the splitted items, you reconstruct the full list:

from itertools import chain, izip
import re

def nonconsumesplit(regex,content):
    outer = re.split(regex,content)
    inner = re.findall(regex,content)+['']
    return [val for pair in zip(outer,inner) for val in pair]

Which results in:

>>> nonconsumesplit('\D[.!?]\s[A-Z]|$|\d[!?]\s|\d[.]\s[A-Z]',content)
['Is this the number ', '3? ', 'The text goes on...', '']
>>> list(nonconsumesplit('\s',content))
['Is', ' ', 'this', ' ', 'the', ' ', 'number', ' ', '3?', ' ', 'The', ' ', 'text', ' ', 'goes', ' ', 'on...', '']

Or you can use a string concatenation:

def nonconsumesplitconcat(regex,content):
    outer = re.split(regex,content)
    inner = re.findall(regex,content)+['']
    return [pair[0]+pair[1] for pair in zip(outer,inner)]

Which results in:

>>> nonconsumesplitconcat('\D[.!?]\s[A-Z]|$|\d[!?]\s|\d[.]\s[A-Z]',content)
['Is this the number 3? ', 'The text goes on...']
>>> nonconsumesplitconcat('\s',content)
['Is ', 'this ', 'the ', 'number ', '3? ', 'The ', 'text ', 'goes ', 'on...']