demo:

import re

inner = 'VARCHAR(30) COLLATE "en_US"'

result = re.sub(
    r'((?: COLLATE.*)?)$',
    r'FOO\1',
    inner
)

print(inner)
print(result)


in all Python versions prior to 3.7:

    VARCHAR(30) COLLATE "en_US"
    VARCHAR(30)FOO COLLATE "en_US"

in Python 3.7.0b2:

    VARCHAR(30) COLLATE "en_US"
    VARCHAR(30)FOO COLLATE "en_US"FOO

platform: Fedora 27 
python build:
Python 3.7.0b2 (default, Mar  5 2018, 09:37:32) 
[GCC 7.2.1 20170915 (Red Hat 7.2.1-2)] on linux

correction, that's fedora 26, not 27

This is intentional change.

Prior to 3.7 re.sub() didn't replace empty matches adjacent to a previous non-empty match. In 3.7 it does. Together with other changes this made all four functions that search multiple matches of the pattern (re.findall(), re.finditer(), re.split() and re.sub()) consistent.

In your example the pattern matches not only from " COLLATE" to the end of input string, but an empty string at the end of input string. If you do not want matching an empty string, just remove the '?' qualifier.

can you point me to the documentation?

also, removing the "?" is not an option for me.   I need the brackets to be placed prior to the "COLLATE" subsection, but unconditionally even if the "COLLATE" section is not present.     Looking at the change the behavior seems wrong to me.   The regexp states, "match the end of the string, plus an optional "COLLATE" clause, into a capturing expression.  replace everything here, e.g. the capturing part as well as the dollar sign part, with a single instance of FOO plus the captured part".   It is entirely unintuitive to me how a second replacement would be occurring here.   I cannot prove it but I think this change is wrong.

I will try to rewrite the expression.

Just see the re.sub() documentation for 3.7. There is also a note in the What's New document, in the "Changes in the Python API" section.

In your case you can just pass 1 as the fourth parameter of re.sub().

for now the quickest solution is to add "count=1" so that it only replaces once.

for those watching this would be the findall() case which is consistent between pythons:

import re

for reg in [
    'VARCHAR(30) COLLATE "en_US"',
    'VARCHAR(30)'
]:

    print(re.findall(r'(?: COLLATE.*)?$', reg))


output (all pythons):

[' COLLATE "en_US"', '']
['']

so yes there are two matches for one and only one for the other.