Q:

How can I check if a list index exists?

Seems as though
if not mylist[1]:
    return False

Doesn't work.

A:

You just have to check if the index you want is in the range of 0 and the length of the list, like this
if 0 <= index < len(list):

it is actually internally evaluated as
if (0 <= index) and (index < len(list)):

So, that condition checks if the index is within the range [0, length of list).
Note: Python supports negative indexing. Quoting Python documentation,

If i or j is negative, the index is relative to the end of the string: len(s) + i or len(s) + j is substituted. But note that -0 is still 0.

It means that whenever you use negative indexing, the value will be added to the length of the list and the result will be used. So, list[-1] would be giving you the element list[-1 + len(list)].
So, if you want to allow negative indexes, then you can simply check if the index doesn't exceed the length of the list, like this
if index < len(list):

Another way to do this is, excepting IndexError, like this
a = []
try:
    a[0]
except IndexError:
    return False
return True

When you are trying to access an element at an invalid index, an IndexError is raised. So, this method works.

Note: The method you mentioned in the question has a problem.
if not mylist[1]:

Lets say 1 is a valid index for mylist, and if it returns a Falsy value. Then not will negate it so the if condition would be evaluated to be Truthy. So, it will return False, even though an element actually present in the list.

A:

In the EAFP style of Python:
try:
    mylist[1]
except IndexError:
    print "Index doesn't exist!"

A:

In the case of integer-indexed lists, I'd simply do
if 1 < len(mylist):
  ...

For dicts, you can of course do
if key in mydict:
  ...

