The argument can easily be modified to yield a proof for the case of $k$ positive.
At first glance Lemma 2 seems to yield four possible outcomes.
Another proof (yielding more information) can be found in [GH].
Specifically, one might hope that a clever application of something like Choquet's theorem would yield the desired conclusion.
The idea of the following proof, which yields both (a) and (b) at one stroke, is due to von Neumann.
This finally yields $f=g$. [Not: “yields that $f=g$”]
The case where $i$ is odd yields to a similar argument.
Go to the list of words starting with: a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
s
t
u
v
w
y
z