If function g satisfies $g(x)+1 = g(x+1)$, can $g$ be defined arbitrarily for $0\leq x<1$? How? To try to understand this concretely, I chose function $\mathcal f$ to be defined by $\mathcal f(x) = x + 1$ and the piecewise function:
$$g(x)= 
\begin{cases}\sin(x), & \text{0 $\le x\lt$ 1} \\x-1, & \text{$x\lt$ 0 or $x \ge$ 1}\end{cases}$$ 
for $\mathcal g$. But I don't seem to have gotten $\mathcal f \circ g$ = $\mathcal g \circ f$ for all $\mathcal x$ (perhaps I may have miscalculated). How do I prove this analytically, and$-$if this specific $\mathcal g(x)$ does not satisfy this condition$-$why not?
My background in math is severely lacking, so try your best to be as simplistic as possible. Thank you.
 A: You appear to be misunderstanding what is being asserted here.  The statement is that the functional equation $g(x) + 1 = g(x+1)$ has solutions where $g(x)$ is defined arbitrarily on the interval $0 \le x < 1$.  On each other interval $n \le x < n+1$ for nonzero integer $n$, this choice determines what $g$ must 
be  in order to make the functional equation true.  If you want to take $g(x) = \sin(x)$ for $0 \le x \le 1$, then $g(x) = n + \sin(x-n)$ for $n \le x < n+1$.
A: The equation
$$
g(x + 1) = g(x) + 1\quad 
$$
is "standardly" understood as
$$
\Delta _{\,x} \,g(x) = 1\quad  \Leftrightarrow \quad \,g(x) = \Delta _{\,x} ^{\,( - 1)} 1 = \sum\nolimits_x 1  = x + c
$$
where $\sum\nolimits_x {} $ indicates the Indefinite Sum
and $\Delta _{\,x}$ the Forward Difference.
Now, the $c$ above is normally taken as a constant, but in the more general terms it can actually
be any periodic function of $x$ with period  (one of the periods) $= 1$, so the "general" solution is
$$
g(x) = x + \pi _1 (x)
$$
For instance, it can be taken as
$$
\pi _1 (x) \in \left\{ {\sin (2\pi x),\;1 - \cos ^{\,2} (\pi x),\;\left\{ x \right\} = x - \left\lfloor x \right\rfloor ,\; \cdots } \right\}
$$
or, as in the answer above,  as  ${\sin \left( {\left\{ x \right\}} \right) - \left\{ x \right\}}$.  
So that is the indefinition /arbitrary choice you are wondering about.
