Is the sum of sine and fractional part periodic? I am trying to to find whether the function $\sin(x)+\{x\}$ is periodic or not, where $\{x\}$ is fractional part of $x$. 
Now $\sin$ and $\{\ \}$ both are periodic functions with periods $2\pi $ and $1$ respectively. We can't use the result of continuous functions:

If $f$ and $g$ are continuous periodic functions with positive periods $T_{1}$ and $T_{2}$ respectively then $f+g$ is periodic iff l.c.m.$(T_{1},T_{2})$ exists and l.c.m.$(T_{1},T_{2})$ is a period of $f+g.$

But in our case first of all $\{\ \}$ is not a continuous function and second, l.c.m.$(2\pi,1)$ is not defined. Then how to check whether $\sin(x)+\{x\}$ is periodic or not and how to determine its period if it is?
 A: Let
$f(x)
=\sin(x)+\{x\}
$.
If $f$ is periodic
with period $t$,
then
$f(x)
=f(x+t)
$
for all $x$,
or
$\sin(x)+\{x\}
=\sin(x+t)+\{x+t\}
$.
Therefore
$\sin(x)+\{x\}
=\frac1{n}\sum_{k=1}^n (\sin(x+kt)+\{x+kt\})
=\frac1{n}\sum_{k=1}^n\sin(x+kt)+\frac1{n}\sum_{k=1}^n\{x+kt\}
=s_n(x)+t_n(x)
$.
Since
$\sum_{k=0}^n \sin(x+kt)
=\frac{\sin((n+1)t/2)\sin(x+nt/2)}{\sin(t/2)}
$,
$s_n(x)
\to 0
$
for
$t
\ne 2m\pi
$
for $m \in \mathbb{Z}$.
I now use the
equidistribution theorem
(https://en.wikipedia.org/wiki/Equidistribution_theorem)
which states that
$$"\lim\limits_{n \to \infty} 
\frac1{n}\sum\limits_{k=1}^n f((x+ka)\bmod 1)
=\int_0^1 f(y)dy
$$
holds for almost all $x$
and any Lebesgue integrable function $f$."
This shows that,
for almost all $x$,
$t_n(x)
\to \int_0^1 x\,dx
=\frac12
$
for almost all $x$.
Therefore,
if $f(x)$ has period $t$,
$f(x)
= \frac12
$
for almost all $x$,
which is obviously false.
A: Since, according to a comment, there seem to be examples of functions with noncommensurable periods having a periodic sum we have to provide an adhoc proof that $f(x):=\sin x+\{x\}$ is not periodic.
Assume that $f$ is periodic with period $T>0$. The function $f$ has a jump discontinuity  at the integers, and is continuous otherwise. This implies that $T\in{\mathbb N}_{\geq1}$. Now $f(\pi)=\{\pi\}$, and $f(\pi+T)=\{\pi\}-\sin T$. Since $\pi$ is irrational $T\notin \pi{\mathbb Z}$, and this implies $\sin T\ne0$. We then would have $f(\pi)\ne f(\pi+T)$, a contradiction.
