Average of periodic functions A function $f$ is periodic if there exists $T \neq 0$ (a period) such that $f(x+T)=f(x)$ for all $x$. Let $f$ be a periodic function with fundamental period $T$ (that is, $T$ is the lowest positive period, supposing that one such exists). Suppose that $f$ is bounded and integrable on the interval $[0, T]$.
in the previous theorem we have proved that:
$$ \lim\limits_{\vert{y}\vert\to\infty}\frac{1}{y}\int_0^yf=\frac{1}{T}\int_0^Tf$$
The quantity $$\frac{1}{T}\int_0^Tf$$  we will call mean of the periodic function
Exercise:
Let $$F(x)=\int_0^xf$$ Show that $F$ is periodic if and only if $f$ has mean zero.
My approach:
First suppose that $F$ is periodic function with the lowest positive period $a$.
Then $$F(x)=F(x+a)$$
$$\int_0^xf(u)\,d(u)=\int_0^{x+a}f(u)\,d(u)=\int_0^af(u)\,d(u)+\int_a^{a+x}f(u)\,d(u)$$
Since $$\int_a^{a+x}f(u)\,d(u)=\int_0^xf(u)\,d(u)$$ it implies that $$\int_0^af(u)\,d(u)=0$$
Second:
Assume that: $$\frac{1}{T}\int_0^Tf=0$$
$$F(x+T)=\int_0^{x+T}f(u)\,d(u)=\int_0^Tf(u)\,d(u)+\int_T^{T+x}f(u)\,d(u)=\int_0^Tf(u)\,d(u)+\int_0^xf(u+T)\,d(u)=\int_0^Tf(u)\,d(u)+\int_0^xf(u)\,d(u)=\int_0^xf(u)\,d(u)=F(x)$$
Please, can anybody check this proof??

copying the answer of @ N. S. to the similar question in here
\begin{align}
F(2T)&=F(T+T)=2F(T)-F(0) \\
F(3T)&=F(2T+T)=F(2T)+F(T)-f(0) \\
F(4T)&=F(3T+T)=F(3T)+F(T)-f(0) \\
\vdots& \\
F(nT)&=F((n-1)T)+F(T)-F(0)\end{align}
Add all these relations and cancel $F(2T), F(3T),.., F((n-1)T)$. You get
$$F(nT)=nF(T)-(n-1)F(0)=(n-1)\left( \frac{n}{n-1}F(T)-F(0) \right)$$
Now, if $F(T) \neq F(0)$ the last bracket goes to $\infty$ or $-\infty$
Now if $F(a)≠F(0)$, then this would be unbounded. But F is bounded since it is continuous periodic function.
