$f$ is periodic ($p$>0) and integrable at $[0,p]$. Prove that $f$ is integrable in any compact interval and $\int_0^{np} f = n \int_0^p f$ Suppose that $f$ is periodic ($p$>0) and integrable at $[0,p]$. Prove that $f$ is integrable in any compact interval and that $$\int_0^{np} f = n \int_0^p f$$
I already proved before that:
If $f$ is integrable in $[a,b]$ and $g(x)=f(x-c)$, then $g$ is integrable $[a+c,b+c]$ and
$$\int^{b+c}_{a+c}g= \int_a^b f$$
How can I use this result to prove the question?
I know that a function is periodic with period $p$ if $f(x)=f(x+p) \ \forall x$
 A: 
If $f$ is integrable in $[a,b]$ and $g(x)=f(x-c)$, then $g$ is integrable $[a+c,b+c]$ and
$$\int^{b+c}_{a+c}g= \int_a^b f$$

Take $a=0$ and $b=c=p$ in the above result. Since $f(x-p)=f(x)$ for any real $x$ by periodicity, you get $$\int_p^{2p} f(x)\,\mathrm d x=\int_p^{2p} f(x-p)\,\mathrm dx=\int_0^p f(x)\,\mathrm dx.$$ By induction, you can prove the same way that $$\int_{mp}^{(m+1)p}f(x)\,\mathrm dx=\int_0^p f(x)\,\mathrm dx$$ for any non-negative integer $m$. Therefore, $$\int_0^{np}f(x)\,\mathrm dx=\sum_{m=0}^{n-1}\int_{mp}^{(m+1)p}f(x)\,\mathrm dx=\sum_{m=0}^{n-1}\int_{0}^{p}f(x)\,\mathrm dx=n\int_{0}^{p}f(x)\,\mathrm dx.$$

As for proving integrability on any compact interval in the first place, consider any pair $a,b$ of real numbers with $a<b$. Let

*

*$m$ be the integer satisfying $mp<a\leq(m+1)p$; and

*$n$ be the integer satisfying $np<b-mp\leq (n+1)p$.

This way, the integration of $|f|$ on $[a,b]$ can be decomposed as follows:
\begin{align*}
\int_a^b|f(x)|\,\mathrm dx&=\int_{a-mp}^{b-mp}|f(x)|\,\mathrm dx\leq\int_0^{b-mp}|f(x)|\,\mathrm dx\\&=\underbrace{\sum_{\ell=0}^{n-1}\int_{\ell p}^{(\ell+1)p}|f(x)|\,\mathrm dx}_{(\diamondsuit)}+\underbrace{\int_{np}^{b-mp}|f(x)|\,\mathrm dx}_{(\clubsuit)},
\end{align*}
where the first equality is due to periodicity, and the weak inequality comes from the fact that $mp<a$, so $0<a-mp$. By periodicity, $$\int_{\ell p}^{(\ell+1)p}|f(x)|\,\mathrm dx=\int_0^p|f(x)|\,\mathrm dx<\infty\quad\text{for every $\ell\in\{0,\ldots,n-1\}$}$$ given that $f$ is integrable on $[0,p]$, so that the term ($\diamondsuit$) is finite. As for the ($\clubsuit$) term, periodicity again implies that $$\int_{np}^{b-mp}|f(x)|\,\mathrm dx=\int_{0}^{b-mp-np}|f(x)|\,\mathrm dx\leq\int_{0}^{p}|f(x)|\,\mathrm dx<\infty,$$ where the weak inequality comes from the fact that $np<b-mp\leq(n+1)p$, so $0<b-mp-np\leq p$.
A: $$\int^{np}_0 f=\sum^n_{k=1}\int^{kp}_{(k-1)p} f(x)\,dx=\sum^n_{k=1}\int^p_0f(x+(k-1)p)\,dx=\sum^n_{k=1}\int^p_0 f(x)\,dx=n\int^p_0 f$$
where we have used the fact that $f$ is $p$ periodic by assumption.
