exercise:
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 $[0, T]$.
Show that $f$ is integrable on any bounded interval.
My approach:
Let $\varepsilon$ be given, because of integrability of $f$ on the interval $[0, T]$ there is a partition $P$ of that interval satisfies following inequality
$U(f,P)-L(f,P)<\varepsilon$
Let take any bounded interval $[c,d]$ and choose a partition of this interval $P'$ so that each subinterval of $P'$ be lesser than of any subinterval of $P$. In this case we have:
$U(f,P')-L(f,P')<\frac{d-c}{a} \varepsilon$ and from here we can conclude that it is integrable on $[c,d]$
My question:
Is this proof valid?
Best Answer
I think I found the acceptable proof with the help of Spivak's book. We need theorem 4 and exercise 12 on the chapter 13 from Spivak's "Calculus"
THEOREM 4: Let $a<c<b$. If $f$ is integrable on $[a, b]$, then $f$ is integrable on $[a, c]$ and on $[c, b]$. Conversely, if $f$ is integrable on $[a, c]$ and on $[c, b]$, then $f$ is integrable on $[a, b]$. Finally, if $f$ is integrable on $[a, b]$, then $$ \int_{a}^{b} f=\int_{a}^{c} f+\int_{c}^{b} f . $$ (The proof was given by the author in his book)
exercise 12
If $a<b<c<d$ and $f$ is integrable on $[a,d]$ prove that $f$ is integrable on $[b,c]$
proof: Theorem 4, applied to $a<b<d$ implies that $f$ is integrable on $[b,d]$. Then Theorem 4, applied to $b<c<d$ implies that $f$ is integrable on $[b,c]$.
Now, because of integrability on $[0,T]$ and periodicity the function $f$ is integrable on $[-T,0], [T,2T], [-2T,-T], [2T,3T], [-3T,-2T]......$
Now, from the above theorems we can conclude that it is integrable on $[-nT,nT]$ $\text{(where n is any natural number)}$ and also integrable on any bounded interval $\blacksquare$