Let's say I've a function $y:[0,\infty]\to\mathbb{R}$ that is periodic in $T$. If I want to know that average of the function over the defined region I need to find:
$$L:=\lim_{n\to\infty}\frac{1}{n-0}\int_0^ny(t)dt$$
But I can also write that as:
$$L=\frac{1}{T}\int_0^Ty(t)dt$$
How can I prove that that is true?
you can write $$ \frac 1n \int_0^n y(t) dt = \frac{1}{n} \sum_{k=1}^{ \left[\frac{n}{T} \right] } \int_{(k-1)T}^{kT} y(t) dt + \frac{1}{n} \int_{ T [n/T] }^n y(t) dt, $$ where $[\cdot]$ stands for the integer part. All integrals in the sum are equal to $\int_0^T y(t) dt$ in view of peridicity and you have $[n/T]$ of them. So we end up with $$ \frac 1n \int_0^n y(t) dt = \frac 1n [n/T] \int_0^T y(t) dt + \frac{1}{n} \int_{ T [n/T] }^n y(t) dt. $$
The last item above converges to 0, and $\frac{1}{n}[n/T]$ converges to $1/T$.