Compute Limit Involving Integral and Periodic function

Question

If $f: \bf R \to \bf R$ is continuous and periodic with period $T$, then show that $$ \frac{1}{t}\int_{a}^{a+t}f(s)ds \to \frac {1}{T}\int_{0}^{T}f(s)ds$$ where $a\in \mathbb{R}$ and $ t \to \infty$

I think its a easy problem but unable in catching the "trick".Any hint?

Answer 1

This is a little sloppy, but gives the idea:

Partition the interval $[a,a+t]$ into $(a, nT, (n+1)T,...., (n+k_t)T, a+t)$, where $n, k_t$ are such that $nT-a < T, a+t-(n+k_t)T <T$.

Let $I= {1 \over T}\int_0^T f$, then ${1 \over t} (\int_a^{a+t} f) = {1 \over t} (\int_a^{nT} f + k_t I +\int_{(n+k)T}^{a+t} f)$.

Now bound the various terms in the integral and compute $\lim_{t \to \infty} {k_t \over t} $.

Hint:

Note that the first and last terms inside the parentheses are bounded, and note that $k_t T \le t < (2+k_t)T$, so $\lim_{t \to \infty} {k_t \over t} T =1$.