# Compute Limit Involving Integral and Periodic function [closed]

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?

## closed as off-topic by user26857, user228113, tired, Daniel W. Farlow, Chris GodsilFeb 19 '16 at 13:38

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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 $\lim_{t \to \infty} {k_t \over t} T =1$.

• Thank you,I'll accept your answer if i don't get any other "better" answer. – Dontknowanything Feb 19 '16 at 9:32