Integration of a Periodic Function Problem: 
$f : \mathbb{R} \to  \mathbb{R}$  is a continuous and periodic function with period $T>0.$
Prove that: 

$$\lim_{n\to +\infty} \int_{a}^{b} f(nx) dx =\frac{b-a}{T} \int_{0}^{T} f(x) dx$$

I tried substituting $nx=t$, but it gave me $\frac{1}{n} \int_{na}^{nb} f(t)dt$, and I don't know what to do. Can anyone give me hints to solve this? Or is there another way to solve this problem?
 A: Assume $a < b$ and $f$ is bounded.
We have $ b = a + \dfrac{n(b-a)}{T} \dfrac{T}{n} = a + \left\lfloor \dfrac{n(b-a)}{T} \right \rfloor \dfrac{T}{n} + r_n\dfrac{T}{n}$ where $0 \leq r_n < 1.$
So we can write for large $n$ $$\int_{a}^{b} f(nx) dx = \sum_{k=1}^{\lfloor\frac{n(b-a)}{T}\rfloor} \int_{a+(k-1)\frac{T}{n}}^{a+k\frac{T}{n}} f(nx)dx + \int_{a+\lfloor \frac{n(b-a)}{T}\rfloor\frac{T}{n}}^{b} f(nx)dx.$$
Since $f(nx)$ is a periodic function of period $\frac{T}{n}$ we have $$\int_{a+(k-1)\frac{T}{n}}^{a+k\frac{T}{n}}f(nx)dx = \int_{0}^{\frac{T}{n}} f(nx)dx = \dfrac{1}{n} \int_{0}^{T}f(x)dx.$$
If $|f(x)| \leq M$ for all $x \in [0,T]$ and hence all $x$ note that $$\left|\int_{a+\lfloor \frac{n(b-a)}{T}\rfloor\frac{T}{n}}^{b} f(nx)dx\right| \leq M (b - (a+\lfloor \frac{n(b-a)}{T}\rfloor\frac{T}{n})$$ and since $\lfloor \frac{n(b-a)}{T}\rfloor\frac{T}{n} \to b-a$  as $n \to \infty$ this term tends to $0$.
Hence the above sum reduces to 
$$\left\lfloor \dfrac{ n (b - a) } {T}\right\rfloor \dfrac{1}{n} \int_{0}^{T} f(x)dx + \delta_n $$ where $\delta_n \to 0$.
Since $$\left\lfloor \dfrac{ n (b - a) } {T}\right\rfloor \dfrac{1}{n} \to \frac{b-a}{T}$$ the result follows.
A: This maybe to late but I am posting a solution of a slightly more general problem whence a solution to the OP follows:

(Féjer) Suppose $f$ is a bounded measurable $T$-periodic function on $\mathbb{R}$ ($T>0$). For any $\phi\in\mathcal{L}_1(\mathbb{R})$ and numeric sequence $\alpha_n\in\mathbb{R}$,
$$
\lim_n\int \phi(x)f(nx+\alpha_n)\,dx=\Big(\frac{1}{T}\int^T_0f\Big)\int \phi \tag{1}\label{one}
$$
Proof:
Let $M=\sup_x|f(x)|$. Suppose $\phi=\mathbb{1}_{[a,b]}$ for $-\infty<a<b<\infty$. Since
$$
bn+\alpha_n= an+\alpha_n+\Big\lfloor \frac{n(b-a)}{T}\Big\rfloor T + r_n
$$
where $0\leq r_n<T$, we get from the periodicity of $f$ that
$$
\begin{align}
\int \phi(x)f(nx+\alpha_n)\,dx &= \int^b_a f(nx + \alpha_n)\,dx =\frac{1}{n}\int^{nb+\alpha_n}_{na+\alpha_n}f(u)\,du\\
&=\Big[\frac{n(b-a)}{T}\Big]\Big(\frac{1}{n}\int^T_0f \Big) + \frac{1}{n}E_n
\end{align}
$$
where $|E_n|\leq \Big|\int^T_0 f\Big|\leq TM$ for all $n$. Passing to the limit gives $\eqref{one}$ for intervals, and by linearity, for any step function.
Since step functions are dense in $L_1$, given $\varepsilon>0$, there is a step function $s$ such that $\|\phi-s\|_1<\varepsilon/M$. Hence
Let $I_n\phi$ denote the integral on the left hand side of $\eqref{one}$. Then
$$ |I_n\phi-I_ns|\leq M\|\phi-s\|_1<\varepsilon$$
and
$$\Big|\Big(\frac{1}{T}\int^T_0 f\Big)\int (\phi-s)\Big|< \varepsilon$$
Putting things together, leads to the conclusion of the problem.

The OP hollows from the particular case $\phi=\mathbb{1}_{[a,b]}$ and $a_n=0$.
A: Hint: notice that, as $n$ grows larger, the interval $n b - n a$ can be split into $k$ subintervals of size $T$, with $k = \text{floor}(\frac{n b -n a}{T})$. Using periodicity, split the integral over this interval as a sum of integrals over $(0,T)$ and a remainder, then take the limit.
