Prove that $$\lim_{n\to \infty} \int_{a}^{b}f(nx)dx=\frac{b-a}{T}\int_{0}^{T}f(t)dt.$$ Given that $f:\mathbb{R}\to \mathbb{R}$ and is continuous and periodic with period $T>0.$ Also $a<b.$
I observe that if we let $k=\frac{n(b-a)}{T}$ then $$\int_{na}^{na+T}f(t)dt+\int_{na+T}^{na+2T}f(t)dt+....+\int_{na+(k-1)T}^{nb}f(t)dt=n\int_{a}^{b}f(nx)dx.$$ Thus $$\int_{a}^{b}f(nx)dx=\frac{k}{n}\int_{0}^{T}f(t)dt=\frac{b-a}{T}\int_{0}^{T}f(t)dt.$$ I don't understand why would the author put a limit since the final answer does not seem to depend on $n.$ Please explain where am I making a mistake.
The mistake here is that the $k$ you define $k = \frac{n(b-a)}{T}$ is not always an integer, but you assume that it's an integer when you split the integral into $k$ integrals. Instead define $k = \left\lfloor\frac{n(b-a)}{T}\right\rfloor$ which is explicitly an integer. Then you are left with $k$ integrals over one period plus an extra bit over $\int_{na + kT}^{nb}$. This leads to
$$\frac{k}{n}\int_0^T f(t){\rm d}t + \frac{1}{n}\int_{na+kT}^{nb}f(t){\rm d}t = \int_a^b f(nt){\rm d}t$$
What's missing now is just to show that the second part vanish (note that a continuous and periodic function is bounded and use $|\int_c^d g(t){\rm d}t| \leq |g|_{\rm max}(d-c)$) and that $\frac{k}{n} \to \frac{b-a}{T}$ as $n\to\infty$ (use $x-1\leq \lfloor x\rfloor \leq x$).