Given $ f: [a, b] \to \mathbb R $ and $ n \in \mathbb N $, assume $ h = \frac {b-a} {n} $ and the average of the numbers $ f ( a + h ), \ldots, f (a + nh) = f (b) $ given by $$ M (f; n) = \frac {1}{n} \sum_{j=1}^n f (a + kh) $$ Show that if $f$ is integrable in $[a,b]$, then $$ \lim_{n \to \infty} M (f; n) = \frac{1} {b-a} \int_a^bf(x)\,dx$$
For this reason, the number $ \frac{1}{b-a}\int_a^bf(x) \, dx $ is called the average of $ f $ in $[a,b]$.
Proof
We take the partition $P$ with $t_0 = a$, $t_1 = a + \frac{b-a}{n}$, … ,$t_k = a + \frac{{k(b-a)}}{n}$, … , $t_n = b$ with $\Delta t_{k-1} = \frac{b-a}{n}$, we dither the partition taking in each $[t_{k-1},t_k]$ the point $t_k$ we have with that $$\lim_{n \to \infty} \frac{b-a}{n} \sum_{k=1}^nf(a+kh)= \int_a^bf(x) \, dx$$ and so on $$\lim_{n \to \infty} M(f,n) = \frac{1}{b-a}\int_a^bf(x) \, dx$$
I found this proof and I don't think I understood it very well.
I know from class that $S(f; P)= \sum_{k=1}^n f(\alpha_k)(t_k-t_{k-1})$
where $P={}$dotted partition (sorry I don't know the English term) and $(t_k-t_{k-1})= \Delta t_k$ and in this case $(t_k-t_{k-1})= \Delta t_k = \frac{b-a}{n}$
then it would look like this $S(f, P)= \sum_{k=1}^n f(a+kh)\Delta t_k$? then the delta was removed from the sum and the limit was added to get the integral and we were like this $$\lim_{n \to \infty} \frac{b-a}{n} \sum_{k=1}^nf(a+kh)= \int_a^bf(x)\,dx$$
and we can do this because $f$ is integrable so it is also valid that $L= \lim_{\left|P\right| \to 0} S(f;P)$ but if it were that $n$ would tend to zero and not to infinity right?
And if that's what was done
To finish $$\lim_{n \to \infty} \frac{b-a}{n}\sum_{k=1}^nf(a+kh)= (b-a)\lim_{n \to \infty} \frac {1}{n}\sum_{k=1}^nf(a+kh)=\int_a^bf(x) \, dx $$ we divide by $b-a$ and we get $$\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^nf(a+kh)=\frac{1}{b-a} \int_a^bf (x)\,dx=\lim_{n \to \infty} M(f,n)=\frac{1}{b-a} \int_a^bf(x)\,dx$$
So how much did I understand correctly?
Thank's in advance for any help.