I have a question regarding the Riemann approximation. Suppose $a_{n,i}:=a_i=i/n$ is a sequence such that $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^na_i=\int_0^1x\text{d} x.$$ The first suggestion is that $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^na_{\pi(i)}=\int_0^1x\text{d} x$$ holds true due to absolute convergence, where $\pi$ denotes a permutation of $\{1,\ldots,n\}$. Another question is if we can say something about $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^na_ia_{\pi(i)}.$$ My idea was to use the Cauchy product, but I'm not really getting anywhere. For now, the convergence of the sum should be clear, since $$ \sum_{i=1}^n a_ib_i\leq \sum_{i=1}^n a_i\sum_{i=1}^n b_i.$$ But can one determine the limit and is the limit independent of the permutation $\pi$?
Thanks a lot already!
Update: It seems like the answer to the question above is that the Riemann sum above depends on the permutation $\pi$ and has no integral representation for general $\pi$. Therefore, I am wondering if the answer changes if we change the setup of this problem. Lets suppose we have distinct coordinates $0= x_1<x_2,\ldots ,x_{n-1}<x_n=1$ with $\max_{\substack{i,j=1,\ldots,n \\ i\neq j}}|x_i-x_j|\leq C_n$, where $C_n\rightarrow 0$ converges to zero with $n\rightarrow \infty$. Note that we do not impose that the coordinates have an order such as $x_1\leq \ldots < x_n$.
I think that we have $$ \lim_{n\rightarrow \infty}\frac{1}{n}\sum_{i=1}^n x_i = \int_0^1 x\text{ d}x.$$
Let $\pi$ be a permutation on $\{1,\ldots,n\}$. Is there anything we can say about $$\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{i=1}^nx_ix_{\pi(i)}?$$
I can also open a new question but since it is related to the one above, I'll keep it as an update of the old one.
The limit depends on $\pi$.
$\pi(i) = i$, the limit is $\int_0^1 x^2\mathrm dx=\frac13$.
$\pi(i) = (n-i)$ the limit is $\int_0^1 x(1-x)\mathrm d x = \frac16$.
for any $\pi$ the limits is between $\frac16$ and $\frac13$.