Summation of integrals over odd and even parts Let the function $f: [0,1]\rightarrow \mathbb{R}$ be continuous.
For $n\in \mathbb{P}$, partition $[0,1]$ into $2n$ equal pieces, and add up the integrals over the odd and even parts separately. 
$$A_{n}=\sum_{k=1}^{n}\int_{\frac{2k-2}{2n}}^{\frac{2k-1}{2n}}f(x)dx$$
$$B_{n}=\sum_{k=1}^{n}\int_{\frac{2k-1}{2n}}^{\frac{2k}{2n}}f(x)dx$$
Prove the following:
$$\lim_{n\rightarrow \infty }A_{n}=\lim_{n\rightarrow \infty }B_{n}=\frac{1}{2}\int_{0}^{1}f(x)dx$$
This is a problem from my analysis class. I am not sure how to approach the problem. Here are some of my thoughts:
Since the function is continuous on a compact set $[0,1]$, it is therefore integrable, and the integral over the closed interval $[0,1]$ is well-defined.
It is easy to show that $$\forall n\in \mathbb{P},A_{n}+B_{n}=\int_{0}^{1}f(x)dx$$
Therefore it suffices to show
$$\lim_{n\rightarrow \infty }A_{n}=\lim_{n\rightarrow \infty }B_{n}$$
Any suggestions on what to do next?
Thanks!
 A: Since $f$ is a continuous function, we can apply the Mean Value Theorem for Integrals to find points $\xi_k \in [\frac{2k-2}{2n},\frac{2k-1}{2n}]$ and  $\xi_k' \in [\frac{2k-1}{2n},\frac{2k}{2n}]$ such that
$$A_{n}=\sum_{k=1}^{n}\int_{\frac{2k-2}{2n}}^{\frac{2k-1}{2n}}f(x)dx = \frac{1}{2n}\sum_{k=1}^{n} f(\xi_k), \\\ B_{n}=\sum_{k=1}^{n}\int_{\frac{2k-1}{2n}}^{\frac{2k}{2n}}f(x)dx = \frac{1}{2n}\sum_{k=1}^{n} f(\xi_k').$$
where $|\xi_k - \xi_k'| \leq \frac{1}{n}$.
As $f$ is uniformly continuous on $[0,1]$, for each $\epsilon > 0$ there exists $\delta >0$ such that for all sufficiently large integers $n$ we have
$$|\xi_k - \xi_k'| \leq \frac{1}{n}< \delta, \\\ |f(\xi_k) - f(\xi_k')| < 2\epsilon $$.
Hence, for sufficiently large $n$ it follows that
$$|A_n - B_n| \leq \frac{1}{2n}\sum_{k=1}^{n} |f(\xi_k)-f(\xi_k')|<\frac{1}{2n}n(2\epsilon)= \epsilon,$$ and
$$\lim_{n\rightarrow \infty }[A_{n}-B_{n}]=0.$$
Since $A_n+ B_n = \int_{0}^{1}f(x)dx$, it follows that
$$\lim_{n\rightarrow \infty }A_{n}=\lim_{n\rightarrow \infty }B_{n}= \frac{1}{2}\int_{0}^{1}f(x)dx.$$
