Riemann Intergral proof question I am currently studying for a grad test and I have came across one problem that I cannot even begin to solve. I have thrown definitions and theorems at it and just keep running into road blocks. A little help or maybe a shove in the right direction would be appreciated.
The question reads as the following.
Let f be continuous on interval [0,1]. Prove that $\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^nf(\frac{k}{n})= \int_0^1f(x)dx$
Again any help would be greatly appreciate. I have attempted throwing the definition of continuity at it cause it states that it is continuous. I then used a uniform partition  and stated my $M_k$, $m_k$ based on the equation. I had my partion looking like this $P=(\frac{1}{n}+\frac{2}{n}+...+\frac{n}{n})$. From here it was not apparent where to go. 
 A: Let $\epsilon>0$. According to Heine-Cantor theorem, there is a $\delta>0$, such if $|x_1-x_2|<\delta$, then $|f(x_1)-f(x_2)|<\epsilon$. Let's take $N$ large enough, so ${1\over N}>\delta$, and let $n>N$. For every $1\le k\le n$, and for every $x\in [{k-1\over n},{k\over n}]$, we got $|{k\over n}-x|\le {1\over n}<\delta$, and so, $|f({k\over n})-f(x)|<\epsilon$. Or, in another words, $f({k\over n})-\epsilon<f(x)<f({k\over n})+\epsilon$. If we take its integral, we get ${f({k\over n})-\epsilon\over n}\le\int_{k-1\over n}^{k\over n}f(x)dx\le {f({k\over n})+\epsilon\over n}$. Sum it for every $1\le k\le n$, and you get ${1\over n}\sum_{k=1}^n{f({k\over n})-\epsilon\le\int_0^1 f\le {1\over n}\sum_{k=1}^n f({k\over n})+\epsilon}$ (or $|\int_0^1 f(x)dx-{1\over n}\sum_{k=1}^n f({k\over n})|\le\epsilon$). Now, by the defination of limit, $\lim_{n\to\infty}{1\over n}\sum_{k=1}^n f({k\over n})=\int_0^1 f(x)dx$
A: Let $P = \{x_1, ..., x_n \}$ be a partition of $[0, 1]$.
Now, clearly, $L(P, f) \leq \frac{1}{n} \sum_{j=1}^n f(\frac{j}{n}) \leq U(P,f)$ and given $\epsilon > 0$ there exists a partition $P$ such that $U(P,f) - L(P,f) < \epsilon$.  Now complete the proof on your own.
