# Riemann sums with equidistant sample points converge to the integral

Suppose $f \in R(x)$ on $[0,1]$. Define $$a_n = \frac 1n \sum_{k=1}^n f(\frac kn)$$ for all n. Prove $\{a_n\}_{n=1}^\infty$ converges to $\int_0^1 fdt$.

Honestly I've been trying to tackle this problem for away and am not really even sure how to begin, so any help would be greatly appreciated!

What definition of Riemann integral are you using?

For me, it's a number $s$ such that for any $\epsilon >0$ there is a $\delta$ such that the Riemann sum on a mesh finer than $\delta$ is within $\epsilon$ of $s$.

In that case, you need to show that for any $\epsilon$ there is a $N$ such that $\left| a_n - \int_0^1fdt \right| < \epsilon$ whenever $n>N$. So just choose $N$ such that $1/N < \delta$ in the definition of the Riemann integral, since your sequence is a sequence of Riemann sums.

EDIT: Let me spell out some more details given your comments. The sum that you have evaluates $f$ at $\frac{k}{n}$ and multiplies by $\frac{1}{n}$ and then sums. So, that's like dividing $[0,1]$ in to $n$ equal sized parts of $\frac{1}{n}$ and marking the right endpoint of each part of the partition.

So, given $\epsilon$, we know there is a $\delta$ such that taking any partition $P$ with $\mu(P)<\delta$ and any way of marking it, the Riemann sum is within $\epsilon$ of $\int_0^1fdt$. So, if we take $N$ very large, so that $\frac{1}{N} < \delta$, then with $n\geq N$ the partition $P_n$ that cuts $[0,1]$ in to equal sized pieces of size $\frac{1}{n}$ is such that taking the Riemann sum with any marking over that partition is within $\epsilon$ of $\int_0^1fdt$. So, in particular, if we take right endpoints in the partition, we get the sum $$\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k}{n}\right)$$ is within $\epsilon$ of the integral.

So, given $\epsilon$, we can find a $N$ such that for all $n\geq N$ we have $|a_n - \int_0^1fdt|<\epsilon$, which shows that $a_n$ converges to the integral.

• Yeah that is the definition of the Reimann sum I would be using for this problem. In all honestly I'm sort of confused by the entire concept of Reimann Sums because I'm not entirely sure how to choose the necessary marking of each partition. Which makes it difficult to decide what real number $\delta$ should be. – jlang Nov 6 '14 at 7:02
• There is a number A such that for each $\epsilon > 0$ there is $\delta > 0$ such that for any partition P of $[a,b$] with $\mu(P) < \delta$, regardless of how P is marked, $|S(P,f) - A| \le \epsilon$. If the condition is satisfied, then A = $\int_a^b fdx$. – jlang Nov 6 '14 at 7:09