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I am trying to prove the following:

Suppose $f$ is integrable on $[0,1]$ and define

$a_n = \frac{1}{n} \sum_{k=1}^n f \bigg(\frac{k}{n}\bigg)$

for all n. Prove $\{a_n\}_{n=1}^\infty$ converges to $\int_0^1 f dt$.

My strategy was to show the Riemann sum is equal to the limit of $\{a_n\}_{n=1}^\infty$, and I feel like I didn't exactly show that explicitly. I'm wondering if anyone can point out where I'm wrong or maybe redirect me. Here's my attempt:

Suppose $f$ is integrable on $[0,1]$ and define \begin{align*} a_n = \frac{1}{n} \sum_{i=1}^n f \bigg(\frac{i}{n}\bigg) \end{align*} Now let $\{P_n\}_{n=1}^\infty$ be a sequence of marked partitions of $[0,1]$ with $P_n = \big\{\frac{i}{n} : 0 < i \leq n\big\}$. Mark $P_n$ by letting $t_i = \frac{i}{n}$ for all $n$. Then since $f$ is integrable we have \begin{align*} S(P_n,f) = \sum_{i=1}^n f(t_i)\frac{1}{n} = \sum_{i=1}^n f \bigg(\frac{i}{n}\bigg)\frac{1}{n} = \int_0^1 f dt.\\ \end{align*} But, \begin{align*} \lim_{n \rightarrow \infty} a_n &= \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^n f \bigg(\frac{i}{n}\bigg)\\ & = \lim_{n \rightarrow \infty} \sum_{i=1}^n f \bigg(\frac{i}{n}\bigg) \frac{1}{n} = S(P_n,f) \end{align*} So we have $\lim_{n \rightarrow \infty} a_n = \int_0^1 f dt$.

Thank you!

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  • $\begingroup$ What is the definition of Riemann integrability you are working with? $\endgroup$ – Henricus V. Mar 28 '16 at 2:42
  • $\begingroup$ I am using the sequential definition. Essentially it says f is Riemann integrable on [a,b] iff for each sequence of marked partitions with its mesh converging to zero the sequence of Riemann sums converges to $\int_a^b f dx$ $\endgroup$ – Justin Mar 28 '16 at 2:45
  • $\begingroup$ Note that $S(P_n,f)$ does not equal the integral, but it's limit does. Your answer is fine (it's somewhat overcomplicated). You just needed to show that the sequence was really a Riemann sum (or a sequence of Riemann sums). $\endgroup$ – YoTengoUnLCD Mar 28 '16 at 3:12

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