# Proof using Darboux's Definition: $\int_0^1f=\lim_{n\to \infty}\frac1n\sum_{i=1}^{n}f\left(\frac in\right)$

Let $f:[0,1]\to \mathbb{R}$ be Darboux integrable. I ask for a proof of $$\int_0^1f=\lim_{n\to \infty}\frac1n\sum_{i=1}^{n}f\left(\frac in\right)$$ where the integral in the left hand side is the Darboux Integral of $f$. I know how to prove this when the integral is the Riemann Integral or when $f$ is monotone. I also know that the Darboux Integral is equivalent to the Riemann Integral. But can the above be proven directly without going into Riemann Sums?

For completeness here are the definitions: If $\mathcal{P}=\left\{a=x_0<...<x_n=b\right\}$ partitions $[a,b]$ and $f:[a,b]\to \mathbb{R}$ is bounded we define $$M_i(f)=\sup_{x\in [x_{{i-1}},x_i]}f(x)\text{ and }m_i(f)=\inf_{x\in [x_{{i-1}},x_i]}f(x)$$ The Darboux sums are: $$U_{f,\mathcal{P}}:=\sum\limits_{i=1}^{n}{M_i(f)\left( x_i-x_{i-1} \right)}\text{ and } L_{f,\mathcal{P}}:=\sum\limits_{i=1}^{n}{m_i(f)\left( x_i-x_{i-1} \right)}$$ The Darboux upper and lower integrals are $$\overline{\int\limits_a^b}f:=\inf_{\mathcal{P}}U_{f,\mathcal{P}}\text{ and } \underline{\int\limits_{a}^{b}}f:=\sup_{\mathcal{Q}}L_{f,\mathcal{Q}}$$ If the two coincide, $f$ is integrable and their common value is the Darboux integral of $f$.

We also have this criterion:

$f$ is integrable iff $\forall \epsilon>0$ there exists a partition $\mathcal{P}$ of $[a,b]$ such as that $U_{f,\mathcal{P}}-L_{f,\mathcal{P}}<\epsilon$

which can be written equivalently as

$f$ is integrable iff there exists a sequence of partitions $\mathcal{P}_n$ of $[a,b]$ such as that $U_{f,\mathcal{P}_n}-L_{f,\mathcal{P}_n}\to 0$. In that case $$\int\limits_{a}^{b}f=\lim_{n\to \infty}U_{f,\mathcal{P}_n}=\lim_{n\to \infty}L_{f,\mathcal{P}_n}$$

Proof if $f$ is monotone: WLOG $f$ is increasing. Then $$\mathcal{P}_n=\left\{ 0=x_0<...<x_i=\frac{i}{n}<...<x_n=1 \right\}$$ partitions $[0,1]$. Obviously, $M_i(f)=f(\frac in)$ and so $$\int\limits_{0}^{1}f=\lim_{n\to \infty}U_{f,\mathcal{P}_n}=\lim_{n\to \infty}\sum\limits_{i=1}^{n}{M_i(f)\left( x_i-x_{i-1} \right)}=\lim_{n\to \infty}\sum\limits_{i=1}^{n}{f(\frac in)\left(\frac1n\right)}$$

I will say it again: I don't want Riemann sums to be used, only upper and lower sums as above.

What happens if you compare $f(i/n)$ to $M_i(f)$ and $m_i(f)$ for $x_i=i/n$? – Alex R. Jan 4 '13 at 19:41
@Alex I can't believe it's that simple. You have $m_i<f(\frac in)<M_i\implies L_{f,\mathcal{P}_n}<\sum_{i=1}^n\frac1n f(\frac in)<U_{f,\mathcal{P}_n}$ and by letting $n\to \infty$ we have the result. I believe this is correct. Why don't you post it as an answer? – Nameless Jan 4 '13 at 19:45
@ThomasAndrews Yeah I mistook $\exists$ for $\forall$. Can you explain the point with the refinements? – Nameless Jan 4 '13 at 20:59
@ThomasAndrews Here is what I think. Let $\mathcal{P}_n$ be paritions so that $U_{f,P_n}-L_{f,P_n}\to 0$ and $\mathcal{Q}_n=P_n\cup\left\{0,\frac1n,...,1\right\}$. Then $U_{f,Q_n}-L_{f,Q_n}\to 0$. Can we show that $U_{f,Q_n}\ge \sum\limits_{i=1}^{n}{f(\frac in)\left(\frac1n\right)}$? – Nameless Jan 4 '13 at 21:14