# 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.

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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
You are allowed to answer your own questions. You'll learn more than me if you do :). –  Alex R. Jan 4 '13 at 20:04
It's not clear that by "letting $n\to\infty$ we have the result." Because the limit as $n\to\infty$ of the left-most side is not $\underline{\int}$, but rather, possibly some value $\leq \underline{\int}$. You need to instead deal with refinements of the sequences $\{0,1/n,2/n,..,1\}$ to get the result –  Thomas Andrews Jan 4 '13 at 20:11
But the limit of the left side is not for all partitions, just the partions of this form. If we had all partitions, I'd agree, but we don't. So the squeeze theorem doesn't work. @Nameless –  Thomas Andrews Jan 4 '13 at 20:22