Theorem. If $f:[a,b]\to\mathbb{R}$ be a Riemann integrable function on $[a,b]$ then show that for all $\varepsilon>0$ there exists $\delta(\varepsilon)>0$ such that for all partition $P_\varepsilon$ with $\lVert P_\varepsilon\rVert<\delta(\varepsilon)$ we will have $U(f;P_\varepsilon)-L(f;P_\varepsilon)<\varepsilon$.

Proof. The proof of the part that for all $\varepsilon>0$ there exists partition $P_\varepsilon$ such that $U(f;P_\varepsilon)-L(f;P_\varepsilon)<\varepsilon$ is easy and will not be discussed here. I focus mainly on the existence of $\delta(\varepsilon)$.

Fix $\varepsilon>0$ and observe that for all $\varepsilon>0$ there exists $P_\varepsilon$ and $Q_\varepsilon$ such that, $$U(f)\le U(f;P_\varepsilon)<U(f)+\varepsilon$$$$L(f)\ge L(f;Q_\varepsilon)>L(f)-\varepsilon$$where $U(f)$ and $L(f)$ are respectively upper and lower Darboux intergrals of $f$ on $[a,b]$.

Now consider the sets, $$\mathcal{U}\langle\varepsilon\rangle:=\{\lVert P_\varepsilon\rVert:U(f)\le U(f;P_\varepsilon)<U(f)+\varepsilon\}$$ $$\mathcal{L}\langle\varepsilon\rangle:=\{\lVert Q_\varepsilon\rVert:L(f)\ge L(f;Q_\varepsilon)>L(f)-\varepsilon\}$$Then observe that both $\sup\mathcal{U}\langle\varepsilon\rangle$ and $\sup\mathcal{L}\langle\varepsilon\rangle$ exists (because both sets are bounded above by $(b-a)$) and they are positive since $P_\varepsilon, Q_\varepsilon$'s are partitions. We choose, $$0<\delta(\varepsilon)<\min\left(\sup\mathcal{U}\langle\varepsilon\rangle,\sup\mathcal{L}\langle\varepsilon\rangle\right)$$Now observe that for all $P_\varepsilon$ such that $\lVert P_\varepsilon\rVert<\delta(\varepsilon)$, $\lVert P_\varepsilon\rVert\in \mathcal{U}\langle\varepsilon\rangle\cap\mathcal{L}\langle\varepsilon\rangle$ (such $P_\varepsilon$'s exist because if $\lVert P_\varepsilon\rVert\in \mathcal{U}\langle\varepsilon\rangle$ and $\lVert Q_\varepsilon\rVert\in \mathcal{L}\langle\varepsilon\rangle$ then $\lVert P_\varepsilon\cup Q_\varepsilon\rVert\in \mathcal{U}\langle\varepsilon\rangle\cap\mathcal{L}\langle\varepsilon\rangle$).

This completes the proof.


  1. Is my proof of the theorem correct?

  2. Is there any part of the written proof that isn't explained clearly?

  • $\begingroup$ How do you define $||P_\epsilon||$? $\endgroup$
    – Elliot
    Aug 11, 2016 at 16:26
  • $\begingroup$ @Elliot: $\lVert P_\varepsilon\rVert:=\max\{|x_k-x_{k-1}|:k\in\{1,2,\ldots,n\}\}$ where $P_\varepsilon:=\{a=x_0,x_1,\ldots,x_n=b\}$. $\endgroup$
    – user170039
    Aug 11, 2016 at 16:30
  • $\begingroup$ I'm assuming $U(f):=\sum_{j=1}^nf(x_j^*)(x_j-x_{j-1})$, where $x_j^*=\sup\{x\in[x_{j-1},x_j]|f(x)\geq f(y)\forall y\in[x_{j-1},x_j]\}$, so then how do you define $U(f;P_\epsilon)$? $\endgroup$
    – Elliot
    Aug 11, 2016 at 16:34
  • $\begingroup$ @Elliot: Please see this pdf. My notations have similar meaning. $\endgroup$
    – user170039
    Aug 11, 2016 at 17:05
  • $\begingroup$ ...or, you may see this page also. $\endgroup$
    – user170039
    Aug 11, 2016 at 17:08

1 Answer 1


The proof is incorrect. There is circular reasoning to the extent that partitions in the class $\mathcal{U}\langle\varepsilon\rangle\cap\mathcal{L}\langle\varepsilon\rangle$ already satisfy the desired inequality. You need to show that you can produce $\delta(\epsilon)$ such that the inequality is satisfied for any partition with $\lVert P \rVert < \delta(\epsilon).$

Your argument implicitly uses the assumption that, for any partition $P,$ the condition $\lVert P \rVert < \lVert P_\epsilon \rVert$ implies that $U(P,f) - L(P,f) < U(P_\epsilon,f) - L(P_\epsilon,f).$ This would be true if $P$ were a refinement of $P_\epsilon$, but not simply because the norm is smaller. Recall that the norm of a partition is the length of the longest sub-interval.

For a counterexample to the assumption, consider the Riemann integrable function $f:[0,1] \to \mathbb{R}$ where

$$f(x) = \begin{cases} 0, \,\,\, 0 \leqslant x < 1/2 \\1 , \,\,\, 1/2 \leqslant x \leqslant 1 \end{cases}$$

Consider the partitions, $P_\epsilon = (0,1/2- \delta, 1/2+\delta,1)$ and $P = (0,1/2- \delta', 1/2+\delta',1).$

Choose $\delta < 1/6$ and $\delta' = 1/6.$


$$\lVert P_\epsilon \rVert = \max(2\delta, 1/2 - \delta) > 1/3, \\ \lVert P \rVert = \max(2\delta', 1/2 - \delta') = 1/3$$


$\lVert P \rVert < \lVert P_\epsilon \rVert ,$ but

$$U(P_\epsilon,f) - L(P_\epsilon,f) = 2\delta < 1/3, \\ U(P,f) - L(P,f) = 2\delta' = 1/3,$$

and $U(P,f) - L(P,f) > U(P_\epsilon,f) - L(P_\epsilon,f).$

Valid Proof

Suppose $f$ is Riemann integrable with integral equal to $I$. Given any $\epsilon >0$ there exists a partition $P_\epsilon$ with $U(P_\epsilon,f) < I + \epsilon/4$ and $L(P_\epsilon,f) > I - \epsilon/4.$

Let $D = \sup\{|f(x)-f(y)|:x,y \in [a,b] \}$ denote the maximum oscillation of $f$ and let $\delta = \epsilon/(4mD)$ where $m$ is the number of points in the partition $P_\epsilon$.

Now let $P$ be any partition with $\lVert P \rVert<\delta$. Form the common refinement $Q = P \cup P_\epsilon$.

Since $Q$ is a refinement of $P_\epsilon$ we have

$$L(P_\epsilon,f) < L(Q,f) \leqslant U(Q,f) < U(P_\epsilon,f).$$

We can see that terms of the upper sums $U(P,f)$ and $U(Q,f)$ differ in at most $m$ sub-intervals where the deviation is bounded by $\delta D$ and

$$U(P,f) < U(Q,f) + m D \delta = U(Q,f) + \epsilon/4.$$

It follows that

$$U(P,f) < U(Q,f) + \epsilon/4 < U(P_\epsilon,f) + \epsilon/4 < I + \epsilon/2.$$

By a similar argument, we can show $L(P,f)> I - \epsilon/2$.

Hence, $U(P,f)-L(P,f) < \epsilon$.


You must log in to answer this question.