Let $f: [a,b] \rightarrow \mathbb{R}$ be continuous. Suppose $(P_n)$ is a sequence of partitions with mesh$(P_n) \rightarrow 0$ as $n \rightarrow \infty$. Prove that $\int_{a}^{b} f(x) dx = \lim_{n \to \infty} U(f, P_n) = \lim_{n \to \infty} L(f, P_n)$.


  • mesh$(P) = \max\{|x_i - x_{i-1}| : i = 1,2,3,\ldots, n\}$ where $P$ is any partition of $[a,b]$.

  • $U(f , P) = \sum_{i=1}^{n} M_i(x_i - x_{i-1})$ where $M_i = \sup\{f(x) : x \in [x_{i-1}, x_i]\}$

  • $L(f , P) = \sum_{i=1}^{n} m_i(x_i - x_{i-1})$ where $m_i = \inf\{f(x) : x \in [x_{i-1}, x_i]\}$

  • $U(f) = \inf\{U(f , P) :$ P is a partition of $[a,b]\}$

  • $L(f) = \sup\{L(f , P) :$ P is a partition of $[a,b]\}$

  • $f$ is integrable on $[a,b]$ iff $U(f) = L(f)$.

I am pretty lost on how to prove this. I think we can prove this statement if we can first show that $\lim_{n \to \infty} [U(f, P_n) - L(f, P_n)] = 0$. From there, if we can show $\lim_{n \to \infty} U(f, P_n) = U(f)$ and $\lim_{n \to \infty} L(f, P_n) = L(f)$, then since $f$ is continuous on $[a,b]$ it must also be integrable on $[a,b]$ and hence, $U(f) = L(f)$. I am also confused on where to use the mesh of $P_n$. Again, since $f$ is continuous on $[a,b]$, we know that $f$ is uniformly continuous on $[a,b]$. So do we somehow use the mesh of $P_n$ to find an $N \in \mathbb{N}$ such that $x_i - x_{i-1} < \delta$? Any help is appreciated!

  • $\begingroup$ $f$ is Riemann integrable on $[a,b]$ if $U(f) = L(f)$. $\endgroup$
    – ATLKing
    Jan 26, 2017 at 4:15
  • $\begingroup$ Use the fact that $f$ is uniformly continuous to show that for any $\epsilon>0$ there is some $P$ such that $U(f,P)-L(f,P) < \epsilon$. $\endgroup$
    – copper.hat
    Jan 26, 2017 at 4:41

1 Answer 1


This is true for any Riemann integrable function:

Let $K=\max f$ on $[a,b]$, and choose a partiton $P$ such that $U(f,P) - L(f,P) <\epsilon.\ $ Suppose $P$ has $N$ points. Now let $P_n$ be one of the partitions in the sequence with mesh $\delta<\epsilon /NK.$ Consider $Q=P\cup P_n$ which is a refinement of $P_n$. Certainly

$\tag1U(f,Q) - L(f,Q) <\epsilon. $

Furthermore, since $Q$ is obtained by adding $N$ points to $P_n,$ we also have

$\tag2 U(f,P_n)-U(f,Q)<2NK\delta <2\epsilon.$

The same reasoning shows that

$\tag3 L(f,Q)-L(f,P_n)<2\epsilon.$

Thus, using $1), 2)$ and $3), $

$U(f,P_n)-L(f,P_n)=U(f,P_n)-U(f,Q) + U(f,Q) - L(f,Q) + L(f,Q) - L(f,P_n)<5\epsilon, $

and this completes the proof.

  • $\begingroup$ Thank you for a nice proof! Some questions. $Q$ is also a refinement of $P$, is it not? Also, could you explain more in detail why inequality $(2)$ (and $(3)$) holds? What is $2NK\delta$? I am moreover confused by your construction of $\delta$. Should it not depend on $n$, since we are demanding that $\delta\to0$ as $n\to\infty$? $\endgroup$
    – sunny
    Apr 25 at 12:11
  • $\begingroup$ This is a $\delta-\epsilon$ proof so by definition $\delta$ should depend only on $\epsilon.$ But $P$ is a fixed partition so $N$ is constant. As $K$ is also constant, $\delta$ depends only on $\epsilon.$ Hint for inequality $(2):$ do it by adding a single point to $P_n$, then for a known value of $N, $ say $N=3$ first. Drawing a picture will help. $\endgroup$ Apr 25 at 23:08

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