First definition of Riemann integrable function. Let $f:[a,b] \to \mathbb{R}$ be a bounded function and $P=\{x_0,x_1,\dots, x_n\}$ partition of $[a,b]$. Define $U(P,f):=\sum \limits_{i=1}^{n}M_i\Delta x_i$ and $L(P,f):=\sum \limits_{i=1}^{n}m_i\Delta x_i$ where $M_i=\sup\limits_{[x_{i-1},x_i]} f(x),\quad m_i=\inf\limits_{[x_{i-1},x_i]} f(x), \Delta x_i=x_i-x_{i-1}.$ Let $\inf \limits_{P}U(P,f)=I^*$ and $\sup \limits_{P}L(P,f)=I_*$. If $I^*=I_*=I$ then we called $f(x)$ is Riemann integrable function on $[a,b]$ with integral $I$.

Second definition of Riemann integrable function. Let $P=\{x_0,x_1,\dots, x_n\}$ is a partition of $[a,b]$ with $\xi_i\in [x_{i-1},x_i]$ and we define Riemann-integral sum $\sigma(P):=\sum \limits_{i=1}^{n}f(\xi_i)\Delta x_i$ and $\lVert P\rVert=\max\limits_{i}\Delta x_i$. If the following limit $\lim \limits_{\lVert P\rVert\to 0}\sigma(P)$ exists and has value $L$ we say that $f(x)$ is Riemann integrable function on $[a,b]$ with integral $L$.

The first definition is from Rudin's PMA book but in other books I met the second definition. But I can't prove the equivalence of these definitions for a couple days. Can anyone show to me a strict and rigorous proof? I would be very grateful for your help!

P.S. Happy New Year! :)


2 Answers 2


That the first condition implies the first is immediate, since (using your notation) you always have $m_i \le f(\xi_i) \le M_i$, so the sums in the second definition are caught between the $L$ und $U$ sums.

Edit in response to a comment an additional explanation is necessary here. For this direction it suffices to show that $I^* = lim_{||P||\rightarrow 0} L(f,P)$ and $I_* = lim_{||P||\rightarrow 0} U(f,P)$ Since both parts are similar it suffices to show, e.g., the first equality.

First it is easy to see that for partitions $P\subset P^\prime$ we have $L(f,P)\le L(f,P^\prime)$. A remaining hurdle is that for two partitions we do not necessarily know that one is a subset of the other one. This is resolved by looking at common refinements:

Assume $P$ satisfies $L(f,P) > I^* - \varepsilon$ and $Q$ is an arbitrary partition. We need to show that then there is a refinement $Q^\prime$ of $Q$ such that $L(f,Q^\prime)\ge L(f,P)$ (and, consequently, $L(f,Q^\prime)>I^*-\varepsilon$).

For $Q^\prime$ one can choose the common refinement $R$: if $P=\{x_1,\ldots x_n \}$ and $Q=\{y_1,\ldots y_m \}$ then we just let $R = P\cup Q$. Since this is a refinement of both $P$ and $Q$ we have both $L(f,R)\ge L(f,P)$ as well as $L(f,R)\ge L(f,Q)$

Second edit: the original version was not correct:

For the other direction it suffices to show that if the function is integrable in the sense of the second definition then both $I_*$ and $I^*$ agree with the of the sums from the second definition. Since the reasoning is the same in both cases I'll just look at $I_*$.

So fix $\varepsilon >0$ and a given partition $P$ such that $$|L - \sum_{i=1}^n f(\xi_i)\Delta x_i |< \varepsilon$$ if only the partition is fine enough.

Choose such a partition $P=\{x_0,\dots x_n\}$ and to $[x_{i-1},x_{i}]$ choose $\eta_i\in[x_{i-i},x_{i}]$ such that for $m_i:=\inf \{ f(x):x\in [x_{i-1},x_i]\} $ we have $$0\le f(\eta_i)-m_i\le \frac{\varepsilon}{2n}$$

Then \begin{eqnarray} | L -\sum_{i=1}^n m_i \Delta x_i| & = & |L- \sum_{i=1}^n f(\eta_i)\Delta x_i + \sum_{i=1}^n f(\eta_i)\Delta x_i -\sum_{i=1}^n m_i\Delta x_i| \\ &\le & |L- \sum_{i=1}^n f(\eta_i)\Delta x_i| + \sum_{i=1}^n | f(\eta_i) - m_i|\Delta x_i \\ & < & \frac{\varepsilon}{2} + \sum_{i=1}^n \frac{\varepsilon}{2n}=\varepsilon \end{eqnarray}

If you 'see' that $0 <L -I_*< L -\sum_{i}m_i \Delta x_i$ then you are done here, otherwise it follows easily from the last estimate that the $\sum_i m_i \Delta x_i$ are, for any partition which is fine enough, $\varepsilon $ close to the fixed real number $L$, which of course implies that the $\sup$ over these sums exists and equals $L$ (here you need to use again the fact that you will approach the $\sup$, if it exists, if the width of the partitions goes to $0$).

  • 1
    $\begingroup$ "That the first condition implies the first is immediate" Are you sure that here all right? $\endgroup$
    – RFZ
    Dec 31, 2015 at 9:47
  • $\begingroup$ I have question to your first direction: $(1)\Rightarrow (2)$ I noted that $m_i\leqslant f(\xi_i)\leqslant M_i$ then $L(P,f)\leqslant \sigma(P)\leqslant U(P,f)$. We know that $\inf \limits_{P}U(P,f)=\sup \limits_{P}L(P,f)=I$. How to conclude from above that $\lim \limits_{\lVert P\rVert\to 0}\sigma(P)=I$? Sorry but it's not obvious to me. The limit operation confuses me :( $\endgroup$
    – RFZ
    Dec 31, 2015 at 10:55
  • 1
    $\begingroup$ We know that $1)\sup \limits_{P}L(P,f)=I$ and $2)$ if $P_1\subset P_2$ then $L(P_1,f)\leqslant L(P_2,f)$. Right? I want to prove that $\lim \limits_{\lVert P\rVert\to 0}L(P,f)=I$. Let $\epsilon>0$ be given then by definition of supremum exists some partition $P_1$ such that $I-\epsilon< L(P_1,f)<I+\epsilon$. Let's $\delta:=\lVert P_1\rVert\ $ and for any partition $P$ such that $P_1\subset P$ we have $\lVert P\rVert<\delta$ and $I-\epsilon< L(P,f)<I+\epsilon$. But how to prove the same for partitions $Q$ such that $\lVert Q\rVert<\delta$ and $Q\cap P_1 \neq P_1$? $\endgroup$
    – RFZ
    Dec 31, 2015 at 11:05
  • 2
    $\begingroup$ @RFZ I see your point and will check. The reason it works is that, for arbitrary partions $P,Q$, for their so called common refinement (the partition $R$ created from the points in $P$, $Q$) you can apply the inequlity you noted for the case $P_1\subset P_2$. But you are right, some argument is still missing here. Will think about it and update. $\endgroup$
    – Thomas
    Dec 31, 2015 at 11:13
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    $\begingroup$ Now, if so, I'm not so sure how your method works. As you've done, let's only discuss lower sums. You're taking some partition $P$ with a lower sum within $\epsilon$ of the integral and showing that all partitions which are refinements of it are within $\epsilon$ the integral. However, where exactly does the mesh $||P||$ come into play? What you need to prove is the implication $||P|| < \delta \Longrightarrow I - L(f,P) < \epsilon$. $\endgroup$ Jun 15, 2016 at 23:56

This is explained very nicely in Apostol's Mathematical Analysis. In this book Apostol uses Riemann sums to define the definite integral, but he does not use the limit based on $||P|| \to 0$. Rather he uses the concept of finer partition.

Thus if $P, P'$ are partitions of $[a, b]$ then $P'$ is said to be finer than $P$ if $P \subseteq P'$. Adding more points to an existing partition makes it finer. The other concept is called norm of a partition which is defined as the length of largest sub-interval made by the partition. When you add points to a partition to make it finer, then the norm can only decrease (or remain unchanged). Thus finer partitions corresponds to partitions with smaller norms, but the converse does not hold.

It appears from Apostol's presentation that dealing with a limit where norm of partition tends to $0$ is difficult compared to dealing with limit when partitions become finer and finer. And he uses the following definition of Riemann integral:

Let $f$ be bounded on $[a, b]$ and let $P = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n}\}$ be a partition of $[a, b]$. A sum of the form $$S(P, f) = \sum_{i = 1}^{n}f(\xi_{i})(x_{i} - x_{i - 1})$$ where $\xi_{i}$ is any point in the interval $[x_{i - 1}, x_{i}]$ is called a Riemann sum of $f$ over partition $P$. A number $I$ said to be the Riemann integral of $f$ over $[a, b]$ and we write $$\int_{a}^{b}f(x)\,dx = I$$ if for every $\epsilon > 0$ there is a partition $P_{\epsilon}$ of $[a, b]$ such that $$|S(P, f) - I| < \epsilon$$ for partitions $P$ of $[a, b]$ which are finer than $P_{\epsilon}$.

It is now easy to show the equivalence of this definition of Riemann integral with the definition based on Darboux sums (the first definition of your question). This is because the Riemann sums are always sandwiched between the upper and lower Darboux sums and as partitions get finer and finer the lower sums increase and upper sums decrease. If these Darboux sums have same limit (i.e. supremum of lower sums is equal to infimum of upper sums) then obviously the Riemann sums also tend to same limit as partitions become finer and finer. On the other hand if $M_{i}, m_{i}$ are the supremum and infimum of $f$ on a sub-interval $[x_{i - 1}, x_{i}]$ generated by partition then it is easy to choose points $\xi_{i}, \xi'_{i}$ in this sub-interval such that $f(\xi_{i})$ is near $M_{i}$ and $f(\xi'_{i})$ is near $m_{i}$. Due to this we can a find a Riemann sum close to $U(P, f)$ and another Riemann sum close to $L(P, f)$. So if Riemann sums tends to a limit when partitions get finer and finer then the upper and lower Darboux sums also tend to the same limit.

Next Apostol shows the equivalence of his definition (mentioned above) with the definition based on limit of type $||P|| \to 0$ in his exercise. This proof is also available as an answer here.

Also note that in general (i.e. in the context of Riemann-Stieltjes integral) these two definitions are not equivalent and the one which uses the concept of finer partitions is more inclusive than the one based on norm. See this answer for more details.


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