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Let ${\displaystyle f:[a,b]\rightarrow \mathbb {R} }$ be a function defined on a closed interval ${\displaystyle [a,b]}$ of the real numbers, $\mathbb {R}$ , and ${\displaystyle P=\left\{[x_{0},x_{1}],[x_{1},x_{2}],\dots ,[x_{n-1},x_{n}]\right\}},$ be a partition, where ${\displaystyle a=x_{0}<x_{1}<x_{2}<\cdots <x_{n}=b}$. A Riemann sum $S$ of $f$ with partition P is defined as ${\displaystyle S=\sum _{i=1}^{n}f(x_{i}^{*})\,\Delta x_{i}}$ where ${\displaystyle \Delta x_{i}=x_{i}-x_{i-1}}$ and ${\displaystyle x_{i}^{*}\in [x_{i-1},x_{i}]}$.

  • If ${\displaystyle x_{i}^{*}=x_{i-1}}$ for all $i$, then $S$ is called a left Riemann sum.
  • If $x_{i}^{*}=x_{i}$ for all $i$, then $S$ is called a right Riemann sum.

I was wondering if 'left Riemann sums converge to $a\in\mathbb{R}$' is equivalent to 'right Riemann sums converge to $a$. Is it possible that one limit exists and the other does not, or they converge to different limits?

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    You did not specify a method of "convergence". Perhaps you mean as the mesh $|P| := \max{x_j-x_{j-1} : 1 \le j \le n}$ goes to zero. But do you mean for all sequences of partions with mesh going to zero? Or merely for one such sequence? @CSquared ... how about an example where limits of left and right sums both exist, but are different? – GEdgar Jun 10 '21 at 12:00
  • Yes, I mean as the mesh goes to zero for left and right sums, respectively. –  Jun 10 '21 at 12:09
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    @GEdgar thank you for challenging me. it seems that the difference between the two sums is $\frac{b-a}{n}(f(b)-f(a))$ which goes to $0$ as $n$ tends to infinity (using the usual partition of $[a,b]$). so if one exists, then the other must also exist and converge to the same limit. – C Squared Jun 10 '21 at 14:59

2 Answers2

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A bounded function $f:[a,b] \to \mathbb{R}$ is said to be Cauchy integrable (using left endpoints) with integral $I$ if for every $\epsilon > 0$ there is a $\delta > 0$ such that for every partition $P$ with points $a = x_0 < x_1 < \ldots < x_n = b$ and $\|P\| < \delta$, then

$$\left|\sum_{j=1}^n f(x_{j-1})(x_j - x_{j-1})\right| < \epsilon$$

Gillespie proved that for bounded functions the Cauchy and the Riemann definitions of integrability are equivalent. Hence, convergence of left Riemann sums implies convergence of right Riemann sums to the same limit. By considering $f(-x)$ this can also be framed starting with the convergence of right Riemann sums.

This is not a trivial result and the proof by Gillespie is somewhat difficult to follow. An elementary proof was given by Schneider, but is also not easy reading.

RRL
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$\newcommand{\Riem}[1]{\mathtt{#1}}\newcommand{\Left}{\Riem{Left}}\newcommand{\Right}{\Riem{Right}}$Too long for a comment, but possibly of interest. In general, letting $\Left(f, P)$ denote the left Riemann sum of $f$ with respect to a partition $P$ and $\Right(f, P)$ the right, we have $$ \Right(f, P) - \Left(f, P) = \sum_{i=1}^{n} [f(x_{i}) - f(x_{i-1})]\, \Delta x_{i}. $$ To construct a (non-integrable) function and a sequence of partitions whose mesh decreases to $0$ but the left and right Riemann sums approach different limits, we can define $f$ on $[0, 1]$ by $f(x) = 0$ for $x$ rational and $f(x) = 1$ for $x$ irrational, and choose a sequence of partitions whose points are alternately rational and irrational, but for which the "rational-to-irrational intervals" asymptotically have total length $1$ and whose "irrational-to-rational intervals" therefore asymptotically have total length $0$, so that $\Right(f, P) - \Left(f, P)$ does not converge to $0$.

For instance, if $n = 2k$ is an even positive integer we could pick our partition $P_{n} = (x_{i})_{i=0}^{n}$ by taking $x_{2j} = 2j/n$ for $0 \leq j \leq k$ and $x_{2j-1} = x_{2j} - 1/(\sqrt{2}\, n!)$ for $1 \leq j \leq k$. For each $j$ with $1 \leq j \leq k$, we have \begin{align*} [f(x_{2j-1}) - f(x_{2j-2})]\, \Delta x_{2j-1} = \Delta x_{2j-1} &= \frac{2}{n} - \frac{1}{\sqrt{2}\,n!}, \\ [f(x_{2j}) - f(x_{2j-1})]\, \Delta x_{2j} = -\Delta x_{2j} &= -\frac{1}{\sqrt{2}\,n!}. \end{align*} Consequently, $\Right(f, P_{n}) - \Left(f, P_{n}) = 1 - \frac{k\sqrt{2}}{n!} \to 1$ as $n \to \infty$.

Here, the left Riemann sums and right Riemann sums converge, but not to the same limit. Modifying this idea, it's not difficult to construct for this function and interval a sequence of partitions whose mesh decreases to $0$ but the left Riemann sums do not converge and the right Riemann sums do not converge.

On the other hand, in elementary calculus, one often uses equal-length partitions, for which $\Delta x_{i} = (b - a)/n$ is the same for each $i$. For such a partition, whether or not $f$ is integrable on $[a, b]$, the difference is a telescoping sum: $$ \Right(f, P) - \Left(f, P) = \frac{b - a}{n} \sum_{i=1}^{n} [f(x_{i}) - f(x_{i-1})] = \frac{b - a}{n} [f(b) - f(a)], $$ and this converges to $0$ as $n \to \infty$, i.e., the left Riemann sum (for these partitions) converges if and only if the right Riemann sum converges, and if either exists the limits are equal. (Again, existence of such a limit does not guarantee $f$ is integrable!)

Andrew D. Hwang
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  • So to definie a definite integral, we need to consider arbitrary Riemann sums where we take any points from subintervals? –  Jun 11 '21 at 06:32
  • Yes, in that looking at one particular Riemann sum over one sequence of partitions is not enough to establish integrability. – Andrew D. Hwang Jun 11 '21 at 13:01
  • But the left and right Riemann sums don't really converge since they don't converge over all partitions. Apparently, this can't happen https://math.stackexchange.com/questions/326197/the-equivalence-between-cauchy-integral-and-riemann-integral-for-bounded-functio/352476#352476 – JLA Feb 18 '24 at 17:25