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Suppose we have a function $f:[a,b]\to\mathbb{R}$. The Riemann integral $\int$ of the function is defined somewhat like this:

For any tagged partition $\dot{P}=(P,t_P)$, (where $P$ is the partition, and $t_P=\{t_i\}$ the corresponding tagging) of $[a,b]$, denote the Riemann sum of $f$ corresponding to $\dot{P}$ by $S(f,\dot{P})$. If $S(f,\dot{P})$ approaches a definite value $I$ as $||\dot{P}||\to0$, then $$\int f = I$$

By the definition, $\dot{P}$ runs over the set of all tagged partitions of [$a,b]$, taking into account any partitioning and any tagging for that particular partition. Is this much freedom of choice really necessary? From this question, I know that if we specify the choices for both $P$ and $t$, the integral is no longer equivalent to the Riemann integral (In the link, $P$ is always an equipartition, and $t$ is the set of the left endpoints of $P$). What if we fix exactly one of $P,t$? Is the resulting "integral" still equivalent to the Riemann integral?

For example, if I define another integral $\int_1$ similarly (but specifying choice of $P$):

For $n\in\mathbb{N}$ define $P_n$ to be the equipartition of $[a,b]$ into $n$ subintervals of equal length, and for any tagging $t_{P_n}$ of $P_n$ denote the corresponding tagged partition as $\dot{P}_{n,t}$. If $S(f,\dot{P}_{n,t})$ approaches a definite value $I_1$ as $n\to \infty$ then $$\int_1f = I_1$$

I can similarly define some other $\int_2$ by considering all partitions, but specifying $t$, for example, by taking the mid-points of each subinterval of a given partition. Will this $\int_1$ or $\int_2$ be equivalent to $\int$?

More or less, what is the advantage of simultaneously varying $P$ as well as $t$, which couldn't be achieved by doing that one at a time?

[Most probably my definitions are not extremely precise, I apologise for those]

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  • $\begingroup$ So in $I_1$ the tags can wander freely within each $P_n$, whereas in $I_2$, for each partition $P$ there is just one tagging? $\endgroup$
    – zhw.
    Jan 6, 2017 at 19:02
  • $\begingroup$ @zhw that's exactly what I'm saying $\endgroup$ Jan 6, 2017 at 19:31
  • $\begingroup$ It is easy to show that your first definition with equipartitioning and arbitrary tagging is equivalent to Riemann's definition. But second definition appears a bit tricky to handle. $\endgroup$
    – Paramanand Singh
    Jan 7, 2017 at 0:05
  • $\begingroup$ +1 for a very good question. Normally students don't bother to question definitions given in textbook whereas you dare (in positive sense) to think alternatives to the standard definition all by yourself. $\endgroup$
    – Paramanand Singh
    Jan 7, 2017 at 3:26

2 Answers 2

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I expand my comment as an answer below.


First of all it is obvious that if a function $f$ is Riemann integrable according to the standard definition (the one which involves choice of partition as well as tags) then it is also Riemann integrable according to the two alternative definitions suggested by OP.

Further we know that the theory of Riemann integration is conveniently studied using Darboux sums instead of Riemann sums and in this connection we have the following result (proof is available in almost any real-analysis textbook):

Criterion for Riemann Integrability: A function $f$ defined and bounded on $[a, b]$ is Riemann integrable on $[a, b]$ if and only if for every $\epsilon > 0$ there is a partition $P_{\epsilon}$ of $[a, b]$ such that $$U(f, P_{\epsilon}) - L(f, P_{\epsilon}) < \epsilon$$ where for any partition $$P = \{a = x_{0}, x_{1}, x_{2}, \ldots, x_{n} = b\}, x_{i - 1} < x_{i}, i = 1, 2, \ldots, n$$ the upper Darboux sum for $f$ (denoted by $U(f, P)$) and lower Darboux sum (denoted by $L(f, P)$) is given by $$U(f, P) = \sum_{i = 1}^{n}M_{i}(x_{i} - x_{i - 1}),\, L(f, P) = \sum_{i = 1}^{n}m_{i}(x_{i} - x_{i - 1})$$ with $$M_{i} = \sup\,\{f(x), x\in [x_{i - 1}, x_{i}]\},\,m_{i} = \inf\,\{f(x), x\in [x_{i - 1}, x_{i}]\}$$

Using this result it is possible to prove that if $f$ is Riemann integrable according to the first definition suggested by OP then it is Riemann integrable according to the standard accepted definition of Riemann integrability.

Let's fix an $\epsilon > 0$. Let $P_{n} = \{x_{0}, x_{1}, \ldots, x_{n}\}$ denote equipartition of $[a, b]$ so that $x_{i} = a + i(b - a)/n$. Let us assume that a function $f$ defined and bounded on $[a, b]$ is Riemann integrable according to the first definition given by OP. Then there exists a number $I$ and a positive integer $N$ such that $$|S(f, P_{n}) - I| < \frac{\epsilon}{4}$$ for all $n > N$. Here $S(f, P_{n})$ represents a Riemann sum for $f$ over equipartition $P_{n}$ with any arbitrary set of tags (the notation $S(f, P)$ does not explicitly indicate the dependence on tags and it will be augmented later to show this dependence). The important thing to note here is that $N$ depends only on $\epsilon$ and not on the choice of tags with partition $P_{n}$.

In what follows let us fix a positive integer $n > N$. Since $M_{i} = \sup\,\{f(x), x \in [x_{i - 1}, x_{i}]\}$ it follows that there is a point $t_{i} \in [x_{i - 1}, x_{i}]$ such that $$M_{i} - f(t_{i}) < \frac{\epsilon}{4(b - a)}$$ and hence $$U(f,P_{n}) = \sum_{i = 1}^{n}M_{i}(x_{i} - x_{i - 1}) < \sum_{i = 1}^{n}f(t_{i})(x_{i} - x_{i - 1}) + \frac{\epsilon}{4} = S(f, P_{n}, t) + \frac{\epsilon}{4}$$ where $S(f, P, t)$ is a Riemann sum for $f$ over partition $P$ and tags $t_{i}$.

In a similar manner we can show that there is another set of tags $t_{i}'$ for partition $P_{n}$ such that $$L(f, P_{n}) > S(f, P_{n}, t') - \frac{\epsilon}{4}$$ We now have \begin{align} U(f, P_{n}) - L(f, P_{n}) &= |U(f, P_{n}) - L(f, P_{n})|\notag\\ &= |U(f, P_{n}) - S(f, P_{n}, t)\notag\\ &\,\,\,\,\,\,\,\, + S(f, P_{n}, t) - I + I - S(f, P_{n}, t')\notag\\ &\,\,\,\,\,\,\,\, + S(f, P_{n}, t') - L(f, P_{n})|\notag\\ &\leq |U(f, P_{n}) - S(f, P_{n}, t)|\notag\\ &\,\,\,\,\,\,\,\,+|S(f, P_{n}, t) - I|\notag\\ &\,\,\,\,\,\,\,\,+|I - S(f, P_{n}, t')|\notag\\ &\,\,\,\,\,\,\,\,+|S(f, P_{n}, t') - L(f, P_{n})|\notag\\ &< \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon}{4}\notag\\ &= \epsilon\notag \end{align} Therefore by criterion for Riemann integrability $f$ is Riemann integrable on $[a, b]$ according to the standard definition of Riemann integrability.


On the other hand the second definition suggested by OP (which fixes the tags for a given partition) does not seem amenable to an analysis like the one given in preceding paragraphs. I have not been able to figure out a counter-example to show that this definition is not equivalent to the standard definition. However it does appear that the choice of tags is essential to link the Riemann sums with upper and lower Darboux sums and this is probably an essential ingredient in the definition of Riemann integral.


Update: In the proof presented above I nowhere use the fact that $P_{n} $ is a partition of specific form where all sub-intervals are of equal length. In particular we don't assume anything about norm of $P_{n} $. All I assume is that there is a sequence of such partitions for which the Riemann sums converge to $I$. What is assumed is that the number $I$ does not depend on choice of tags, but the speed of convergence may depend on choice of tags. This will make $N$ in the proof to be dependent on the choice of tags. But this is not a problem as we are dealing with only two choices of tags so that we can have $N_{1},N_{2}$ in place of $N$ and choose $N=\max(N_{1},N_{2})$.

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  • $\begingroup$ I have a question. Let $f$ be any function. For $n\in\mathbb{N}$ define $P_n$ to be the equipartition of $[a,b]$ into $n$ subintervals of equal length, and for any tagging $t_{P_n}$ of $P_n$ denote the corresponding tagged partition as $\dot{P}_{n,t}$. If the limit of $S(f,\dot{P}_{n,t})$ exists as $n\to \infty$, is it possible to show that this limit must be always the same, and in such a case $f$ is integrable and the limit is $\int f$ ? See my question math.stackexchange.com/questions/3182682/… $\endgroup$
    – user312396
    Apr 10, 2019 at 18:11
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You can't define a tagging without saying what partition is being tagged, so the notion of choosing the tagging first is not viable.

Now the wuestion is whether one could define an integral based on just the partitioning, without the tagging, and have it come out equivalent to the Reimann integral. If you consider fucntions which have different behavior on the rationals than on the irrationals, you will see that this will not work.

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  • $\begingroup$ I am sorry, there were some typos on my part. I understand what you want to say, can you re-read the question now? I edited it. $\endgroup$ Jan 6, 2017 at 18:24

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