2
$\begingroup$

In my textbook, we define the the existence of an integral to be satisfied if and only if the supremum of the lower sum (minimum sum) is equal to the infimum of the upper sum (maximum sum). Moreover, the unique number $L \leq \int f \leq U$ is defined to be the integral of $f$. Clearly a Riemann sum is such that $L \leq R \leq U$ --- so why do we say that $R$ is only approximately equal to $\int f$ ? Isn't it equal?

Edit: Let me clarify that I understand why Riemann sums should only be considered approximately equal to the integral. My question rather is semantic. To me it seems justified using the definitions of the text (given above) to prove what it is not --- equal.

Edit for Context:

The author seems to be saying that 'Riemann sum' of the $g'(x_i)$ yielded by the MVT is equal to the integral

$\endgroup$
  • 3
    $\begingroup$ A given $R$ does not in general satisfy $L \leq R \leq U$. It satisfies $L_P \leq R \leq U_P$, where $L_P$ and $U_P$ are the lower and upper sums for the partition used for that Riemann sum. Then $L$ is the supremum of $L_P$ and $U$ is the infimum of $U_P$ over all partitions $P$. $\endgroup$ – Bungo Jul 5 at 18:25
  • $\begingroup$ The Riemann sum converges to the integral as the number of partitions tends to $\infty$ and the size of the partition element tends to zero. $\endgroup$ – Akash Gaur Jul 5 at 18:26
  • $\begingroup$ @ Bungo. Could you respond to my screenshot edit? $\endgroup$ – user_hello1 Jul 5 at 18:36
  • 1
    $\begingroup$ Yes, what the author (I assume Spivak?) is doing is carefully constructing a Riemann sum that actually equals the integral. Good observation! But in general, arbitrary Riemann sums only approximate the integral. $\endgroup$ – Bungo Jul 5 at 18:40
  • $\begingroup$ This construction by the mean value theorem is actually something that Riemann did in his original paper to motivate his integral definition. See maths.tcd.ie/pub/HistMath/People/Riemann/Trig $\endgroup$ – LutzL Jul 5 at 19:42
1
$\begingroup$

As far as I can see, the problem is that you've misquoted the definition of the integral. You wrote "the unique number $L\leq\int f\leq U$" but it should be "the unique number such that for all lower sums $L$ and all upper sums $U$ (i.e., obtained from all partitions), $L\leq\int f\leq U$". Your $R$, being a Riemann sum for some particular partition, will satisfy $L\leq\int f\leq U$ for the lower and upper sums obtained from that partition, but usually not for all lower and upper sums.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.