I am not getting the difference between "lower sum" and "Reimann lower sum" . Lemme explain through an example. (i know i am explaining it wrong, but i need to do it so u guys can correct me where i am wrong as i am not getting it).

Let f(x) be defined as :

f(x) = $ \left\{ \begin{array}{l l} \mid x \mid & \quad \text{if x $\neq$0}\\\\\\\ 1 & \quad \text {if x=0 } \end{array} \right.\]$

now, if we do lower Riemann sum of this function on the interval [-1,1] subdivided into 2 equal intervals,then we will have the partition as {0} such that :

-1 < 0 < 1.

so,we will do the Riemann lower sum as :

f(-1) (1) + f(1) (1) = 2

it can also be taken as :

f(0) (1) + f(0) (1) =2


f(-1) (1) + f(0) (1) =2


f(0) (1) + f(1) (1) =2 (as the infimum of f(x) in all these cases will be 1 for f(1),f(-1),f(0), so they can be written either way)

But then i came across the statement "The lower sum with respect to this partition, on the other hand, is truly zero.Therefore this particular lower sum can not be a Riemann sum." Why is it "0"???

having that said,i also saw that the supremum of f(x) between these 2 subintervals is also "1" . Thus the infimum and supremum is the same for the function. Is that possible??? I guess its not possible.

My question is whats the difference between a "lower sum" and a "Riemann lower sum" and why a continuous bounded function only has Riemann upper and lower sums??? [i know the fact that a continuous bounded function will have a max and min in all the subintervals it is divided into, but how to relate this fact with it being Riemann upper and lower sums???). I know its an easy one, if any of u guys has time to explain it plz do...thanks.


There's no difference between a "lower sum" and a "lower Riemann sum." There is a difference between a "left (Riemann) sum" and a "lower (Riemann) sum." In particular, the left Riemann sum uses the leftmost point of each interval, while the lower Riemann sum uses the point on each interval at which the function is at a minimum.

(Of course for an increasing function the two are the same.)

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  • $\begingroup$ Thanks for your feedback. one question...the text i have been reading, its written for the function which i have described above that "The lower sum with respect to the partition[-1,1], on the other hand, is truly zero. Therefore this particular lower sum can not be a Riemann sum.". If there is no difference between "lower sum" and "Riemann lower sum", what does this sentence given in the text mean???Sorry i could not make out the meaning of it, can u help me out in this regard plz...thanks.... $\endgroup$ – under root Jun 15 '13 at 11:07
  • $\begingroup$ I don't know what it's trying to say. Maybe it's a typo or maybe there's something from context that's missing. What is the text? $\endgroup$ – Daniel McLaury Jun 16 '13 at 3:52
  • $\begingroup$ i am providing the link here : mathcs.org/analysis/reals/integ/answers/ul_sum2.html $\endgroup$ – under root Jun 16 '13 at 18:15
  • $\begingroup$ Ah. Well, they define "upper sum" and "Riemann sum" here: mathcs.org/analysis/reals/integ/riemann.html So far as I know these aren't standard definitions for these terms. $\endgroup$ – Daniel McLaury Jun 16 '13 at 23:14
  • $\begingroup$ i have completed all this page and exercises in it and i guess there infact is a differnce between the two terms..but i am not sure... will explore it more and see what comes out... thanks for your feedback :)) $\endgroup$ – under root Jun 17 '13 at 0:49

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