# Difference between “Lower sum” and “Riemann lower sum”

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

or

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

or

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.