# Why do we use supremum instead of maximum in Riemann Integration?

Context: I've been learning about how we define the Riemann definite integral, and how we use supremum and infimum to define both the rectangle heights ($M_k,m_k$) and finding the value at which the greatest value of the lower sum ($L(f,P)$) is equal to the lowest value of the upper sum ($U(f,P)$), and that's the Riemann integral.

Question: Why do we use supremum/infimum instead of maximum/minimum? i.e. instead of $$m_k = \inf \{f(x): x_{k-1} \leq x \leq x_k \}$$ Use, $$m_k = \min \{f(x): x_{k-1} \leq x \leq x_k \}$$ Similarly for $M_i$, and instead of $$L_a^b := \sup \{L(f,P): \text{P is a partition of} [a,b] \}$$ Use maximum etc.

This website provides some material on the definition of the Riemann integral;http://math.feld.cvut.cz/mt/txtd/1/txe3da1a.htm

• How do you know the minimum exists?
– user856
Feb 25, 2018 at 4:50
• The maximum or minimum of a set must be an element of the set. There's no guarantee than an infinite set of reals has a maximum or minimum. It will always have an infimum and supremum, however. Feb 25, 2018 at 4:51

The maximum might not exist.

For instance, if $f$ is a continuous function, then we know by the extreme value theorem that on any part of the partition $[x_i, x_{i+1}]$, both the imfimum and supremum are achieved, so it could be a min/max, but if $f$ is not continuous then the minimum and maximum might not exist and the definition of a Riemann integrable function wants to include these cases.

Example: Consider $f(x) = x$ on $(0,1)$ and $f(x) = 2$ for $x = 0, 1$. Then on $[1, \epsilon] = [x_0, x_1]$, $f$ has an infimum but no minimum (it does have a maximum!) but on $[x_{n-1}, x_n] = [1-\epsilon, 1]$, $f$ has a supremum but no maximum (but it does have a minimum!).

Likewise, in defining the integral as the supremum of the areas of the step functions below the function $f$, if $f$ is a continuous function that is not constant, then any approximation by steps below $f$ will always have area strictly less than the area of $f$ and so we need to define the area as a supremum in order for the definition to be what we expect it to be.

• In your first paragraph when you said the infimum and supremum might not exist, how does using them allow the Riemann integrable function to include these cases? Feb 25, 2018 at 6:27
• I apologize... for a Riemann integrable function $f$, $f$ is always assumed to be bounded so the infimum/supremum always exists... but might not be attain and so the maximum might not exist. I have fixed what I wrote. Feb 25, 2018 at 6:46
• But on $[1-\epsilon,1]$ f has maximum ! it is $2$ which is attained at $x=1$ we could define function is following way $f(x) = x$ $x \in (0,1)$ and $f(0)=2$ and $f(1) = 1/2$ then in the case left most interval doesn't have minimum and rightmost interval doesn't have maximum Jan 3, 2020 at 11:40

The maximum of a set need not necessarily exist. In $\mathbb{R}$, for exemple, consider $(a,b)$. This interval has no maximum and minimum. Any lower sums or upper sums (of a partition $P$) form sets of real numbers.