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
Best Answer
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.