In chapter $13$ in Spivak Calculus in the beginning of Integrals chapter he introduced a new definition which is as follow :

Suppose $f$ is bounded on $[a,b]$ and $P$ is a partition of $[a,b]$ . Let

$m_{i}=inf\{ f(x):t_{i-1}\le x \le t_{i}\}$

$M_{i}=sup\{ f(x):t_{i-1}\le x \le t_{i}\}$

then the Lower sum denoted by $L(f,P) =\sum_{i=1}^{n} m_{i}(t_{i}-t_{i-1})$

and the Upper sum denoted by $U(f,P)=\sum_{i=1}^{n} M_{i}(t_{i}-t_{i-1})$

My question is as follows :Is the using of $inf$ and $sup$ insted of $max$ and $min$ indicates that the function may not be well defined at the end points of the interval which is $a$ and $b$ $?$

because I read that the function to be riemann integrable must be defined on the whole interval $[a,b]$ so does the definition in Spivak mean something else or not


1 Answer 1


The definition uses $\inf$ and $\sup$ because the sets $\{f(x)\mid t_{i-1}\leqslant x\leqslant t_i\}$ is bounded (since $f$ is bounded) and non-empty, and therefore they have a supremum and and an infimum. But those sets don't have to have a maximum or a minimum. For instance, if $a=0$, $b=1$, $P=\{0,1\}$, and$$f(x)=\begin{cases}\cos(\pi x)&\text{ if }x\in[0,1]\setminus\Bbb Q\\0&\text{ otherwise,}\end{cases}$$then $\{f(x)\mid0\leqslant x\leqslant1\}$ has no maximum or minimum. But its supremum is $1$ and its infimum is $-1$.

  • $\begingroup$ @BernardPan I've edited my answer. Thank you. $\endgroup$ Aug 15 at 22:09
  • 2
    $\begingroup$ or another example (I'm sure you know this; this is just for OP) $f(x)=x$ for $x\in (0,1)$ and $f(0)=f(1)=\frac{1}{2}$. This is a bounded function on $[0,1]$ with no minimum/maximum, but infimum/supremum of $0,1$ respectively. $\endgroup$
    – peek-a-boo
    Aug 15 at 22:11
  • $\begingroup$ @JoséCarlosSantos No problem. Very nice construction! $\endgroup$ Aug 15 at 22:12
  • $\begingroup$ @JoséCarlosSantos So we use $inf$ and $sup$ because sometimes the function can't have a $max$ or $min$ but the function must be defined on the whole interval $[a,b]$ to be riemann integrable on the interval $[a,b]$, is that right ? $\endgroup$ Aug 15 at 22:27
  • $\begingroup$ Yes, that is correct. $\endgroup$ Aug 15 at 22:29

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