First, let $g$ be bounded on $[a,b]$. Now, assume $\exists P$, a partition, such that $L(g,P)=U(g,P)$.

I am told the correct answer to the question "describe $g$" is that $g$ is continuous on $[a,b]$.

Yet, this confuses me. Since we are talking about Riemann integrability, wouldn't we exclude the case where $g$ simply has a finite number of discontinuities?

Lastly, do we know $g(x)=c$, a constant?

Thank you in advance for any help!

I was asked for the exact definition of Lower and Upper Sum:

A partition $P$ of $[a,b]$ is a finite (we'll say it contains $n$ points) set of points from $[a,b]$ that includes both $a$ and $b$. The notational convention is to always list the points of a partition in increasing order.

For each subinterval $[x_{k-1},x_k]$ of $P$, let

$m_k$ be the infimum of the subinterval. $M_k$ is the supremum of the subinterval.

The lower sum of $f$ with respect to $P$ is given by $L(f,P)=\sum\limits_{k=1}^nm_k(x_k-x_{k-1})$.

Similarly, the upper sum is: $U(f,P)=\sum\limits_{k=1}^nM_k(x_k-x_{k-1})$.

  • $\begingroup$ Take $P=\{x_1= a, x_2, \dots , x_n=b\}$ to be the partition on your hypothesis, what does $L(g,P)=U(g,P)$ implies? Use your definitions. $\endgroup$
    – Luis Vera
    Apr 26, 2016 at 17:22
  • $\begingroup$ Could it imply that mk = Mk? $\endgroup$
    – user322548
    Apr 26, 2016 at 17:25
  • $\begingroup$ @EthanZell Can you prove that? $\endgroup$
    – Mambo
    Apr 26, 2016 at 17:27
  • $\begingroup$ Perhaps. If you use the assumption that L(g,P)=U(g,P), then set their respective sums equal to one another. Then use the fact that the series is telescoping to show that mk =Mk. Really, I'm just spitballing here but hopefully that makes sense. $\endgroup$
    – user322548
    Apr 26, 2016 at 17:29
  • $\begingroup$ If you can, what can you say about $g$ from that? $\endgroup$
    – Luis Vera
    Apr 26, 2016 at 17:33

1 Answer 1


The thing is

For each subinterval $[x_{k-1},x_k]$ of $P$, let

that the closed subintervals of the partition are used to define $m_k$ and $M_k$, and each closed subinterval intersects its neighbours. If the open - or half-open - subintervals of the partition were used, $g$ could have discontinuities at the partition points, but using the closed subintervals rules that out.

To obtain the conclusion - that $g$ is in fact constant, not only continuous - compute

\begin{align} 0 &= U(g,P) - L(g,P)\\ &= \sum_{k = 1}^n M_k(x_k - x_{k-1}) - \sum_{k = 1}^n m_k (x_k - x_{k-1}) \\ &= \sum_{k = 1}^n (M_k - m_k)\cdot (x_k - x_{k-1}) \end{align}

and note that all terms are non-negative since $m_k \leqslant M_k$ and $x_{k-1} < x_k$ for all $k$. A sum of non-negative terms can only be $0$ if each term is $0$, and since $x_k - x_{k-1} > 0$, it must be that $m_k = M_k$ for every $k$. But that means $g$ is constant on each closed subinterval $[x_{k-1},x_k]$. And since neighbouring subintervals intersect, the value $g$ attains must be the same on all subintervals, i.e. $g$ is (globally) constant.


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