Let $a,b \in \mathbb{R}$, $a<b$, $f :[a,b] \to \mathbb{R}$ is monotonically decreasing.

Use the def. of integrability with Darboux sums, prove that $f$ is integrable on $[a,b]$.

I know that the definition of integrability with Darboux sums says that if the lower darboux sum equals the upper darboux sum, then the function is integrable. But, if the function is monotonically decreasing then how can the infimum = supremum.

  • $\begingroup$ Can you prove that a monotone function can only have countably many discontinuities? If so, then you can delete a small set to make the set of discontinuities only finite, and then integrate the continuous functions on each piece. Then prove that the deletion can be refined to make the error small. (There is a more direct proof but I think it is somewhat more complicated than this one.) $\endgroup$
    – Ian
    Dec 17, 2014 at 4:44
  • $\begingroup$ This is theorem 6.9 in Principles of mathematical analysis, 3rd ed. $\endgroup$
    – hrkrshnn
    Dec 17, 2014 at 5:28

1 Answer 1


To show that $f$ is Darboux integrable you need to show that $\sup_P L(f,P) = \inf_P U(f,P)$, where $P$ are the partitions of $[a,b]$. Since we always have $L(f,P) \le U(f,P)$, it is sufficient to show that we can make $U(f,P)-L(f,P)$ as small as we like.

If $f$ was continuous, we could use uniform continuity to show this quickly. For general non increasing $f$ we need to find a similar uniformity while taking care around points where $f$ 'drops' too much.

The idea of the proof is straightforward but the notation is cumbersome.

Let $o(x) = \lim_{y \downarrow x} f(y) - \lim_{y \uparrow x} f(y)$. Since $f$ is non increasing, we have $o(x) \ge 0 $. It is not hard to show that the set of points for which $o(x) >0$ is at most countable, and furthermore, $f(a)-f(b) \ge \sum_k o(x_k)$ where $x_k$ are the points at which $o(x_k) >0$.

Note that $|f|$ is bounded by some $B$. Also note that if $I$ is an interval, and $f$ is non increasing, $\sup_{y \in I} f(y) -\inf_{y \in I} f(y) \le f(\inf I)-f(\sup I)$.

Let $\epsilon>0$ and choose $N$ such that $\sum_{k >N} o(x_k) < \epsilon$. Choose $\delta>0$ such that $4 \delta N B < \epsilon$ and the intervals $[x_k-\delta,x_k+\delta]$ do not intersect for $k=1,...N$. Let $I_k = [x_k-\delta,x_k+\delta]$, then we have $(\sup_{y \in I_k} f(y)-\inf_{y \in I_k} f(y)) l(I_k) < {1 \over N}\epsilon$. Let $P_1 $ be the partition consisting of the points $x_k-\delta,x_k+\delta$, for $k =1,...N$.

Let $J_k$ be the remaining intervals. We may take the $J_k$ to be closed (and hence compact). Note that $o(x) < \epsilon$ for all $x \in \cup_k J_k$ (this is the 'uniformity' condition that we need). By definition, for each such $x$ we may find an open interval $U_x$ containing $x$ such that $f(\inf U_x) -f(\sup U_x) < \epsilon$. Since the $J_k$ are compact, they are covered by a finite number of such sets. In particular, we find a partition $\Pi_k$ of each $J_k$ such that for any neighbouring pair of points $x,y \in \Pi_k$ ($x<y$), we have $f(x)-f(y) < \epsilon$.

Now let $P$ be the partition consisting of the union of $P_1$ and the $\Pi_k$. Let ${\cal I}$ be the collection of intervals $I$ in the partition $P$. Note that $|{\cal I} \setminus {\cal U}| = N$. Let ${\cal U} \subset {\cal I}$ (u for 'uniform') be the intervals $I$ contained in the $\Pi_k$. Then we have \begin{eqnarray} U(f,P)-L(f,P) &=& \sum_{I \in {\cal I}} (\sup_{y \in I} f(y) - \inf_{y \in I} f(y)) l(I) \\ &=& \sum_{I \in {\cal U}} (\sup_{y \in I} f(y) - \inf_{y \in I} f(y)) l(I) + \sum_{I \notin {\cal U}} (\sup_{y \in I} f(y) - \inf_{y \in I} f(y)) l(I) \\ &\le& \sum_{I \in {\cal U}} (\sup_{y \in I} f(y) - \inf_{y \in I} f(y)) l(I) + N {1 \over N} \epsilon \\ &\le& \epsilon \sum_{I \in {\cal U}} l(I) + N {1 \over N} \epsilon \\ &\le& \epsilon (b-a) + \epsilon \end{eqnarray} Hence we can make $U(f,P)-L(f,P)$ as small as we like and so we conclude that $f$ is Darboux integrable.


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