# Darboux integrability implies Riemann integrability

I have searched the site for posts regarding Darboux integrability $$\implies$$ Riemann integrability, but haven't found any that specifically adress this question.

My definition of Darboux integrability: Let $$f$$ be defined and bounded on $$[a,b]$$, then $$f$$ is Darboux integrable if for all $$\epsilon >0$$ there exists a partition $$P$$ of $$[a,b]$$ such that $$U(f,P)-L(f,P)<\epsilon$$ (where $$U$$ and $$L$$ are the upper and lower Riemann sums respectively).

My definition of Riemann integrability: Let $$f$$ be defined and bounded on $$[a,b]$$, then $$f$$ is Riemann integrable if $$\lim_{N\to\infty} \sum\limits_{k=1}^{N} f(c_k)(x_{k}-x_{k-1})$$ has the same limit for all sequences of partitions $$P_N$$ and all choices of $$c_k\in[x_{k-1},x_{k}]$$.

If my definitions are correct, it seems that Darboux integrability only requires one partition to fulfil the epsilon-inequality, whereas Riemann integrability requires all sequences of partitions to be fulfilled. How can this lead to an implication nevertheless?

• You want $f$ to be bounded I think. Also $N\to \infty$ is not what you want: you want the mesh size of $P$ to $\to 0.$
– zhw.
May 8, 2019 at 0:34
• The key here is that Darboux definition uses refinement of partitions and Riemann's definition uses norm of partition. Luckily both concepts are equivalent for Darboux and Riemann sums. See for example this answer : math.stackexchange.com/a/2047959/72031 May 9, 2019 at 4:30

If $$U(f,P)-L(f,P)<\epsilon$$, for a given partition $$P$$, then the inequality holds for $$any$$ refinement of $$P$$, since if $$P\subseteq P'$$, we have

$$L(f,P)\le L(f,P')\le U(f,P')\le U(f,P).$$

And since for any partition $$P=\{a,x_1,\cdots, x_{n-2},b\}$$,

and any $$\{x^*_i:x_i\le x^*_i\le x_{i+1};\ 1\le i\le n\},$$

we have $$L(f,P)\le \sum^{n-1}_{i=1}f(x^*_i)(x_{i+1}-x_i)\le U(f,P),$$

it follows that Darboux integrability implies Riemann integrability.

edit: Assuming that the OP wants to show that if bounded function $$f$$ on $$[a, b]$$ is Darboux integrable then for each given $$\epsilon> 0$$, there is a $$\delta > 0$$ such that

mesh$$P < \delta \Rightarrow U(f, P) − L(f, P) < \epsilon,$$ here is a sketch:

Suppose we have a partition $$P_0=\{a = x_0 < \cdots < x_m = b\}$$ such that $$U(f,P_0)-L(f,P_0)<\epsilon.$$ Let $$P=\{a = t_0 < t_1 < \cdots < t_n = b\}$$ be any other partition, fine enough so that at most one member of $$P_0$$ lies between any two members of $$P$$.

Now, form $$Q=P\cup P_0$$ and note that in the difference $$L(f,Q)-L(f,P)$$, the only terms that remain correspond to elements of $$Q$$ of the form $$[t_{n_k}\le x_{n_k}\le t_{n_{k+1}}]$$.

Then, with $$m_k,m'_k,m''_k$$ the mimima of $$f$$ on $$[t_{n_k},x_{n_k}], [x_{n_k}, t_{n_{k+1}}]$$ and $$[t_{n_k}, t_{n_{k+1}}],\$$ respectively and $$M$$ an upper bound on $$|f|$$,

$$L(f, Q) − L(f, P)=\sum _k[(m_k(x_{n_k}-t_{n_k})+m'_k(t_{n_{k+1}}-x_{n_k}))-m''_k(t_{n_{k+1}}-t_{n_k})]\le \sum_k[M(t_{n_{k+1}}-t_{n_k})-m_k''((t_{n_{k+1}}-t_{n_k}))]\le MK(\text{mesh P})$$ where $$K$$ is the upper limit of the sum.

Similarly, $$U(f,P)-U(f,Q)\le MK(\text{mesh P})$$.

The result now follows from the string of inequalities:

$$U(f, P) − L(f, P) ≤ U(f, P) − U(f, Q) + U(f, Q) − L(f, Q) + L(f, Q) − L(f, P)$$.

• How have you shown the $$\lim_{N\to\infty} \sum\limits_{k=1}^{N} f(c_k)(x_{k}-x_{k-1})$$ condition?
– zhw.
May 8, 2019 at 0:35
• @zhw. I haven't shown it. I said " it follows that.." assuming the OP can finish the proof, which is routine. Or am I missing something? May 8, 2019 at 1:37
• I don't think it's routine. First, note that the OP hasn't stated the condition correctly, as I stated in a comment.
– zhw.
May 8, 2019 at 3:31
• @zhw. Now I am not even sure what the OP wants to prove. But I think he/she wants to show the content of my edit. May 8, 2019 at 4:16

Here's a direct solution in case you don't want to end up repeating the second half of Riemann-Lebesgue's proof.

You can use the partition $$P$$ for which $$U(f,P)-L(f,P)<\epsilon$$ to argue that for all partitions $$Q$$ with mesh$$(Q) = \|Q\|< \delta_P$$, the Riemann sum will be somewhere close to $$U(f, P)$$ and $$L(f, P)$$.

Assume that $$P$$ has $$N$$ points $$\{p_1=a, p_2, \cdots, p_N=b\}$$, partition $$Q$$ has $$N'$$ points $$\{q_1=a, \cdots, q_{N'}=b\}$$ and $$|f| \leq M$$ in $$[a, b]$$ (this should hold for some $$M$$ or Darboux integral won't be well-defined).

Now take $$\delta_P< \min(\|p\|, \frac{\epsilon}{MN})$$. Now for each $$i \leq N'$$, either $$[q_i, q_{i+1}] \subset [p_j, p_{j+1}]$$ (for some $$j\leq N$$), or $$p_{j-1} \leq q_i \leq p_j \leq q_{i+1} \leq q_{j+1}$$. The latter can happen at most $$N$$ times, so the area under $$f$$ for such cases can be at most $$N \times M \times \delta_P \leq \epsilon$$. For the rest, the sum happens to be nicely sandwiched by $$U(f,P)$$ and $$L(f, P)$$, proving the Riemann integrability.

• Hi. Sorry, I could not see why for the rest, the sum happens to be nicely bounded. Isn't it possible that the sup is negative? May 26, 2020 at 6:12
• @10understanding Because for the rest of the cases, we will have $[q_i, q_{i+1}] \subset [p_j, p{j+1}]$ (for some $i$, $j$) entirely. Since the latter is sandwiched by $U$ and $L$, so will the former. The point is to isolate the cases where $[q_i, q_{i+1}]$ cannot be contained within a single $[p_j, p_{j+1}]$ interval and bound them. The rest are already bounded.
– PkT
Aug 16, 2020 at 0:19
• This is beautiful, thanks! Oct 26, 2021 at 22:31

Hint

One has anyway $$\lim_{\|P\|\to 0} L(f,P)=\sup_P L(f,P)$$ and $$\lim_{\|P\|\rightarrow 0} U(f,P)=\inf_P U(f,P)$$ If for all $$\varepsilon >0$$ there exists a partition $$P$$ such that $$U(f,P)-L(f,P)<\varepsilon$$, those limits have the same value.

Then $$\lim_{\|P\|\to 0} S(f,P)$$ exists and is equal to that value since $$L(f,P) \le S(f,P) \le U(f,P)$$ for every, however tagged, partition $$P$$.

The sequential version of the last limit is valid too but one has to suppose $$\,\|P_N\| \to 0\,$$ of course.

For convenience I give a proof of the statement $$\lim_{\Vert P\Vert\to 0} L(f,P)=\sup_P L(f,P)$$ By definition of supremum, if $$\varepsilon>0$$, then there exists a partition $$Q$$ such that $$L(f,Q)>\sup_P L(f,P)-\frac \varepsilon 2$$If $$P$$ is any partition and $$r$$ is the number of the inner points of $$Q$$, then the set $$P \cup Q$$ is obtained by adding to $$P$$ a maximum of $$r$$ points.

Note that $$L(f,P\cup Q)\ge L(f,P)$$ since $$P\cup Q$$ is a refinement of $$P$$.

If only one point $$c$$ is added and $$[\alpha,\beta]$$ is the subinterval of $$P$$ containing $$c$$, then $$0 \le L(f,P\cup Q)-L(f,P)=\inf_{[\alpha,c]} f \cdot (c-\alpha)+\inf_{[c,\beta]} f \cdot (\beta-c)-\inf_{[\alpha,\beta]} f \cdot (\beta-\alpha)$$$$\le M \cdot (c-\alpha)+M \cdot (\beta-c)+M \cdot (\beta-\alpha)=2M \cdot (\beta-\alpha)\le 2M \cdot \Vert P\Vert$$ where $$M=\sup_{[a,b]} |f|$$ Note that $$\inf_{[\alpha,\beta]} f\ge -M$$By induction one gets $$0 \le L(f,P\cup Q)-L(f,P)\le2Mr \cdot \Vert P\Vert$$ whichever number of points is added to $$P$$.

From $$L(f,P\cup Q)\ge L(f,Q)$$ since $$P\cup Q$$ is a refinement of $$Q$$, it follows that $$0 \le \sup_P L(f,P)-L(f,P)<\frac \varepsilon 2+2Mr \cdot \Vert P\Vert$$ so $$0 \le \sup_P L(f,P)-L(f,P)<\varepsilon$$ if $$\Vert P\Vert < \frac {\varepsilon}{4Mr}$$

• May you explain more explicitly why $\displaystyle\lim_{\|P\|\to0}L(f,P)=\displaystyle\sup_P L(f,P)$ ? I was thinking of proving with each side $\leq$ and $\geq$ the other. I saw $\displaystyle\lim_{\|P\|\to0}L(f,P)=\limsup_P\sum_{R}\inf_Rf|R|=\limsup_P s_P$. Easily see that sequence $\{s_P\}$ is bounded, but I’m struggle to find it’s limit. And that’s just for the “$\leq$” direction, then what about the “$\geq$” direction, how can we show it ? Besides, is this the way of proof that you hinted ? And, would it be easier if we only considered the uniform partition ? Oct 3, 2023 at 20:29
• @PermQi See my addendum. My proof is standard. I doubt you can find another very different one. Oct 15, 2023 at 11:37
• On the right of $$\inf_{[\alpha,c]} f \cdot (c-\alpha)+\inf_{[c,\beta]} f \cdot (\beta-c)-\inf_{[\alpha,\beta]} f \cdot (\beta-\alpha)$$$$\le M \cdot (c-\alpha)+M \cdot (\beta-c)+M \cdot (\beta-\alpha)$$ why the third term is $+M$, but not $-M$ ? Oct 15, 2023 at 12:29
• @PermQi It's explained a little further ... Note that $$\inf_{[\alpha,\beta]} f\ge -M$$ Oct 15, 2023 at 22:53
• I realize your proof is actually the same to the hint of exercise 10.6(b),(c) in the book Analysis of Manifolds of Munkres. So how do you apply this method to the n-dimensions case ? Oct 17, 2023 at 4:26

A function is Riemann integrable if $$\sup_P\{L(f,P)\}=\underline{\int_a}^b f(x) =\overline{\int_a^b} f(x)=\inf_P\{U(f,P)\},$$ or equivalently, $$\displaystyle \inf_P\{U(f,P)\}-\sup_P\{L(f,P)\}<\epsilon$$ for any $$\epsilon>0.$$

Let $$\epsilon>0.$$ There exists $$P$$ such that $$U(f,P)-L(f,P)<\epsilon.$$ Then since $$L(f,P)\leq \sup_P\{L(f,P)\}\leq \inf_P\{U(f,P)\}\leq U(f,P),$$ clearly $$\displaystyle \inf_P\{U(f,P)\}-\sup_P\{L(f,P)\}<\epsilon$$ holds.

One way to see your definition of Riemann integrability is not correct is to look at $$[0,1]$$ and define $$f$$ to equal $$0$$ on $$[0,1/2],$$ $$f=1$$ on $$(1/2,1].$$ We know this function gives $$\int_0^1 f =1/2$$ using the Darboux definition.

But suppose $$P_N=\{k/(2N): k=0,\dots,N-1\}\cup \{1\}.$$ Define $$c_k= k/(2N),k=1,\dots,N-1,$$ $$c_N=1/2.$$ Then

$$\sum_{k=1}^{N}f(c_k)(x_k-x_{k-1}) = 0$$

for all $$N.$$