Darboux integrability implies Riemann integrability

Question

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?

Answer 1

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:

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

Answer 2

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.

Answer 3

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.

Addendum

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 $$\Vert P\Vert < \frac {\varepsilon}{4Mr}$$

Answer 4

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.

Answer 5

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 $N.$