# Help with using a Cauchy criterion for Riemann Integrability to show that a continuous function is Riemann Integrable

Let $f:[a,b]\rightarrow\mathbb{R}$. A tagged partition, $\mathcal{P}$ of $[a,b]$ is a set of ordered pairs defined as $$\mathcal{P}:=\{([x_{k−1},x_k]),t_k)\}^n_{k=0},$$ where $a=x_0<...<x_n=b$ and the "tags" $t_k∈[x_{k−1},x_k]$, where $\mathbb{P}_{[a,b]}$ is the set of all tagged partitions over $[a,b]$. $$\|\mathcal{P}\|:=\sup\{x_k-x_{k-1}|1\leq k\leq n\}$$ is the mesh of the partition. The Riemann sum of $f$ over $[a,b]$ w.r.t $\mathcal{P}$ is defined as $$S(f,\mathcal{P}):=\sum\limits^n_{k=1}f(t_k)(x_k−x_{k−1})$$ and $f$ is said to be Riemann integrable with $\int_a^bf=L$ iff $$(\forall\epsilon>0)(\exists\delta>0)(\forall \mathcal{P}\in\mathbb{P}_{[a,b]})\bigg(\|\mathcal{P}\|<\delta\Rightarrow |S(f,\mathcal{P})-L|<\epsilon\bigg)$$

A Cauchy integrability criterion which is a straightforward exercise says that $f$ Riemann integrable on $[a,b]$ iff $$(\forall\epsilon>0)(\exists\delta>0)(\forall \mathcal{P}\in\mathbb{P}_{[a,b]})(\forall \mathcal{Q}\in\mathbb{P}_{[a,b]})\bigg(\|\mathcal{P}\|<\delta\wedge\|Q\|<\delta\Rightarrow |S(f,\mathcal{P})-S(f,\mathcal{Q})|<\epsilon\bigg).$$

My question is that how can we use the above definitions and Cauchy criterion to show that if $f$ is continuous on $[a,b]$ then $f$ is intgrable on $[a,b]$?

If we use the Darboux definition for integrability which uses the upper and lower Riemann sums, then we have an analogous Cauachy integrabilityly condition with the difference of the upper and lower sums. With this and the fact that a continuous function is uniformly continuous on a compact set it is a very simple task to show that continuity is a sufficient condition for integrability. However in this situation we do not have much flexibility as we need to take two different partitions so there is not much we can do with $|S(f,\mathcal{P})-S(f,\mathcal{Q})|$, and taking the common refinement partition of $\mathcal{P}\cup\mathcal{Q}$ leads nowhere. So how do we in this setup up and not using upper and lower sums show that continuity implies integrability? So any input and assistance is greatly needed and appreciated. Also if you have other proofs that don't use the Darboux definition of integrability it will be most welcomed.

• Well, doesn't it suffice to show $$\text{Daroux} \Longleftrightarrow \text{Cauchy}$$ and then use the proof using the Darboux definition? – MathematicsStudent1122 Apr 7 '17 at 19:21

We can use the following:

Lemma. If $f$ is continuous on $[a,b]$ then for any $\epsilon > 0$ there exists $\delta > 0$ such that for any partition $P$ of $[a,b]$ with $\|P\| < \delta$ and any refinement $R$ of $P$, we have $|S(f,P) - S(f,R)| < \epsilon$.

Applying the lemma, we can show that if $f$ is continuous, then the Cauchy criterion is satisfied. That is, for any $\epsilon >0$ there exists $\delta > 0$ such that if $P$ and $Q$ are any partitions satisfying $\|P\|, \|Q\| < \delta$, then $|S(f,P) - S(f,Q)| < \epsilon.$

To see this, let $R = P \cup Q$ be a common refinement and take $\delta$ as specified in the lemma such that if $\|P\|, \|Q\| < \delta$, we have $|S(f,P) - S(f,R)| < \epsilon/2$ and $|S(f,Q) - S(f,R)| < \epsilon/2$. Whence, it follows that

$$|S(f,P) - S(f,Q)| \leqslant |S(f,P) - S(f,R)| + |S(f,Q) - S(f,R)| < \epsilon/2 + \epsilon/2 = \epsilon.$$

It remains to prove the lemma.

Since $[a,b]$ is compact, $f$ is uniformly continuous and for any $\epsilon > 0$ there exists $\delta >0$ such that if $|x - y| < \delta$, then $|f(x) - f(y)| < \epsilon/(b-a)$. Suppose $\|P\| < \delta$ and $R$ is a refinement of $P$. Any subinterval $[x_{j-1}, x_{j}]$ of $P$ can be decomposed as the union of subintervals of $R$,

$$[x_{j-1},x_{j}] = \bigcup_{k=1}^{n_j}[y_{j,k-1}, y_{j,k}],$$

and

\begin{align}\left|S(f,P) - S(f,R)\right| &= \left|\sum_{j=1}^n f(\xi_j)(x_j - x_{j-1}) - \sum_{j=1}^n \sum_{k=1}^{n_j}f(\eta_{j,k})(y_{j,k} - y_{j,k-1})\right| \\ &\leqslant \sum_{j=1}^n \sum_{k=1}^{n_j}|f(\xi_j) - f(\eta_{j,k})|(y_{j,k} - y_{j,k-1}) \\ &\leqslant \sum_{j=1}^n \sum_{k=1}^{n_j}\frac{\epsilon}{b-a}(y_{j,k} - y_{j,k-1}) \\ &= \epsilon\end{align}

• Thank you, this is what I was looking for. – user152874 Apr 8 '17 at 7:22
• Just a quick question shouldn't it be $$|S(f,\mathcal{P})-S(f,\mathcal{R})|=|\sum\limits_{j=1}^n f(\xi_j)\sum\limits_{k=1}^{n_j}(y_{j,k}-y_{j,k-1})-\sum\limits_{j=1}^{n}\sum\limits_{k=1}^{n_j}f(\eta_{j,k})(y_{j,k}-y_{j,k-1})|?$$ – user152874 Apr 8 '17 at 7:42
• Yes, that step is true, but we can move $f(\xi_j)$ which does not depend on $k$ inside the sum over $k$. Then we apply the triangle inequality and estimate $|f(\xi_j) - f(\eta_{j,k}| < \epsilon/(b-a)$ since both intermediate points are inside the interval $[x_{j-1},x_j]$ with length less than $\|P\| < \delta$. – RRL Apr 8 '17 at 7:54
• Thanks, but just to confirm also you had $(y_{j,k-1}-y_{j,k})$ so shouldn't it be $(y_{j,k}-y_{j,k-1})$? Thanks again. – user152874 Apr 8 '17 at 7:58
• That's a typographical error. Will fix. Thanks – RRL Apr 8 '17 at 8:00