Riemann integrability of the following function.

Question

I have the following assignment problem on which I have been stuck for over a week now.

Let $f: [a,b] \rightarrow \mathbb{R}$ be function such that there is a number $K>0$ such that

$\sum_{j=1}^{n}|f(a_j)-f(a_{j-1})| \leq K$

whenever $\{a=a_o < a_1 < ... < a_n = b\}$ is a partition of $[a,b]$. Show that $f$ is integrable on $[a,b]$.

I have tried using the Cauchy criterion to establish integrability. But I don't know how to proceed. Here's what I tried. Begin with two arbitrary tagged partitions $P = \{[x_{k-1},x_k]\}_{k=1}^{n}$ and $Q = \{[y_{k-1},y_k]\}_{k=1}^{n}$ of $[a,b]$. We obviously have

$\begin{align*} |R(f,P) - R(f,Q)| &= |\sum_{k=1}^{n}f(x_k^*)(x_k - x_{k-1}) - \sum_{k=1}^{n}f(y_k^*)(y_k - y_{k-1})|\\ &\leq |\sum_{k=1}^{n}f(x_k^*)(x_k - x_{k-1})| + |\sum_{k=1}^{n}f(y_k^*)(y_k - y_{k-1})|\\ &\leq \sum_{k=1}^{n}|f(x_k^*)|\cdot||P|| + \sum_{k=1}^{n}|f(y_k^*)|\cdot||Q|| \end{align*}$

Here, $x_k^*$ and $y_k^*$ are the tags of $P$ and $Q$, and $||P||$ and $||Q||$ are their norms. Now I don't know how to link the last line above with the condition given in the problem. Any hints would be very helpful.

Answer 1

Here is a direct approach which avoids any mention of functions of bounded variation.

Let $\epsilon>0$ be pre-assigned and let $n$ be a positive integer such that $K/n<\epsilon/3$. Let us define a partition $$P_n=\{x_0,x_1,\dots,x_n\} $$ by $$x_j=a+j\cdot\frac{b-a} {n} $$ so that each subinterval $[x_{j-1},x_j]$ is of the same length $(b-a) /n$.

Let us observe that if the function $f$ satisfies the conditions given in question then it is bounded on $[a, b] $. And then we can define $$M_j=\sup\, \{f(x) \mid x\in[x_{j-1},x_j]\}, m_j=\inf\, \{f(x) \mid x\in[x_{j-1},x_j]\} $$ Also let the points $a_j, b_j\in[x_{j-1},x_j]$ be such that $$0\leq M_j-f(a_j) <\frac{\epsilon} {3(b-a)},0\leq f(b_j) - m_j<\frac{\epsilon} {3(b-a)}$$ Such points exists because of the way $M_j, m_j$ are defined. We now consider the difference between upper sum $$U(f, P_n) =\sum_{j=1}^{n}M_j(x_j-x_{j-1})$$ and the lower sum $$L(f, P_n) =\sum_{j=1}^{n}m_j(x_j-x_{j-1})$$ We have \begin{align} U(f, P_n) - L(f, P_n) & =\sum_{j=1}^n (M_j-m_j) (x_j-x_{j-1})\notag\\ &=\left|\sum_{j=1}^n(M_j-f(a_j)+f(b_j)-m_j+f(a_j)-f(b_j))(x_j-x_{j-1})\right|\notag\\ &\leq\sum_{j=1}^n (M_j-f(a_j)) (x_j-x_{j-1})+\sum_{j=1}^{n}(f(b_j)-m_j)(x_j-x_{j-1})+\sum_{j=1}^n|f(a_j)-f(b_j)|(x_j-x_{j-1})\notag\\ &<\frac{\epsilon}{3(b-a)}\cdot(b-a)+\frac{\epsilon}{3(b-a)}\cdot(b-a) +\frac{K} {n} \notag\\ &<\epsilon \end{align} By Darboux criterion $f$ is Riemann integrable on $[a, b] $.

Answer 2

Regardless on how you define the Riemann integral ((1) through refinement of partitions or (2) through smallness if partition size $\|P\|$) one effective way to verify existence of the Riemann integral in the setting you describe is by comparing the upper and lower Darboux sums for a common partition $\mathcal{P}=\{a=x_0<\ldots <x_n=b\}$.

From your posting, it seems that you are defining the integral is by method (2), that is:

A bounded function $f$ on $[a,b]$ is Riemann integrable if the limit $$\lim_{\|P\|\rightarrow0}S(f,\mathcal{P},\{t^*_j\})=\lim_{\|P\|\rightarrow0}\sum^n_{j=1}f(t_k)(x_{j-1}-x_j)$$ exists and is independent of the tags $\tau=\{t_k: x_{k-1}\leq t_k\leq x_j, \, 1\leq k\leq n\}$.

Meaning that there exists a number $A$ such that for any $\varepsilon>0$, there is $\delta>0$ such that for any partition $P$ with $\|P\|<\delta$ and any set of tags $\tau$ adapted to $P$ $$|S(f,P,\tau)-A|<\delta$$

Let's proceed then in within this framework.

For a subinterval $I_j:=[x_{j-1},x_j]$ of the partition $\mathcal{P}$ denote $M_j=\sup\{f(t): t\in I_j\}$ and $m_j=\inf\{f(t):t\in I_j\}$. The Darboux sums related the partition $\mathcal{P}$ are defined as \begin{align} U(f,\mathcal{P})=\sum^n_{j=1}M_j(x_j-x_{j-1}),\qquad L(f,\mathcal{P})=\sum^n_{j=1}m_j(x_j-x_{j-1}) \end{align} Darboux's criteria in your framework states that the existence of the integral $A=\int^b_af$ is equivalent to $$\lim_{\|\mathcal{P}\|\rightarrow0}U(f,\mathcal{P})-L(f,\mathcal{P})=0$$

Suppose then that $f$ satisfies $\sum^n_{j=1}|f(x_j)-f(x_{j-1})|\leq K$ for any partition $\mathcal{P}=\{a=x_0<\ldots<x_n=b\}$. For the moment, fix a partition $\mathcal{P}$. Notice that for each subinterval $I_j-[x_{j-1},x_j]$ $$0\leq M_j-m_j=\sup\{|f(t)-f(s): t,s\in I_j\}$$ Given $\varepsilon>0$ choose for each subinterval $I_j$ points $t_j,s_j\in I_j$ such that $$M_j-m_j -\tfrac{\varepsilon}{2n}<|f(t_j)-f(s_j)|$$ This can e done by definition of the supremum. Observe that $$(M_j-m_j)(x_j-x_{j-1})\leq \big(|f(t_j)-f(s_j)|+\tfrac{\varepsilon}{2n}\big)(x_j-x_{j-1})<\big(|f(t_j)-f(s_j)|+\tfrac{\varepsilon}{2n}\big)\|P\|$$ Adding over $j$ yields $$U(f,\mathcal{P})-L(f,\mathcal{P})\leq \Big(\sum^n_{j=1}|f(t_j)-f(s_j)|\Big)\|\mathcal{P}\|+\tfrac{\varepsilon}{2}\|\mathcal{P}\| $$ By considering the augmented partition $\mathcal{P}'=\mathcal{P}\cup\{s_j,t_j:1\leq j\leq n\}=\{a=x'_0<x'_1<\ldots<x'_{n'}=b\}$ and noticing that $x_{j-1}\leq s_j,t_j\leq x_j$ it is easy to see that $$\sum^n_{j=1}|f(t_j)-f(s_j)|\leq \sum^{n'}_{j=1}|f(x'_j)-f(x'_{j-1})|\leq K$$

Finally, by choosing partitions $\mathcal{P}$ for which $\|\mathcal{P}\|<1$ and $\|\mathcal{P}\|<\tfrac{\varepsilon}{2K}$ we obtain that $$U(f,\mathcal{P})-L(f,\mathcal{P})<\varepsilon$$

This shows that $$\lim_{\|\mathcal{P}\|\rightarrow0}U(f,\mathcal{P})-L(f,\mathcal{P})=0$$

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Notes:

• A separate but similar proof can be obtained in the framework (1) of partition refinements.

• It is well known (but requires some work) that frameworks (1) and (2) are equivalent.

• The condition $\sum^n_{j=1}|f(x_j)-f(x_{j-1})|\leq K$ for all partitions $\mathcal{P}=\{a=x_0<\ldots < x_n=b\}$ is known as saying that $f$ has finite variation on $[a,b]$. This means that $f$ can be expressed as the difference of two bounded monotone nondecreasing functions. Knowledge of tis would simplify the proof considerably as bounded monotone functions are easily seen to be Riemann integrable.