7. Riemann Integration

Let $L_1$ and $L_2$ be two real numbers satisfying the definition of Riemann integrability and let $\epsilon >0$ be arbitrary. Then there exists $\delta >0$ such that $|S(f;\dot{\mathcal{P}})-L_1|<\epsilon /2$ and $|S(f;\dot{\mathcal{P}})-L_2|<\epsilon /2$, for all sampled partitions $\dot{\mathcal{P}}$ with $\Vert \dot{\mathcal{P}}\Vert <\delta$. Then, if $\Vert \dot{\mathcal{P}}\Vert <\delta$ it holds that $$\begin{array}{rl}|L_1-L_2|& \le |S(f;\dot{\mathcal{P}})-L_1|+|S(f;\dot{\mathcal{P}})-L_2|\\ & <\epsilon .\end{array}$$ By Theorem 2.2.7 this proves that $L_1=L_2$.

Let $f:[a,b]:\to \mathbb{R}$ be such that $f(x)=C$ for all $x\in [a,b]$ and let $\dot{\mathcal{P}}=\{([x_{k-1},x_k],t_k)\}$ be a sampled partition of $[a,b]$. Then $$\begin{array}{rl}S(f;\dot{\mathcal{P}})& =\sum _{k=1}^{n}f(t_k)(x_k-x_{k-1})\\ & =C\sum _{k=1}^n(x_k-x_{k-1})\\ & =C(x_n-x_0)\\ & =C(b-a).\end{array}$$ Hence, with $L=C(b-a)$, we obtain that $|S(f;\dot{\mathcal{P}})-L|=0<\epsilon$ for any $\epsilon >0$ and therefore ${\int }_a^{b}f=C(b-a)$. This proves that $f$ is Riemann integrable.

We consider the special case that $[a,b]=[0,1]$, the general case is similar. Let $\dot{\mathcal{Q}}=\{([x_{k-1},x_k],q_k)\}$ be a sampled partition of $[0,1]$ chosen so that $q_k=\frac{1}{2}(x_k+x_{k-1})$, i.e., $q_k$ is the midpoint of the interval $[x_{k-1},x_k]$. Then $$\begin{array}{rl}S(f;\dot{\mathcal{Q}})& =\sum _{k=1}^{n}f(q_k)(x_k-x_{k-1})\\ & =\sum _{k=1}^n\frac{1}{2}(x_k+x_{k-1})(x_k-x_{k-1})\\ & =\frac{1}{2}\sum _{k=1}^n(x_k^2-x_{k-1}^2)\\ & =\frac{1}{2}(x_n^2-x_0^2)\\ & =\frac{1}{2}(1^2-0^2)\\ & =\frac{1}{2}.\end{array}$$ Now let $\dot{\mathcal{P}}=\{([x_{k-1},x_k]),t_k{\}}_{k=1}^n$ be an arbitrary sampled partition of $[0,1]$ and suppose that $\Vert \dot{\mathcal{P}}\Vert <\delta$, so that $(x_k-x_{k-1})\le \delta$ for all $k=1,2,\dots ,n$. If $\dot{\mathcal{Q}}=\{([x_{k-1},x_k],q_k){\}}_{k=1}^n$ is the corresponding midpoint sampled partition then $|t_k-q_k|<\delta$. Therefore, $$\begin{array}{rl}|S(f;\dot{\mathcal{P}})-S(f;\dot{\mathcal{Q}})|& =|\sum _{k=1}^n{t}_k(x_k-x_{k-1})-q_k(x_k-x_{k-1})|\\ & \le \sum _{k=1}^n|t_k-q_k|(x_k-x_{k-1})\\ & <\delta (1-0)\\ & =\delta .\end{array}$$ Hence, we have proved that for arbitrary $\dot{\mathcal{P}}$ that satisfies $\Vert \dot{\mathcal{P}}\Vert <\delta$ it holds that $|S(f;\dot{\mathcal{P}})-1/2|<\delta$. Hence, given $\epsilon >0$ we let $\delta =\epsilon$ and then if $\Vert \dot{\mathcal{P}}\Vert <\delta$ then $|S(f;\dot{\mathcal{P}})-1/2|<\epsilon$. Therefore, ${\int }_0^{1}f=\frac{1}{2}$.

Let $$. Suppose that $$ except at one point $$. Let $$ be a sampled partition. We consider mutually exclusive cases. First, if $$ and $$ for all $$ then $$. If $$ for some $$ then $$ If $$ for some $$ then necessarily $$ and then $$ Hence, in any case, by the triangle inequality $$ where $$. Let $$ be arbitrary. Then there exists $$ such that $$ for all partitions $$ such that $$. Let $$. Then if $$ then $$ This proves that $$ and $$. Now suppose by induction that if $$ for all $$ except at a $$ number of points in $$ then $$ and $$. Now suppose that $$ is such that $$ for all $$ except at the points $$. Define the function $$ by $$ for all $$ except at $$ and define $$. Then $$ and $$ differ at the points $$. Then by the induction hypothesis, $$ and $$. Now $$ and $$ differ at the point $$ and therefore $$ and $$. This ends the proof.

If $$ then $$ for all $$ and then clearly $$, so assume that $$. Let $$ be given. Then there exists $$ such that if $$ then $$. Now for any partition $$, it holds that $$. Therefore, if $$ then $$ To prove (b), it is easy to see that $$. Given $$ there exists $$ such that $$ and $$, whenever $$. Therefore, if $$ we have that $$ To prove (c), let $$ be arbitrary and let $$ be such that if $$ then $$ Now, by assumption, $$ and therefore $$ Therefore, $$ Since $$ is arbitrary, we can choose $$ and then passing to the limit we deduce that $$.

Let $$ and put $$. There exists $$ such that if $$ then $$ and therefore $$. Suppose by contradiction that $$ is unbounded on $$. Let $$ be a partition of $$, with sets $$, and with $$. Then $$ is unbounded on some $$, i.e., for any $$ there exists $$ such that $$. Choose samples in $$ by asking that $$ for $$ and $$ is such that $$ Therefore, (using $$ implies that $$) $$ This is a contradiction and thus $$ is bounded on $$.

Let $$ be arbitrary and let $$. By definition of $$, the set $$ is finite, say consisting of $$ elements. Let $$ and let $$ be a sampled partition of $$ with $$. We can separate the partition $$ into two sampled partitions $$ and $$ where $$ has samples in the the set $$ and $$ has no samples in $$. Then the number of intervals in $$ can be at most $$, which occurs when all the elements of $$ are samples and they are at the endpoints of the subintervals of $$. Therefore, the total length of the intervals in $$ can be at most $$. Now $$ for every sample $$ in $$ and therefore $$. For samples $$ in $$ we have that $$. Therefore, since the sum of the lengths of the subintervals of $$ is $$, it follows that $$. Hence $$. Thus $$.

To show that $$ is not in $$, we must show that there exists $$ such that for all $$ there exists sampled partitions $$ and $$ with norm less than $$ but $$. To that end, let $$, and let $$ be arbitrary. Let $$ be sufficiently large so that $$. Let $$ be a sampled partition of $$ with intervals all of equal length $$ and let the samples of $$ be rational numbers. Similarly, let $$ be a partition of $$ with intervals all of equal length $$ and with samples irrational numbers. Then $$ and $$, and therefore $$.

If $$ then let $$. Then clearly $$ for all $$. Now suppose the converse and let $$ be arbitrary. Let $$ and $$ satisfy the conditions of the theorem, with $$. Now, there exists $$ such that if $$ then $$ and $$ For any sampled partition $$ it holds that $$, and therefore $$ If $$ is another sampled partition with $$ then also $$ Subtracting the two inequalities $$-$$, we deduce that $$ Therefore, since $$ it follows that $$ By the Cauchy criterion, this proves that $$.

Let $$ be the intervals where $$ is constant, and let $$ be the constant values taken by $$ on the intervals $$, respectively. Then it is not hard to see that $$. Then $$ is the sum of Riemann integrable functions and therefore is also Riemann integrable. Moreover, $$.

Let $$ be arbitrary. Since $$ is uniformly continuous on $$ there exists $$ such that if $$ then $$. Let $$ be sufficiently large so that $$. Partition $$ into $$ subintervals of equal length $$, and denote them by $$, where $$ and $$ for $$. Then for $$ it holds that $$. For $$ define $$. Therefore, for any $$ it holds that $$. Since $$, it holds that $$ for all $$.

Suppose that $$ is continuous. Let $$ be arbitrary and let $$. Then there exists a step-function $$ such that $$ for all $$. In other words, for all $$ it holds that $$ The functions $$ and $$ are Riemann integrable integrable on $$, and $$. Hence, by the Cauchy criterion, $$ is Riemann integrable.

Assume that $$ is increasing and that $$ (if $$ then $$ is the zero function which is clearly integrable). Let $$ be arbitrary. Let $$ be such that $$. Partition $$ into subintervals of equal length $$, and as usual let $$ denote the resulting points of the partition. On each subinterval $$, it holds that $$ for all $$ since $$ is increasing. Let $$ be the step-function whose constant value on the interval $$ is $$ and similarly let $$ be the step-function whose constant value on the interval $$ is $$, for all $$. Then $$ for all $$. Both $$ and $$ are Riemann integrable and $$ Hence by the Squeeze theorem for integrals (Theorem 7.2.3), $$.

Assume for simplicity that $$. Let $$ be arbitrary. Then there exists $$ such that if $$ then $$. For any $$, with intervals $$ for $$, there exists, by the Mean Value Theorem applied to $$ on $$, a point $$ such that $$. Therefore, $$ where $$ has the same intervals as $$ but with samples $$. Therefore, if $$ then $$ Hence, for any $$ we have that $$ and this shows that $$.

For any $$ such that $$ it holds that $$ and therefore $$. Since $$ is Riemann integrable on $$ it is bounded and therefore $$ for all $$. In particular, $$ for all $$ and thus $$, and thus $$