Real Analysis: Integrals

Riemann Integrals

Motivation: A Different Formalization of The Integral

The Darboux integral is not the only way to formalize the concept of the area under the curve. The Riemann integral is another and more famous way to do just that. While the Darboux integral is better suited for proving theorems related to integration, the Riemann integral is better suited to both estimating the area under a curve with a finite sum as well as calculating its exact value by evaluating a limit. The equivalence of the Darboux and Riemann integrals allows us to pick the one we want depending on our needs.

The Riemann Integral

A tagged subdivision of a closed interval $[a, b]$ is a finite interval subdivision $P$ along with a set of values $T = \{t_1, \ldots, t_n\}$ called tags where $t_i \in p_i$ for each $p_i \in P.$

The Riemann sum of $f$ with respect to a tagged subdivision $(P, T)$ is

$$\sum\limits_{i=1}^n f(t_i) \mu(p_i).$$

The Riemann integral of $f$ from $a$ to $b$ is defined as 

$$\int\limits_a^b f(x) \, dx = S$$

if for every $\varepsilon > 0$ there exists a $\delta > 0$  such that

$$\left| \sum\limits_{i=1}^n f(t_i)\mu(p_i)  - S\right| < \varepsilon \text{ for all } (T, P) \text{ where } \max\limits_{p_i \in P}(\mu(p_i)) < \delta.$$

The main difference between the Riemann integral and the Darboux integral is that the Riemann integral allows for the values of $f$ to be arbitrarily selected within each subinterval $p_i,$ whereas the Darboux integral only considers the infimum and supremum of $f$ on each $p_i.$ In turn, the Riemann integral explicitly requires that the width of the largest interval tend toward zero in a limit-like "epsilon-delta" definition, while the Darboux integral simply considers the infimum and supremum of all possible upper and lower Darboux sums.

Equivalence of Riemann and Darboux Integrals

The Riemann integral and Darboux integral are equivalent formulations of the area under a curve. That is to say, $f$ is Riemann integrable on $[a, b]$ if and only if it is Darboux integrable on $[a, b],$ and the values of the two integrals are always the same. This important property allows the two to be interchanged at will, depending on whichever is more useful for a particular situation. As mentioned above, whereas Darboux integrals make proving theorems easier, Riemann integrals make calculating values easier, a topic we now address.

Left, Right, and Midpoint Riemann Sums

Once the Riemann-integrability of a function over an interval has been established, the calculation of the actual value of the integral is easier with the Riemann integral than with the Darboux integral. The main idea is to tie tags and subinterval widths to the number of samples. In particular, if $N$ subintervals are all $\frac{1}{N}$ the length of the primary interval and the tags are all located at identical spots within each subinterval, then the sum becomes a function of $N,$ the limit of which equals the definite integral.

Let $f$ be integrable on $[a, b],$ and let $P$ be a finite subdivision of $[a, b]$ consisting of $N$ subintervals each of length $\frac{b-a}{N}.$ If $T$ is a set of tags such that $t_i = \inf(p_i)$ for each $p_i \in P,$ then the left Riemann sum is given by

$$\sum\limits_{i=1}^N f(t_i) \mu(p_i) = \sum\limits_{i=1}^N f(\inf(p_i)) \dfrac{b-a}{N}.$$

The term "left" is derived from the fact that each tag is the infimum of its respective subinterval, which puts it on the lefthand side of the subinterval on the x-axis. Likewise, if instead $t_i = \sup(p_i)$ for each $p_i \in P,$ then the right Riemann sum is given by

$$\sum\limits_{i=1}^N f(t_i) \mu(p_i) = \sum\limits_{i=1}^N f(\sup(p_i)) \dfrac{b-a}{N},$$

the term "right" being derived in the same way. Finally, if $t_i = \frac{\inf(p_i) + \sup(p_i)}{2},$ then the midpoint Riemann sum is given by

$$\sum\limits_{i=1}^N f(t_i) \mu(p_i) = \sum\limits_{i=1}^N f\left(\frac{\inf(p_i) + \sup(p_i)}{2}\right) \dfrac{b-a}{N}.$$

These three particular forms of Riemann sums are valuable for their geometric intuitiveness, and they feature prominently in every modern calculus textbook. The coupling of the width of all subintervals to the number of intervals allows the definite integral to be calculated by taking the limit as $N$ tends towards infinity. In other words,

$$\int\limits_a^b f(x) , dx = \lim\limits_{N \rightarrow \infty} \sum\limits_{i=1}^N f(t_i) \dfrac{b-a}{N}.$$

Riemann or Darboux?

The Riemann integral is more widely known than the Darboux integral for two reasons. The first is that Riemann developed his integral decades before Darboux developed his. The other is that Riemann sums are more amenable to different kinds of concrete calculations that appear in calculus than are Darboux sums. Left, right, and midpoint Riemann sums are easy to compute, while Darboux sums require analyzing the integrand for infima and suprema. 

The mathematical equivalence of the two integrals and the relative fame of Riemann have lead to the Darboux name often being elided or often replaced with the Riemann name. It is not uncommon to see Darboux integrals referred to as Riemann integrals. Likewise, the notation $\mathscr{R}([a, b])$ for the set of integrable functions on the interval $[a, b]$ comes from the Riemann name, and there is no standard notation for the set of Darboux integrable functions.

Problems

  1. Show that Riemann integrability implies Darboux integrability.

    Let $f$ be Riemann integrable on $[a, b].$ Then there is some value $S$ such that for every $\frac{\varepsilon}{4} > 0$ there exists a $\delta > 0$ such that for all tagged finite interval subdivisions $(P, T)$ where $\max\limits_{p_i \in P}(\mu(p_i)) < \delta,$ it is the case that $\left| \sum\limits_{i=1}^{n} f(t_i) \mu(p_i) - S \right| < \frac{\varepsilon}{4}.$

    Pick $N \in \mathbb{N}$ such that $\frac{b-a}{N} < \delta,$ and let $P$ be a finite interval subdivision of $[a, b]$ made up of $N$ subdivisions of equal length such that $\mu(p_i) = \frac{b-a}{N}$ for all $p_i \in P.$

    Pick the tags $T_U = \{t_{U,1}, \ldots, t_{U,N}\}$ such that $f(t_{U,i}) > \sup(f(p_i)) - \frac{\varepsilon}{4(b-a)}.$ Then

    $ U(f, P) = \sum\limits_{i=1}^N \sup(f(p_i))\mu(p_i) \\ U(f, P) = \sum\limits_{i=1}^N \sup(f(p_i))\dfrac{b-a}{N} \\ U(f, P) < \sum\limits_{i=1}^N \left(f(t_{U,i}) + \dfrac{\varepsilon}{4(b-a)}\right)\dfrac{b-a}{N} \\ U(f, P) < \sum\limits_{i=1}^N \left(f(t_{U,i})\dfrac{b-a}{N} + \dfrac{\varepsilon(b-a)}{4(b-a)N}\right) \\ U(f, P) < \sum\limits_{i=1}^N f(t_{U,i})\dfrac{b-a}{N} + \sum\limits_{i=1}^N \dfrac{\varepsilon(b-a)}{4(b-a)N} \\ U(f, P) < \left(S + \dfrac{\varepsilon}{4}\right) + \dfrac{\varepsilon}{4} \\ U(f, P) < S + \dfrac{\varepsilon}{2} $

    Likewise, pick the tags $T_L = \{ t_{L,1}, \ldots t_{L,N} \}$ such that $f(t_{L, i}) < \inf(f(p_i)) + \frac{\varepsilon}{4(b-a)}.$ Then

    $ L(f, P) = \sum\limits_{i=1}^N \inf(f(p_i))\mu(p_i) \\ L(f, P) = \sum\limits_{i=1}^N \inf(f(p_i))\dfrac{b-a}{N} \\ L(f, P) > \sum\limits_{i=1}^N \left(f(t_{L,i}) - \dfrac{\varepsilon}{4(b-a)}\right)\dfrac{b-a}{N} \\ L(f, P) > \sum\limits_{i=1}^N \left(f(t_{L,i})\dfrac{b-a}{N} - \dfrac{\varepsilon(b-a)}{4(b-a)N}\right) \\ L(f, P) > \sum\limits_{i=1}^N f(t_{L,i})\dfrac{b-a}{N} - \sum\limits_{i=1}^N \dfrac{\varepsilon(b-a)}{4(b-a)N} \\ L(f, P) > \left(S - \dfrac{\varepsilon}{4}\right) - \dfrac{\varepsilon}{4} \\ L(f, P) > S - \dfrac{\varepsilon}{2} $

    Therefore $U(f, P) - L(f, P) < \varepsilon,$ and so $f$ is Darboux integrable.

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  2. Consider two finite interval subdivisions $P_1$ and $P_2$ of $[a, b],$ as well as their common refinement $P = P_1 \# P_2.$ Show that there are at least $|P_2| - |P_1| + 1$ subdivisions in $P_2$ that are also in $P.$

    If $p_i \in P_2$ but $p_i \notin P,$ then there is some $p_j \in P_1$ whose endpoint lies in the interior of $p_i.$ Since there are $|P_1| - 1$ endpoints in $(a, b),$ there are at most $|P_1| - 1$ such subintervals $p_i$ that are in $P_2$ but not in $P.$ Therefore there are at least $|P_2| - |P_1| + 1$ subintervals in $P_2$ that are also in $P.$

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  3. Show that Darboux integrability implies Riemann integrability.

    Overview: To prove that Darboux integrability implies Riemann integrability, we need to consider two separate finite interval subdivisions. The first comes from the Darboux integrability condition, where for every $\varepsilon > 0$ there exists a subdivision $P_1$ such that $U(f, P_1) - L(f, P_1) < \varepsilon.$ From this we must somehow conceive of a $\delta > 0$ such that for every tagged finite interval subdivision $(P, T)$ where $\max\limits_{p_i \in P}(\mu(p_i)) < \delta$ we have $\sum\limits_{p_i \in P_2} f(t_i)\mu(p_i) - S < \varepsilon,$ where $S$ is the value of the Darboux integral.

    To find this critical $\delta$ value, we proceed in several steps. First, we consider an arbitrary tagged finite interval subdivision $(P_2, T)$ whose maximum subinterval length is $\delta.$ Then we construct the common refinement $P = P_1 \# P_2.$ From here, we partition $P$ into three categories - those subintervals only in $P_2,$ those both in $P_2$ and $P,$ and those only in $P.$ We then determine upper and lower bounds for the sums concerning only subintervals belonging to each of the three partition sets. This ultimately allows us to bound the Riemann sum over $(P_2, T)$ in terms of the Darboux sums, $\varepsilon,$ and $\delta.$ Finally, we can then require that $\delta$ be chosen below a certain point to ensure that the sum differs from the integral by no more than $\varepsilon,$ thus proving Riemann integrability.

    Part 1: Subdividing $[a, b]$: Let $f$ be a real function that is Darboux integrable on the interval $[a, b].$ Then $f$ is bounded, so there exists a non-negative $M \in \mathbb{R}$ such that $|f(x)| \leq M.$ Furthermore, for any $\frac{\varepsilon}{2} > 0$ there exists a finite interval subdivision $P_1$ such that $U(f, P_1) - L(f, P_1) < \frac{\varepsilon}{2}.$

    Pick another finite interval subdivision $P_2$ such that $\max\limits_{p_i \in P_2}(\mu(p_i)) < \delta.$

    Let $P = P_1 \# P_2.$ Since $P$ is finer than $P_1,$ it follows that $U(f, P) - L(f, P) < \frac{\varepsilon}{2}.$

    Partition $P$ into three subsets:

    • $P_o = P_2 - P$ contains the subintervals that belong to $P_2$ but not to $P.$

    • $P_b = P \cap P_2$ contains the subintervals that belong to both $P_2$ and $P.$

    • $P_s = P - P_2$ contains the subintervals that belong to $P$ but not to $P_2.$

    Part 2: Bounding the Riemann sum: Let $T$ be a set of tags for $P.$ We can see that

    $$\sum\limits_{p_i \in P_b} \inf(f(p_i))\mu(p_i) \leq \sum\limits_{p_i \in P_b} f(t_i)\mu(p_i) \leq \sum\limits_{p_i \in P_b} \sup(f(t_i))\mu(p_i).$$

    As shown in the preceding proof, there are at least $|P_1| - 1$ subintervals in $P_2$ that do not belong to $P,$ i.e. $|P_o| \leq |P_1| - 1.$ From the triangle inequality we can see the following:

    $$ \sum\limits_{p_i \in P_o} f(t_i)\mu(p_i) \leq (|P_o|-1)\delta M. $$

    By definition, we see that $P_2 = P_o \cup P_b.$ Thus

    $$\sum\limits_{p_i \in P_2} f(t_i)\mu(p_i) = \sum\limits_{p_i \in P_o} f(t_i)\mu(p_i) + \sum\limits_{p_i \in P_b} f(t_i)\mu(p_i).$$

    Therefore we can combine the upper and lower bounds on the sums regarding $P_o$ and $P_b$ to get bound for the Riemann sum involving $P_2:$

    $$ -(|P_1|-1)\delta M + \sum\limits_{p_i \in P_b} \inf(f(p_i))\mu(p_i) \leq \sum\limits_{p_i \in P_2} f(t_i) \mu(p_i) \leq \sum\limits_{p_i \in P_b} \sup(f(p_i))\mu(p_i) + (|P_1|-1)\delta M. $$

    Part 3: Bounding the sums concerning $P_b$: Next, consider $P_s.$ Because there are at most $|P_1| - 1$ intervals in $P_2$ that are not in $P,$ the sum of the lengths of the subintervals in $P_s$ is at most $(|P_1| - 1)\delta.$ Therefore

    $$ \sum\limits_{p_i \in P_s} \sup(f(p_i))\mu(p_i) \geq \sum\limits_{p_i \in P_s} -M\mu(p_i) \geq -M(|P_1|-1)\delta. $$

    Since $P_b = P - P_s,$ we can split the sum as follows:

    $$ \sum\limits_{p_i \in P_b} \sup(f(p_i))\mu(p_i) = \sum\limits_{p_i \in P} \sup(f(p_i))\mu(p_i) - \sum\limits_{p_i \in P_s} \sup(f(p_i))\mu(p_i) $$

    Note that $U(f, P) = \sum\limits_{p_i \in P} \sup(f(p_i))\mu(p_i).$ Substituting this equality and the bound for $\sum\limits_{p_i \in P_s} \sup(f(p_i))\mu(p_i)$ gives the following:

    $$ \sum\limits_{p_i \in P_b} \sup(f(p_i))\mu(p_i) \leq U(f, P) + M(P_1 - 1)\delta. $$

    We can apply the same reasoning to the lower bounds:

    $$ \sum\limits_{p_i \in P_s} \inf(f(p_i))\mu(p_i) \leq \sum\limits_{p_i \in P_s} M\mu(p_i) \leq M(|P_1|-1)\delta $$

    Again, applying the fact that $P_b = P - P_s,$ we can split the sum as follows:

    $$ \sum\limits_{p_i \in P_b} \inf(f(p_i))\mu(p_i) = \sum\limits_{p_i \in P} \inf(f(p_i))\mu(p_i) - \sum\limits_{p_i \in P_s} \inf(f(p_i))\mu(p_i) $$

    Note that $L(f, P) = \sum\limits_{p_i \in P} \inf(f(p_i)).$ Substituting this equality and the bound for $\sum\limits_{p_i \in P_s} \sup(f(p_i))\mu(p_i)$ gives the following:

    $$ \sum\limits_{p_i \in P_b} \inf(f(p_i))\mu(p_i) \geq L(f, P) - M(|P_1| - 1)\delta $$

    Part 4: Bounding the Riemann sum in terms of $\delta$ and the Darboux sums: Combining these upper and lower bounds into the bounds for the Riemann sum gives us the following:

    $$ -2M(|P_1| - 1)\delta + L(f, P) \leq \sum\limits_{p_i \in P_2} f(t_i)\mu(p_i) \leq U(f, P) + 2M(|P_1| - 1)\delta. $$

    Because $L(f, P) \leq \int\limits_a^b f(x) \, dx \leq U(f, P),$ we can rearrange this inequality to show that

    $$ \left|\sum\limits_{p_i \in P_2} f(t_i)\mu(p_i) - \int\limits_a^b f(x) \, dx\right| \leq U(f, P) - L(f, P) + 4M(|P_1| - 1)\delta. $$

    We can now apply the original epsilon bound:

    $$ \left|\sum\limits_{p_i \in P_2} f(t_i)\mu(p_i) - \int\limits_a^b f(x) \, dx\right| \leq \dfrac{\varepsilon}{2} + 4M(|P_1| - 1)\delta. $$

    Pick $\delta < \dfrac{\varepsilon}{8M(|P_1|-1)}.$ Then

    $$ \left|\sum\limits_{p_i \in P_2} f(t_i)\mu(p_i) - \int\limits_a^b f(x) \, dx\right| < \varepsilon $$

    for any tagged partition $(P_2, T)$ where $\max\limits_{p_i \in P}(\mu(p_i)) < \delta.$ It follows that $f$ is Riemann integrable on $[a, b]$ and that the Riemann integral is equal to the Darboux integral.

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  4. Riemann sums: Show that if $f \in \mathscr{R}([a, b]),$ and the finite interval subdivisions $P_N$ consists of $N$ equally sized intervals, then

    $$\int\limits_a^b f(x) \, dx = \lim\limits_{N \rightarrow \infty} \sum\limits_{i=1}^N f(t_i) \dfrac{b-a}{N}$$

    regardless of the choice of tags.

    This proves the limits of the left, right, and midpoint Riemann sums are each equal to the integral - all in one go.

    Since $f \in \mathscr{R}([a, b]),$ there is a value $S \in \mathbb{R}$ such that for every $\varepsilon > 0$ there exists a $\delta > 0$ such that for every tagged finite interval subdivision $(P, T)$ where $\max\limits_{p_i \in P}(\mu(p_i)) < \delta,$

    $$\left|\sum\limits_{i=1}^N f(t_i) \mu(p_i) - S\right| < \varepsilon.$$

    This value $S$ is denoted

    $$S = \int\limits_a^b f(x) \, dx.$$

    If $P$ consists of $N$ equally spaced subintervals, then

    $$\mu(p_i) = \dfrac{b-a}{N} \text{ for all } p_i \in P.$$

    Therefore we can focus on sums of the form

    $$\left|\sum\limits_{i=1}^N f(t_i) \dfrac{b-a}{N} - S\right| < \varepsilon.$$

    Given any sequence of sets of tags $T_N, T_{N+1}, \ldots,$ the limit $\lim\limits_{N \rightarrow \infty} \sum\limits_{i=1}^N f(t_{N,i}) \dfrac{b-a}{N},$ where $t_{N,i} \in T_N,$ converges if for each $\varepsilon > 0$ there exists an $N_0 \in \mathbb{N}$ such that $\left|\sum\limits_{i=1}^N f(t_{N,i}) \dfrac{b-a}{N} - S\right| < \varepsilon$ for all $N \geq N_0.$

    Pick $N_0 = \left\lceil\dfrac{b - a}{\delta}\right\rceil + 1.$ Then

    $$\dfrac{b-a}{N} \leq \dfrac{b-a}{N_0} < \delta \text{ for all } N \geq N_0.$$

    By the Riemann-integrability of $f,$ it follows that

    $$\left|\sum\limits_{i=1}^N f(t_{N,i}) \dfrac{b-a}{N} - S\right| < \varepsilon \text{ for all } N \geq N_0.$$

    Therefore

    $$\int\limits_a^b f(x) \, dx = \lim\limits_{N \rightarrow \infty} \sum\limits_{i=1}^N f(t_{N,i}) \dfrac{b-a}{N}.$$

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