comparison between Lebesgue and Riemann Integration

The Riemann and Lebesgue integral are defined in different ways, withthe latter generally perceived as the more general. The aim ofthis article is to clarify this claim by providing a number of,hopefully simple and convincing, examples and arguments. We restrictthis article to a discussion of proper and improper integrals. Forextensions and even more general definitions of the integral we referto http://www.math.vanderbilt.edu/~schectex/ccc/gauge/:

1 Proper Integrals

The Dirichlet Function

Our first example shows that functions exists that are Lebesgueintegrable but not Riemann integrable.

Consider the characteristic function of the rational numbers in$[0,1]$, i.e.,

This function, known as the Dirichlet function, is not Riemannintegrable. To see this, take an arbitrary partition of the interval$[0,1]$. The supremum of $1_{\mathbb{Q}}$ on any interval(with non-empty interior) is $1$, whereas its infimum is $0$. Hence,the upper Riemann sum of $1_{\mathbb{Q}}$ is $1$ while thelower Riemann sum is $0$. Clearly, the upper and lower Riemann sumsconverge to 1 and 0, respectively, in the limit of the size of largestinterval in the partition going to zero. Obviously, these limits arenot the same. As a bounded function is Riemann integrable if and onlyif the upper and lower Riemann sums converge to the same number, theRiemann integral of $1_{\mathbb{Q}}$ cannot exist.

On the other hand, $1_{\mathbb{Q}}$ turns out to beLebesgue integrable, which we now show. Let us enumerate the rationalsin $[0,1]$ as $\{q_0,q_1,\mathrm{\dots }\}$. Now cover each $q_i$ with anopen set $O_i$ of size $ϵ/2^i$. Hence, the set$\{q_0,q_1,\mathrm{\dots }\}$ is contained in the set $\subset G_{ϵ}:={\cup }_{i=0}^{\mathrm{\infty }}{O}_i$. Now it is known that every countable union ofopen sets forms a Lebesgue measurable set. Therefore, $G_{ϵ}$ isalso a Lebesgue measurable set. Consequently, the Lebesgue integralof this set’s characteristic function, i.e.,

exists.

In fact, the integral of $1_{G_{ϵ}}$ is less than or equal to$ϵ{\sum }_{i=0}^{\mathrm{\infty }}{2}^{-i}=ϵ$ as the total length ofthe union of sets $O_i$ is less than or equal to $ϵ$. (Theremight be overlaps between the $O_i$.) Now taking the limit$ϵ\to 0$ it follows that for all $ϵ$,

$$\int 1_{\mathbb{Q}}𝑑x\le \int 1_{G_{ϵ}}𝑑x\le ϵ$$

Since this holds for all $ϵ>0$, the left hand side must be 0.

A ’Worse’ Kind of Dirichlet Function

The above is, arguably, somewhat simple. The only reason that theDirichlet function is Lebesgue, but not Riemann, integrable, is thatits spikes occur on the rationals, a set of numbers which is, incomparison to the irrational numbers, a very small set. By modifyingthe Dirichlet function on a set of measure zero, that is, byremoving its spikes, it becomes the zero function, which is evidentlyRiemann integrable. This reasoning might lead us to conjecture thatit is possible to turn any Lebesgue integrable function into a Riemannintegrable function by modifying it on a set of measure zero. Thisconjecture, however, is false as the next example shows.

Interestingly, the function we seek can be obtained in a more or lessdirect way from the previous example. First cover the rationals byopen sets $O_i$ of length $y_i$, where the sequence of numbers of${\{y_i\}}_{i=0,\mathrm{\dots }}$ is such that $y_i>0$ for all $i$ but such that${\sum }_i{y}_i=1/2$. Observe that $G_y:={\cup }_i{O}_i$ is a measurableset with measure (length) less than or equal to $1/2$. But thisimplies that the complement $G_y^c$ of $G_y$, which contains onlyirrational numbers, has length at least $1/2$.

Observe that the characteristic function $1_{G_y}$ isnowhere continuous on $G_y^c$, as the rationals lie dense in thereals. Since, also, $G_y^c$ has measure at least equalto $1/2$ it is impossible to remove these discontinuity points by amere modification on a countable number of measure zero sets.

Now there is a theorem by Lebesgue stating that a bounded function $f$is Riemann integrable if and only if $f$ is continuous almosteverywhere. Apparently, $1_{G_y}$ is bounded anddiscontinuous on a set with measure larger than $0$. Thus, we mayconclude that $1_{G_y}$ is not Riemann integrable.

To prove that $1_{G_y}$ is Lebesgue integrable followseasily if we approach the subject of integration from a somewhat moreabstract point of view. This is the topic of the next section.

Exchanging Limits and Integrals

Let us first present one further example: the sequence ofcharacteristic functions of ${\cup }_{i=1}^n{O}_i$, where $O_i,i=1\mathrm{\dots }n$ and , are the open sets appearing in thedefinition of the ’worse Dirichlet function’. Clearly, any suchfunction has a finite number of discontinuities, hence is Riemannintegrable. However, these functions converge point-wise to$1_{G_y}$, which is not Riemann integrable.Apparently, sequences of Riemann integrable functions may converge tonon-Riemann integrable functions. Interestingly, the sequence ofintegrals of $1_{{\cup }_{i=1}^n{O}_i}$, i.e. a sequence ofreals, has a limit as it is increasing and bounded by $1/2$.This somewhat disturbing inconsistency is satisfactory resolvedby Lebesgue’s theory of integration.

As a matter of fact, the advantage of Lebesgue integration is perhapsbest appreciated by interpreting this example from a more abstract(functional analysis) point of view. Stated a bit differently, wemight approach the subject not from the bottom up (looking atindividual functions) but from the top down (looking at classes offunctions). In more detail, suppose we are allowed to apply anoperator $T$ to any function that is an element of some functionspace. It would be nice if this space is closed under taking(point-wise) limits. In other words, besides being allowed to apply$T$ to some sequence of functions $f_n$, we are also allowed to apply$T$ to the function $f$ obtained as the limit of $f_n$. It would beeven nicer if ${lim}_{n}T{f}_n$ is the same as $T{lim}_n{f}_n=Tf$.(This is, for instance, useful when it is simple to compute $T{f}_n$for each $n$, but difficult to compute $Tf$, while the latter might bewhat really interests us.)

In the present case, i.e, integration, we perceive the integral of afunction as a (continuous linear) operator. The class of Lebesgueintegrable functions has the desired abstract properties (simpleconditions to check whether the exchange of integral and limit isallowed), whereas the class of Riemann integrable functions does not.

Applying this to the above example, viz. the integration of$1_{G_y}$, we use Lebesgue Dominated Convergence Theorem,which states that when a sequence $\left\{f_n\right\}$ of Lebesguemeasurable functions is bounded by a Lebesgue integrable function, thefunction $f$ obtained as the pointwise limit $f_n$ is also Lebesgueintegrable, and $\int {lim}_n{f}_n=\int f={lim}_n\int f_n$. Since,for all $n$, $1_{{\cup }_i^n{O}_i}$ is bounded and Lebesgueintegrable, $1_{G_y}$ is also Lebesgue integrable,and reversing the (pointwise) limit and the integral is allowed.

1.1 Fubini’s Theorem

Admittedly the function $1_{G_y}$ is rather artificial.A really powerful example of the consequences of being allowed toreverse integral and limit is provided by (the proof of)Fubini’s theorem applied to the rectangle $Q=[a,b]\times [c,d]$.Compare the following two theorems. See, for instance,[Apo69] or [Lan83] for proofs ofthe first theorem, and [Jon00] forthe second theorem.

Theorem 1.1.

Riemann Case. Assume $f$ to be Riemann integrable on $Q$. Assumealso that the one-dimensional function $x\mathrm{\mapsto }f\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{)}$ is Riemannintegrable for almost all $y\mathrm{\in }\mathrm{[}c\mathrm{,}d\mathrm{]}$. Then the function $y\mathrm{\mapsto }{\mathrm{\int }}_a^{b}f\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{)}\mathit{}𝑑x$ is Riemann integrable and ${\mathrm{\int }}_{Q}f\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{)}\mathit{}𝑑x\mathit{}𝑑y\mathrm{=}{\mathrm{\int }}_c^d\mathrm{\left(}{\mathrm{\int }}_a^{b}f\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{)}\mathit{}𝑑x\mathrm{\right)}\mathit{}𝑑y$. Note: bothconditions are satisfied if $f$ is continuous on $Q$.

Theorem 1.2.

Lebesgue Case. Assume that $f$ is Lebesgue integrable on $Q$. Thenthe function $x\mathrm{\mapsto }f\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{)}$ is Lebesgue integrable for almost all$y$ on $\mathrm{[}c\mathrm{,}d\mathrm{]}$. As a consequence, the function $y\mathrm{\mapsto }{\mathrm{\int }}_a^{b}f\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{)}\mathit{}𝑑x$ is Lebesgue integrable and ${\mathrm{\int }}_{Q}f\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{)}\mathit{}𝑑x\mathit{}𝑑y\mathrm{=}{\mathrm{\int }}_c^d\mathrm{\left(}{\mathrm{\int }}_a^{b}f\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{)}\mathit{}𝑑x\mathrm{\right)}\mathit{}𝑑y$.

Observe that the second assumption in the Riemann case hasturned into a consequence in the Lebesgue case. The mainreason behind this difference is precisely that the class of Lebesguemeasurable functions is closed under taking limits (under a boundedcondition), whereas the class of Riemann integrable functions is not.

2 Improper Integrals

From the above the reader may conclude that whenever a function isRiemann integrable, it is Lebesgue integrable. This is true as long weonly include proper integrals. If, on the other hand, we alsoconsider improper integrals the statement is no longer valid.There exist functions whose improper Riemann integral exists, whereasthe Lebesgue integral does not. Concentrating on functions defined onsubsets of ${ℝ}^n$ the situation is as shown by the followingVenn Diagram:

Figure 1: $R$, $RI$, and $L$, are the classes of Riemann, improperRiemann, and Lebesgue integrable functions, respectively.

In the previous we already discussed the inclusion $R\subset L$.

Let us now integrate the function $1/\sqrt{x}$ over $[0,1]$ to showthat some functions exist $RI\cup L$ but are not in $R$. As $1/\sqrt{x}$ is not bounded on $[0,1]$ every upper Riemann sum is infinite. Onthe other hand, both the improper Riemann integral and the Lebesgueintegral exist, and give the same result.

Secondly, $L$ is not contained in $RI$ as follows from the fact thatthe Dirichlet function is not $RI$.

Finally, $RI$ is also not a subset of $L$. Define $f:[0,\mathrm{\infty })\mapsto [-1,1]$ as $1$ on $[0,1)$, $-1/2$ on $[1,2)$, $1/3$on $[2,3)$, $-1/4$ on $[3,4)$, etc. The Riemann integral and theLebesgue integral of $f$ over $[0,n)$ are both equal to${\sum }_{k=1}^n{(-1)}^{k+1}1/k.$ It is well known that thisalternating sum converges, implying the existence of the improperRiemann integral. However, since a function is Lebesgue integrable ifand only if its absolute value is also Lebesgue integrable, $f$ is notin $L$.