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# comparison between Lebesgue and Riemann Integration

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

# 1 Proper Integrals

# The Dirichlet Function

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

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

$1_{{\mathbb{Q}}}(x)=\begin{cases}1,&\text{if $x$ rational},\\ 0,&\text{elsewhere}.\end{cases}$ |

This function, known as the Dirichlet function, is not Riemann integrable. 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 the lower Riemann sum is $0$. Clearly, the upper and lower Riemann sums converge to 1 and 0, respectively, in the limit of the size of largest interval in the partition going to zero. Obviously, these limits are not the same. As a bounded function is Riemann integrable if and only if the upper and lower Riemann sums converge to the same number, the Riemann integral of $1_{{\mathbb{Q}}}$ cannot exist.

On the other hand, $1_{{\mathbb{Q}}}$ turns out to be Lebesgue integrable, which we now show. Let us enumerate the rationals in $[0,1]$ as $\{q_{0},q_{1},\ldots\}$. Now cover each $q_{i}$ with an open set $O_{i}$ of size $\epsilon/2^{i}$. Hence, the set $\{q_{0},q_{1},\ldots\}$ is contained in the set $\subset G_{\epsilon}:=\cup_{{i=0}}^{\infty}O_{i}$. Now it is known that every countable union of open sets forms a Lebesgue measurable set. Therefore, $G_{\epsilon}$ is also a Lebesgue measurable set. Consequently, the Lebesgue integral of this set’s characteristic function, i.e.,

$1_{{G_{\epsilon}}}(x)=\begin{cases}1,&\text{if }x\in O,\\ 0,&\text{elsewhere},\end{cases}$ |

exists.

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

$\int 1_{{\mathbb{Q}}}\,dx\leq\int 1_{{G_{\epsilon}}}\,dx\leq\epsilon$ |

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

# A ’Worse’ Kind of Dirichlet Function

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

Interestingly, the function we seek can be obtained in a more or less direct way from the previous example. First cover the rationals by open sets $O_{i}$ of length $y_{i}$, where the sequence of numbers of $\{y_{i}\}_{{i=0,\ldots}}$ 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 measurable set with measure (length) less than or equal to $1/2$. But this implies that the complement $G^{c}_{y}$ of $G_{y}$, which contains only irrational numbers, has length at least $1/2$.

Observe that the characteristic function $1_{{G_{y}}}$ is nowhere continuous on $G^{c}_{y}$, as the rationals lie dense in the reals. Since, also, $G_{y}^{c}$ has measure at least equal to $1/2$ it is impossible to remove these discontinuity points by a mere 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 almost everywhere. Apparently, $1_{{G_{y}}}$ is bounded and discontinuous on a set with measure larger than $0$. Thus, we may conclude that $1_{{G_{y}}}$ is not Riemann integrable.

To prove that $1_{{G_{y}}}$ is Lebesgue integrable follows easily if we approach the subject of integration from a somewhat more abstract point of view. This is the topic of the next section.

# Exchanging Limits and Integrals

Let us first present one further example: the sequence of
characteristic functions of $\cup_{{i=1}}^{n}O_{i}$, where $O_{i},i=1\ldots n$ *and * $n<\infty$, are the open sets appearing in the
definition of the ’worse Dirichlet function’. Clearly, any such
function has a finite number of discontinuities, hence is Riemann
integrable. However, these functions converge point-wise to
$1_{{G_{y}}}$, which is *not* Riemann integrable.
Apparently, sequences of Riemann integrable functions may converge to
non-Riemann integrable functions. Interestingly, the sequence of
integrals of $1_{{\cup_{{i=1}}^{n}O_{i}}}$, i.e. a sequence of
reals, *has* a limit as it is increasing and bounded by $1/2$.
This somewhat disturbing inconsistency is satisfactory resolved
by Lebesgue’s theory of integration.

As a matter of fact, the advantage of Lebesgue integration is perhaps best appreciated by interpreting this example from a more abstract (functional analysis) point of view. Stated a bit differently, we might approach the subject not from the bottom up (looking at individual functions) but from the top down (looking at classes of functions). In more detail, suppose we are allowed to apply an operator $T$ to any function that is an element of some function space. 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 be even nicer if $\lim_{n}Tf_{n}$ is the same as $T\lim_{n}f_{n}=Tf$. (This is, for instance, useful when it is simple to compute $Tf_{n}$ for each $n$, but difficult to compute $Tf$, while the latter might be what really interests us.)

In the present case, i.e, integration, we perceive the integral of a function as a (continuous linear) operator. The class of Lebesgue integrable functions has the desired abstract properties (simple conditions to check whether the exchange of integral and limit is allowed), 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 Lebesgue measurable functions is bounded by a Lebesgue integrable function, the function $f$ obtained as the pointwise limit $f_{n}$ is also Lebesgue integrable, 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 Lebesgue integrable, $1_{{G_{y}}}$ 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 to reverse 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 of the first theorem, and [Jon00] for the second theorem.

###### Theorem 1.1.

Riemann Case. Assume $f$ to be Riemann integrable on $Q$. Assume also that the one-dimensional function $x\mapsto f(x,y)$ is Riemann integrable for almost all $y\in[c,d]$. Then the function $y\mapsto\int_{a}^{b}f(x,y)\,dx$ is Riemann integrable and $\int_{Q}f(x,y)\,dx\,dy=\int_{c}^{d}\left(\int_{a}^{b}f(x,y)\,dx\right)\,dy$. Note: both conditions are satisfied if $f$ is continuous on $Q$.

###### Theorem 1.2.

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

Observe that the second *assumption* in the Riemann case has
turned into a *consequence* in the Lebesgue case. The main
reason behind this difference is precisely that the class of Lebesgue
measurable functions is closed under taking limits (under a bounded
condition), whereas the class of Riemann integrable functions is not.

# 2 Improper Integrals

From the above the reader may conclude that whenever a function is
Riemann integrable, it is Lebesgue integrable. This is true as long we
only include *proper* integrals. If, on the other hand, we also
consider *improper* integrals the statement is no longer valid.
There exist functions whose improper Riemann integral exists, whereas
the Lebesgue integral does not. Concentrating on functions defined on
subsets of $\mathbb{R}^{n}$ the situation is as shown by the following
Venn Diagram:

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

Let us now integrate the function $1/\sqrt{x}$ over $[0,1]$ to show that 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. On the other hand, both the improper Riemann integral and the Lebesgue integral exist, and give the same result.

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

Finally, $RI$ is also not a subset of $L$. Define $f:[0,\infty)\mapsto[-1,1]$ 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 the Lebesgue integral of $f$ over $[0,n)$ are both equal to $\sum_{{k=1}}^{{n}}(-1)^{{k+1}}1/k.$ It is well known that this alternating sum converges, implying the existence of the improper Riemann integral. However, since a function is Lebesgue integrable if and only if its absolute value is also Lebesgue integrable, $f$ is not in $L$.

# References

- Apo69 T.M. Apostol. Calculus, volume 2. John Wiley & Sons, 1969.
- Jon00 F. Jones. Lebesgue Integration on Euclidean Spaces. Jones and Bartlett, revised edition edition, 2000.
- Lan83 S. Lang. Undergraduate Analysis. Springer-Verlag, 1983.

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## Comments

## Improber Lebesque Integrals

I wonder why improper "Lebesque" Integrals are not defined, but when it comes to define improper integrals they are almost always defined as limits of "Riemann" integrals. Functions like Sin(x)/x multiplied by the characteristic function of the set of irrational numbers would have been integrable in the first sense.

## Re: Improber Lebesque Integrals

Here are the reasons I can think of:

(i) improper integrals fail translation invariance:

\int_{-\infty}^\infty f(x) dx \neq \int_{-\infty}^\infty f(x+t) dx

(ii) it doesn't have an obvious generalization to higher dimensions and other measure spaces (in what way do you take the domain of integration out to infinity?)

(iii) Fubini and limit theorems fail

(iv) In the context of Fourier analysis, where need to perform "improper integrals" when the function f is to be transformed is not in L^1, one can always take _any_ sequence

f_n in L^1 \cap L^2 that converges to f in L^2.

It's common to use the symmetric limit \int_{-T}^T f(x) e^{2\pi i \xi x} \, dx,

T \to \infty, but this restriction is not necessary.

## Re: Improber Lebesque Integrals

> Here are the reasons I can think of:

[why improper Lebesgue integrals are not defined]

>

> (i) improper integrals fail translation invariance:

>

> \int_{-\infty}^\infty f(x) dx \neq

> \int_{-\infty}^\infty f(x+t) dx

>

> (ii) it doesn't have an obvious generalization to higher

> dimensions and other measure spaces (in what way do you take

> the domain of integration out to infinity?)

The same is true for Riemann improper integrals. The reason they are called improper is that their value is defined by the limit of properly defined integrals. On the real line, the limit involves taking each of the integration limits progressively closer to a singularity or to infinity, one by one. There also exist other limiting procedures, e.g. Cauchy principal value. In higher dimensions, there is no limiting procedure quite as natural, but some can still be concocted. For example, on R^2 one could take the limit of integrals over successively larger discs centered at the origin. This would make the functions sin(sqrt(x^2+y^2))/(x^2+y^2)) improper integrable.

The same limiting procedures can be applied when Riemann integrals are replaced by Lebesgue ones. Obviously, the standard theorems of Lebesge integration will not apply to the result, but they would apply to each step in the limit. So the function sin(x)/x times the characteristic function of the irrationals is "improper Lebesgue" integrable as the original poster suggested.

## Re: Improber Lebesque Integrals

I just want to clarify my original comments here;

otherwise there is nothing new here.

> The same is true for Riemann improper integrals.

...

> There also exist other limiting procedures, e.g. Cauchy principal value.

...

> The same limiting procedures can be applied when Riemann integrals are replaced by

> Lebesgue ones.

All the more reason why Lebesgue integrals are usually not defined; there's no coherent and *useful* theory (i.e. equipped with Fubini-type and limit theorems)

that can applied to all (or perhaps even most) instances of "improper integrals".

> Obviously, the standard theorems of Lebesge integration will not apply

> to the result, but they would apply to each step in the limit.

So again there is no real loss in not considering theories of improper integrals

(actually, really just "conditionally convergent" integals), since everything we do with

improper integrals have to be justified step-by-step with proper integrals anyway. Improper integrals are defined in the Riemann case only because the original class of Riemann-integrable functions is too small --- arguably, this is not a problem with the Lebesgue integral.