The Riemann integral makes it so that if we have $|f| \leq |g|$ on $[0,1]$, then integrability of $g$ does not necessarily imply the integrability of $f$. For example, let $f = \chi_\mathbb{Q}$, $g = 1$. Then $g$ is integrable but $f$ is not.

The idea here seems to be that (at least on spaces with finite measure), the Lebesgue integral does a better job at dealing with a lack of regularity that does not occur due to "explosions". Almost always (pun not intended), the Lebesgue integral fails to converge due to a blow-up (i.e $1/x$) or a function having tails that are too large ($1/\sqrt{x}$, away from the origin).

What exactly is it about the Lebesgue integral that prevents singular behavior on finite measure spaces? Holder's inequality tells us that on finite measure spaces, essential boundedness is enough to guarantee us the existence of an integral, but this is in no way true for the Riemann integral. It seems that the catastrophic failure of the Riemann/Darboux integral is this idea that both upper and lower sums need to converge as the partition mesh goes to $0$. In the case of the rationals, for any finite partition, the upper and lower sums are always $0,1$ respectively, meaning convergence does not happen. Does the Lebesgue integral avoid this by only considering a supremum (say over an increasing simple function approximation?)

Edit: The more I think about this, the more I realize that the issue rests with measurability. The Riemann integral for the characteristic of the rationals does not converges because there is no coherent way to assign a Jordan content (strictly speaking it is not a measure) to this set. Specifically, the above monotonicty for converge (RHS converging implies LHS converging) occurs only when both $f,g$ are Lebesgue measurable functions. Thus, I feel that the monotonicity would be valid for the Riemann integral if we were told that both $f,g$ are "Jordan measurable" (whatever that might mean).

  • $\begingroup$ Well, yes. If a function isn't measurable, then it's just like with the Riemann integral, and we're stuck. There's nothing really special about this. The biggest advantage of Lebesgue integral is how it deals with limits. $\endgroup$
    – Jakobian
    May 5, 2019 at 23:40
  • $\begingroup$ @Jakobian fair enough, although I feel that has to do with the fact that there are so many more Leb measurable (and thus integrable functions) than there are Riemann integrable ones. For example, dominated convergence holds for the Riemann integral if the pointwise limit is also a Riemann integrable function, i.e the limiting (pun intended) problem of the Riemann integral is that it works for so few functions. $\endgroup$ May 5, 2019 at 23:56

1 Answer 1


Perhaps you already know most of this, but here are some things to consider.

There is only one definition of Riemann integrability that must be very restrictive for it to work. I am not talking about inproper integrals here. On the other hand, an effective notion of Lebesgue integrability can be defined hierarchically as these restrictive conditions are weakened.

Start with sets of finite measure $E \subset \mathbb{R}$ and bounded functions $f:E \to \mathbb{R}$.

(1) Strictly speaking the Riemann integral is defined for functions on a closed and bounded interval $[a,b]$. Also, it is necessary for the function to be bounded to meet the requirement that there exists $I \in \mathbb{R}$ such that for any $\epsilon > 0$ there exists a partition $P_\epsilon$ of $[a,b]$ such that for any partition $P$ that is a refinement of $P_\epsilon$ and any Riemann sum $S(P,f)$,we have $|S(P,f) - I| < \epsilon$. That $f$ must be bounded is not just an arbitrary part of the definition.

It is, of course, possible to extend the definition to open intervals or even general subsets $E$ of finite measure with $\int_E f$ defined as $\int_a^b f(x) \chi_E(x) \, dx$. Nevetheless, the definition of Riemann integrability can only be met when the measure of the boundary $\partial E$ is $0$, and this is related to the notion of Jordan measurability.

Clearly, there are bounded functions defined on sets of finite measure that are not Riemann integrable -- as with the Dirichlet function you mention -- and this is entirely due to "too much" discontinuity.

(2) Again for bounded functions on sets of finite measure, there always exist lower and upper Lebesgue integrals

$$\underline{\int}_E f = \sup_{\phi \leqslant f} \int_E \phi, \quad \overline{\int_E} f = \inf_{\psi \geqslant f} \int_E \psi,$$

where $\phi$ and $\psi$ are simple functions, and we must have

$$\underline{\int}_E f\leqslant \overline{\int_E} f $$

The most basic definition in this restrictive case is that $f$ is "Lebesgue integrable" on E if

$$\underline{\int}_E f = \overline{\int_E} f$$

There are two important theorems for bounded functions on finite measure sets.

Theorem 1: If a function is Riemann integrable then it is Lebesgue integrable.

Theorem 2: A function is Lebesgue integrable if and only if it is measurable.

An important consequence of Theorem 1 is that the class of Lebesgue integrable functions includes the class of Riemann integrable functions.

An important consequence of Theorem 2 is that, similar to the Riemann integral, there exist bounded functions defined on a set of finite measure that are not Lebesgue integrable. To see this take $E$ as a non-measurable set and consider the function $\chi_E$.

You do raise an interesting question of why the Lebesgue integral is less impacted by the extent of discontinuity as long as we have measrability.

Next consider sets of infinite measure $E \subset \mathbb{R}$ and/or unbounded functions $f:E \to \mathbb{R}$.

Here we cannot even speak of Riemann integrals, yet the Lebesgue integral can be extended. First, we extend to nonnegative functions where the Lebesgue integral can be defined using the previous definition as the supremum of $\int_E g$ over all nonnegative, bounded, measurable functions $g$ with compact support in $E$. In this case the integral may take the value $+\infty$, so satisfaction of this definition alone does not mean that $F$ is Lebesgue integrable. For nonnegative $f$ to be Lebesgue integrable we must have $\int_E f < +\infty$.

The reason for this definition of Lebesgue integrability is to make it possible to extend the definition of the integral further to include general functions. In this case, we consider positive and negative parts $f^+$ and $f^-$ (which are themselves nonnegative functions) and define the Lebesgue integral as

$$\tag{*}\int_E f = \int_Ef^+ - \int_E f^-$$

Since $+\infty - +\infty$ cannot be defined in a meaningful way, this explains why Lebesgue integrability of a nonnegative functions stipulates that the Lebesgue integral is finite. Otherwise, (*) is not well defined. In this way, Lebesgue integrability of a general function $f$ implies that we also have

$$\int_E|f| = \int_Ef^+ + \int_E f^- < +\infty$$

Improper Riemann Integrals

In your question, you cite functions like $x \mapsto 1/x$ on $(0,1]$ and $x \mapsto 1/\sqrt{x}$ on $[1, \infty)$ as examples where the Lebesgue integral "fails". Needless to say, these functions are not Riemann integrable , but we can say that we have defined Lebesgue integrals

$$\int_{(0,1]} \frac{1}{x} = +\infty , \quad \int_{[1,\infty)} \frac{1}{\sqrt{x}} = +\infty$$

We just cannot say these functions are Lebesgue integrable as explained above.

Some of the deficiencies of the Riemann integral can be corrected by introducing the improper Riemann integral. We can even find examples where a function is improperly Riemann integrable but not Lebesgue integrable. Perhaps that should be considered as well in assessing the relative merits of Riemann and Lebesgue integration.

  • $\begingroup$ Nice answer! My hypothesis regarding the Lebesgue integral's compatibility with highly irregular but bounded functions is simply due to the abundance of Lebesgue measurable sets: some axiom of choice type construction is needed to construct a non-measurable set, giving us a very rich class of functions that the Riemann integral/ Jordan content cannot handle. Regarding the improper integrals, you are right, but I would consider $\sin(x)/x$ and the like to be the exception and not the rule. $\endgroup$ May 7, 2019 at 14:40
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    $\begingroup$ @rubikscube09: Thank you. I think your observation is correct. $\endgroup$
    – RRL
    May 8, 2019 at 3:26

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