It is often of interest, both in theory and applications, to be able to pass to the limit under the integral. For instance, a sequence of functions can frequently be constructed that approximate, in a suitable sense, the solution to a problem. Then the integral of the solution function should be the limit of the integrals of the approximations. However, many functions that can be obtained as limits are not Riemann integrable, and so such limit theorems do not hold with the Riemann integral. Therefore, it is of great importance to have a definition of the integral that allows a wider class of functions to be integrated (Rudin 1987).

I'm trying to understand the quote above. It says that there are functions $f(x)=\lim_n f_n(x)$ such that $f(x)$ is not Riemann integrable. However one could calculate its Lebesgue integral. Lebesgue integration is powerful because of the Monotone convergence theorem which states that the limit of a sequence of integrals equals integral of the limit (roughly speaking).

Now I'm interested in the following - if Riemann integral of $f(x)=\lim_n f_n(x)$ does not exist, then what meaning should we attach to the number produced by Lebesgue integral? Area under the curve is defined as Riemann integral. What if a function is not Riemann integrable?

If a function is Riemann integrable, then the values of types both integrals are equal - it's perfectly fine. But why should I treat Lebesgue integral as area? Another thing is that with Lebesgue integral, we only give the lower bound of the 'area' under the curve (supremum of simple functions), so basically the upper bound can be completely different. If the upper bound of the area is greater than lower bound, should we really consider Lebesgue integral as 'area'?

One more thing: If $f$ is a non-negative measurable function on $E$, its Lebesgue integral is defined as $$\int_E f \, d\mu = \sup\left\{\,\int_E s\, d\mu : 0 \le s \le f,\ s\ \text{simple}\,\right\}$$

then is the following true as well?:

$$\sup\left\{\,\int_E s\, d\mu : 0 \le s \le f,\ s\ \text{simple}\,\right\} =\inf\left\{\,\int_E s\, d\mu : f \ge s \ge \infty,\ s\ \text{simple}\,\right\}$$

  • $\begingroup$ Riemann integration might fail since it needs boundedness and almost-everywhere continuity. This is because of the quite concrete choice for the upper and lower sum and the decomposition into Intervals instead of measurable sets. For lebesgue integration even singularities might be fine, which are never okay for Riemann integration. EDIT: There are Riemann-integrable functions that are not Lebesgue integrable. Lebesgue Integral is introduced for nonnegative functions first, it is defined as supremum of the integrals of simple functions which are smaller. $\endgroup$
    – Max
    Oct 11, 2015 at 9:16
  • $\begingroup$ @Max what is an example of a Riemann integrable function which is not Lebesgue integrable? Are you referring to limits of Riemann integrals? The usual definition of Riemann integrable only applies to bounded functions on bounded closed intervals. See eg. math.stackexchange.com/questions/291020/… $\endgroup$
    – Thomas
    Oct 11, 2015 at 9:22
  • $\begingroup$ You are right, what I refer to is actually an improper Riemann Integral. (Boundedness and measurability already imply Lebesgue integrability on a bounded domain by dominated convergence) $\endgroup$
    – Max
    Oct 11, 2015 at 9:27

1 Answer 1


Probably the best way to talk about this is with an example. Let $f_n:[0, 1] \to \mathbb{R}$ where $$ f(x) = \begin{cases} 1 &\mbox{if } x 2^n \in \mathbb{N} \\ 0 & \mbox{otherwise. }\end{cases} \pmod{2} $$

Now, you can see that each $f_n$ is integrable, in both the Riemannian and Lebesgue sense. The limit of the sequence is the function that is $1$ for every value with a terminating binary representation and $0$ elsewhere. This is not Riemann integrable as the lower sums and upper sums don't converge to the same value: the lower sums are always $0$ and the upper sums are always $1$, no matter what partition we choose. However, the limit is Lebesgue integrable, with an integral of $0$.

Now, if I were to ask you "what is the area under the curve?" I'd hope that we could agree that its $0$. To paint it I'd have to make a whole lot of lines, but the lines have $0$ width, so... I feel like the area is $0$. Riemann integration just isn't deep enough to see this fact. So, I'd say they're both area, but one is a better, broader idea of area.

Regarding your final question, can the upper bound differ from the lower? Try to relate the two, you should be pleasantly surprised.

  • $\begingroup$ My point was that in case of Riemann integral we consider both lower and upper bounds for area. If they are equal, we consider this value the area under the graph of $f$. Now in the case of Lebesgue integral we only look at the lower bound and assume that this value is the area without checking the upper bound. Was it your point that they are indeed equal in the case of Lebesgue integral? $\endgroup$ Oct 11, 2015 at 9:49
  • $\begingroup$ Indeed, they are equal. Try to prove it. $\endgroup$
    – user24142
    Oct 11, 2015 at 19:27
  • $\begingroup$ 'Exercise 11 Let $f: {\bf R}^d \rightarrow [0,+\infty]$ be measurable, bounded, and vanishing outside of a set of finite measure. Show that the lower and upper Lebesgue integrals of $f$ agree. ' Link So they are not always equal (the function needs to be bounded, +other conditions). In case upper and lower Lebesgue integral aren't equal, what meaning should I attach to the value of Lebesgue integral? If area, then this notion of area is not as strong as it is in cases when upper and lower bounds are equal. $\endgroup$ Oct 11, 2015 at 19:42
  • $\begingroup$ This is what I don't feel comfortable with. The Lebesgue integral is actually defined as lower Lebesgue integral (according to notation used in link above). If the upper Lebesgue integral can differ, then what is it really? $\endgroup$ Oct 11, 2015 at 21:18
  • $\begingroup$ Whilst its certainly easiest to prove agreement in that case, there is agreement in all cases. Just break the domain of $f$ into $A_i = f^{-1}((2^i, 2^{i-1}])$, apply the easy result to get simple upper bounds $s_i$ such that $\int_{A_i} s_i - f < \epsilon_i$ and $\sum \epsilon_i < \epsilon$ for any chosen $\epsilon$. Lebesgue measure and integration is a better, more general sense of area, volume, etc in every way. $\endgroup$
    – user24142
    Oct 11, 2015 at 22:10

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