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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?:

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

user4205580
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  • 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. – Max Oct 11 '15 at 09:16
  • @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. http://math.stackexchange.com/questions/291020/does-riemann-integrable-imply-lebesgue-integrable – Thomas Oct 11 '15 at 09:22
  • 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) – Max Oct 11 '15 at 09:27

1 Answers1

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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.

user24142
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  • 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? – user4205580 Oct 11 '15 at 09:49
  • Indeed, they are equal. Try to prove it. – user24142 Oct 11 '15 at 19:27
  • '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. – user4205580 Oct 11 '15 at 19:42
  • 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? – user4205580 Oct 11 '15 at 21:18
  • 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. – user24142 Oct 11 '15 at 22:10
  • Okay, that's great then. But you can agree this fact should be emphasised somewhere (now it's hidden because of using only supremum in the definition). – user4205580 Oct 11 '15 at 22:15
  • I think trying to work out the correct exposition of a thing like Lebesgue integration is really non trivial. You probably make hundreds of little choices, and some of them are about clarity of proof, or ease of exposition or any number of other competing claims on the structure you choose. Some of them are going to be wrong and perhaps misleading. However, as the reader or student, you need to be careful. Twice here you've gone from the definition/theorem not explicitly guaranteeing a fact to assuming that its not true. More care will serve you well. Parsimony of belief is key. – user24142 Oct 11 '15 at 22:27
  • Well, I'm still not sure... See here One nice feature of measurable functions is that the lower and upper Lebesgue integrals can match, if one assumes also some boundedness. Terry Tao is considered as one of the best mathematicians alive, so... – user4205580 Oct 12 '15 at 16:44
  • The great thing about mathematics is that you don't need to appeal to authority. I provided a proof. Its valid for any function that's integrable and positive, where the measure space is $\sigma$-additive. Read the proof for Corollary 1.3.14 to see that even Tao agrees. You might have some trouble for the integral on non-$\sigma$-additive measure spaces, but those are generally pretty pathological and low on structure. – user24142 Oct 13 '15 at 07:45
  • Yesterday I've asked a question about it: http://math.stackexchange.com/questions/1476659/are-upper-and-lower-lebesgue-integrals-equal-for-every-f Maybe you can address the objections people have in the comments. Collorary 1.3.14 - do I understand correctly that measurability of $f$ implies that $f$ is bounded? He assumes that $f$ is measurable and at the beginning of the proof concludes that $f$ is bounded. – user4205580 Oct 13 '15 at 09:39
  • This is a consequence of a difference in definition. I call a function simple if there is a countable partition of the space ${B_i}$ such that $B_i\cap B_j=\emptyset$ and $f$ is constant on each $B_i$: the book assumes finiteness. This is one of those choices I was mentioning. He proves that measurability implies that you can cut the domain of $f$ up into regions of finite measure where $f$ is bounded, which is exactly what I did however many comments ago. – user24142 Oct 13 '15 at 19:24
  • Ok. I just don't see why: From the horizontal truncation property and a limiting argument, we may assume that $f, g$ are bounded.. The assumptions are: Let $f, g: \bf R^d \rightarrow [0,+\infty]$ be measurable. Does it follow from the assumptions? I guess not. – user4205580 Oct 13 '15 at 21:21
  • Its doesn't follow that they're bounded, but using truncation and a limiting argument gets you there. – user24142 Oct 13 '15 at 21:26
  • So every measurable $f, g: {\bf R}^d \rightarrow [0,+\infty]$ is bounded. – user4205580 Oct 13 '15 at 21:33
  • No. The truncation property and a limiting argument means that you may as well assume boundedness. You truncate to get a bounded function, and the limiting argument shows that you're getting most of the integral. You need to read the text more closely. – user24142 Oct 13 '15 at 21:37
  • One more thing - He proves that measurability implies that you can cut the domain of $f$ up into regions of finite measure where $f$ is bounded. It means that even though there doesn't exist a real number greater than all the values of $f$, I can slice $f$'s domain and find such a real number for every of these intervals separately (even if I have countably many intervals)? If yes, that sounds like cheating... – user4205580 Oct 14 '15 at 14:14
  • That's exactly what he does. Whether you consider it cheating or not, its a valid argument. – user24142 Oct 15 '15 at 02:50