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Using standard Lebesgue integration, we can write: \begin{equation} \int_{(a,b)}f'(x) d\lambda = f(b) - f(a) \end{equation}

There's no orientation on the left hand side of the equation, yet on the right we take $f(b) - f(a)$ as opposed to $f(a) - f(b)$.

Due to what exactly is that the case? In what piece is the "orientation" $a \rightarrow b$ encoded?

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My thoughts: I'd guess it's due to the fact that the derivative is in some sense defined in a slightly arbitrary way - both $\lim \frac{f(x+h)-f(x)}{h}$ and $\lim \frac{f(x)-f(x+h)}{h}$ make sense. But at the same time, it doesn't seem that arbitrary - it's natural to assume that numerator/enumerator order should be preserved, i.e. $\frac{f(x+h) - f(x)}{(x+h) - (x)}$ is the natural choice. So I think I might be missing something.

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    $\begingroup$ "There's no orientation on the left hand side of the equation" -- how so? The interval $(2,3) = \{x \in \mathbb R: 2<x<3\}$ is rather different from $(3,2) = \{x \in \mathbb R: 3<x<2\} = \emptyset$. $\endgroup$ Jul 24, 2023 at 16:14
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    $\begingroup$ @TorstenSchoeneberg Lebesgue's integral doesn't care about orientation. And what you wrote are two different sets, nothing related to having two different orientations. $\endgroup$
    – NDB
    Jul 24, 2023 at 16:18
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    $\begingroup$ @TorstenSchoeneberg I don't know where you are reading them saying that they can swap $a$ and $b$. They are asking why the FTC presents $a$ and $b$ in that order. It has nothing to do with what you are saying. $\endgroup$
    – NDB
    Jul 24, 2023 at 18:32
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    $\begingroup$ By the way, the easiest case here is: Why is $\int_{(a,b)} 1 d\lambda(x) = b-a$ and not $=a-b$. And as far as I am concerned, that is the definition of the Lebesgue measure. As per @LeeMosher's answer. $\endgroup$ Jul 25, 2023 at 3:18
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    $\begingroup$ Indeed. I object to "$b-a$ vs $a-b$" having nothing to do with the question, it is the whole question. I object to "$\lambda((a,b))=|b-a|$" since this is wrong unless you propose new notation $\endgroup$
    – FShrike
    Jul 25, 2023 at 14:27

4 Answers 4

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If you interpret the expression $\int_{(a,b)}f'(x)dx$ as a Lebesgue or Riemann integral, than the integral itself does not depend on orientation, as Deane's answer explains. The derivative $f'$, however, does depend on orientation. There are a few ways of seeing this.

In the difference quotient $\frac{f(x+h)-f(x)}{(x+h)-x}$, the two subtraction operations encode a sense of orientation, but they do not "cancel out". The numerator is a subtraction in the codomain of $f:\mathbb{R}\to\mathbb{R}$, while the denominator is a subtraction in the domain, and we are only concerned with the orientation of the domain here.

Alternately, we can see precisely where orientation is invoked by instead considering a function $f:M\to\mathbb{R}$ where $M$ is a connected Riemannian $1$-manifold (which happens to be $\mathbb{R}$). The Borel measure $\mu$ compatible with the metric (such as the Lebesgue measure on $\mathbb{R}$) can be defined without choosing an orientation, but this only allows us to integrate functions $M\to\mathbb{R}$. The derivative $df:M\to T^*M$ is a $1$-form on $M$ and there are two possible choices of trivialization $T^*M\to M\times\mathbb{R}$ compatible with the metric. There is a standard choice for $M=\mathbb{R}$, but without such a choice (i.e. an orientation) we cannot view $f'$ as an integrable function.

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    $\begingroup$ Yet another alternative is to use the metric to induce on $M$ the canonical $1$-pseudoform $\omega=\sqrt{g}|dx|$, which allows one to integrate a function $f$ by integrating the $1$-pseudoform $f\omega$, which is equivalent to integrating against the measure $\mu$. But again you cannot integrate $1$-forms, only $1$-pseudoforms and functions. $\endgroup$ Jul 26, 2023 at 6:48
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Although I suggest that you don't overthink this too much, I do like the manner in which you formulated your question, asking how orientation is encoded in the Lebesgue integral.

Intuitively, the difference between the positive and negative orientations on an interval $[a,b]$ can be detected using the displacement from the initial endpoint to the terminal endpoint, which by definition is equal to the terminal endpoint minus the initial endpoint. Under the positive orientation, the initial endpoint is $a$, the terminal endpoint is $b$, and the displacement is the positive quantity $b-a$; whereas under the negative orientation, the initial endpoint is $b$, the terminal endoint is $a$, and the displacement is the negative quantity $a-b$.

One place (the place, perhaps) that the orientation of intervals is encoded is in the formula for lengths of intervals in the real line: for any $a<b$ in $\mathbb R$, the length of the interval $(a,b)$ is, by definition, $b-a$, which just so happens to be equal to the displacement of the positive orientation on the interval $[a,b]$. And, of course, lengths of intervals comes into the definition of the Lebesgue integral when one defines the Lebesgue outer measure, which is then used to define Lebesgue measurable sets and the Lebesgue measure itself, which is then used to define the Lebesgue integral of simple functions, which is then used to define the Lebesgue integral in general.

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  • $\begingroup$ This is wrong. There is no orientation involved in saying that the Lebesgue measure of $[a,b]$ is $|b-a|$. $\endgroup$
    – NDB
    Jul 24, 2023 at 18:24
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    $\begingroup$ I didn't say that the Lebesgue measure of $[a,b]$ is $| b-a |$. I said that for all $a<b \in \mathbb R$, the length of $(a,b)$ is $b-a$. $\endgroup$
    – Lee Mosher
    Jul 24, 2023 at 19:32
  • $\begingroup$ You can then use that, together with the definition of Lebesgue measure, to deduce that the Lebesgue measure of $[a,b]$ is $|b-a|$, if you want to. $\endgroup$
    – Lee Mosher
    Jul 24, 2023 at 19:33
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    $\begingroup$ @NDB There's no orientation needed to describe the measure of some set, but assuming the measure should be non-negative (which is a natural, intuitive demand, as we want to minimize foolery we would get due to $ -1 \cdot -1 = 1$), it does end up being the case that $\int_I 1 d\lambda = \lambda(I) = sup(I) - inf(I) = b - a$ (which can be understood as a consequence of how the real numbers are constructed/defined, particularly the ordering and subtraction operation). $\endgroup$
    – Jake1234
    Jul 25, 2023 at 19:57
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    $\begingroup$ @NDB This special FTC has no concept of a derivative anywhere, and so it's an even more basic observation of what I'm asking in my OP - it's a scenario where on the left hand side, there is no idea of an orientation, and yet for some reason, the right hand side does have a particular order. The orientation is encoded somewhere. Due to this, this seems like a more intrinsic explanation of where the orientation is "encoded", rather than the sign of the derivative - which seems like it should be defined in such a way, so that it agrees with this definition. Though of course that's subjective. $\endgroup$
    – Jake1234
    Jul 25, 2023 at 19:58
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The Riemann or Lebesgue integral of a function $h$ on an interval $[a,b]$ is defined without any orientation on the interval. The concept of orientation is never mentioned when first defining integration.

I rarely see orientation mentioned in an introduction to Lebesgue integration, especially if the fundamental theorem of calculus is not discussed. A measure is always assumed to be positive (or at least nonnegative) and the integral of a measure is always positive. There is no orientation assigned to the support of the measure. Note that when working with abstract measure spaces, there isn't even a well-defined concept of orientation (and therefore no straightforward analogue of the fundamental theorem of calculus).

Orientation appears implicitly when defining the notation of a Riemann integral (or a Lebesgue integral over a connected interval). The integral of $h$ over $[a,b]$, where $a\le b$, is denoted $$\int_a^b h(x)\,dx. $$ But a new convention is also introduced, namely that $$\int_b^a h(x)\,dx = -\int_a^b h(x)\,dx. $$ This convention was introduced specifically to make the equation $$\int_a^b f’(x)\,dx = f(b) - f(a)$$ hold even if $a>b$.

In hindsight, we now recognize that the fundamental theorem of calculus is about the integral of a derivative over an oriented interval. An orientation of an interval is the same as putting an arrow on it, which is equivalent to labeling one point the start point and the other the end point. The fundamental theorem of calculus states simply that the integral of $f$ over an oriented interval is the value of $f$ at the end point minus the value of $f$ at the start point.

Since orientation is needed for a single variable Stokes’ theorem, it is not surprising that it also plays a crucial role in the multivariable Stokes’ theorem (which really should be called the fundamental theorem of calculus).

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In what piece is the "orientation" a→b encoded?

The ordered pair of points $(a,b)$ gave this curve a direction and induced the positive orientation for the point $b$ and negative -- for the $a$, which is used in the RHS (see e.g. https://math.stackexchange.com/a/4697720/73934) as the whole equation is the Stokes theorem of differential geometry.

EDIT following OP's restriction of the integration path to an interval on the real line. That's fine but be careful if your teacher like a commenter @TedShifrin below defines interval $(a,b)$ to be the set $\{x\in \Bbb R: a<x<b\}$. By that definition $(2, -3)$ is a null set. Yet, integral over $(2, -3)$ is not identical 0. They can try to amend their definition for $b<a$ etc. But the right way to specify the direction or "orientation" in this case is to simply decide which point is beginning and which is the end. So, an ordered pair of points (e.g. on a real line)! The OP is right too to try and bring it out of the school calculus into differential geometry context, while the measure theory and Lebesgue vs. Riemann integration are a red herring in this question.

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  • $\begingroup$ @Ted, look under the integral in OP, it's literally $(a,b)$ -- an ordered pair (of the end points of a curve one assumes). Check en.wikipedia.org/wiki/Ordered_pair . And the OP never said "interval", by the way, did you notice? "Sounds like a ChatGPT answer" -- is that a joke? Didn't expect it from a fellow mathematician. What if people start writing it under your answers, would you like that? $\endgroup$
    – rych
    Jul 25, 2023 at 16:55
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    $\begingroup$ No, the interval $(a,b)$ denotes the set $\{x\in \Bbb R: a<x<b\}$. Similarly, $[a,b]$ denotes the closed interval $\{x\in \Bbb R: a\le x\le b\}$. The OP doesn't need to say the word, because it's understood by (almost) all readers. $\endgroup$ Jul 25, 2023 at 17:31
  • $\begingroup$ You use interval but neither OP nor my answer had that word. The differential-forms tag suggested they were looking for other contexts and interpretations as well. Their $a \to b$ notation also could mean a curve going from a point $a$ to a point $b$. Surely their diff. form only depends on the $x$ coordinate, but in the context of DG, points a and b didn't have to be on the real line. Look, the OP is a bit ambiguous, starting from the opening Lebesgue integration. Others answered that well. Erase the differential-forms tag and I'll delete my answer. Are you taking back your ChatGPT comment? $\endgroup$
    – rych
    Jul 25, 2023 at 18:16
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    $\begingroup$ @Jake1234 I strongly disapprove of the edit that was made to your question. If I were you, I would roll it back so that the question is notated the way you wrote it. A surprisingly thought-provoking question, by the way. $\endgroup$
    – David K
    Jul 26, 2023 at 1:15
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    $\begingroup$ @rych I think you have some confusion over what the OP is asking. The point of Lebesgue integral is that we integrate over a set. In this case the set is the interval $(a,b)=\{x|a<x<b\}$. But sets don't have orientation, and thus we have OP's question. $\endgroup$ Jul 26, 2023 at 13:58

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