Bernhard Riemann proved that if $(a_n)$ is a sequence in $\mathbb{R}$, then the sum of the infinite series $\Sigma_{n=1}^\infty a_n$ stays the same regardless of how you rearrange the terms if and only if the series $\Sigma_{n=1}^\infty |a_n|$ is convergent. I’d like to see if something analogous for integrals is true.

My question is, for what functions $f:[a,b]\rightarrow\mathbb{R}$ is it true that $\int_a^b f(g(x)) dx = \int_a^b f(x) dx$ for all bijective functions $g:[a,b]\rightarrow[a,b]$?

Or is that too stringent a condition to be interesting, and do we need to impose some conditions on $g$ to get a more meaningful result?

  • $\begingroup$ Bijections can still ‘stretch’ space in some ways that make this vacuously false. I think the thing you might be looking for is unitary operators! $\endgroup$ May 27, 2019 at 21:52
  • $\begingroup$ Could you explain the relevance of unitary operators? $\endgroup$ May 27, 2019 at 21:53
  • $\begingroup$ As I mentioned, bijections can “stretch” space so you can always construct a $g$ such that the condition on $f$ you mention does not hold, but unitary operators do not stretch space — they’re basically bijective isometries. Being an isometry is sort of the best characterisation we have of something that doesn’t stretch space; the volume of an area after the application of a unitary operator is the same as the volume before. $\endgroup$ May 27, 2019 at 21:55
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    $\begingroup$ The way that the question is currently stated, I believe only (almost everywhere) constant functions $f$ can ever satisfy the condition, since if they took a different value on some interval to another then $g$ could simply stretch the amount of the domain spent in one interval and shrink the amount in the other and effect a change in the total value of the integral. $\endgroup$ May 27, 2019 at 22:00

1 Answer 1


An isometry on a Banach space is a linear map $T$ such that $||Tf||=||f||$ for all $f$ in the space, where $||\cdot||$ is just whichever norm the Banach space is equipped with. (In particular, a unitary operator is a bijective isometry.)

In the sequence space $\ell^1$ of sequences whose sums converge, the norm $||\cdot ||_{\ell^1}$ of a sequence is just the sum of its terms. One basis we can apply to the Banach space is the set of sequences indexed by $n\in \mathbb{N}$ that are zero everywhere except for a $1$ in the $n$th place (similarly to how we deal with finite-dimensional vector spaces, although this is a Schauder basis and is a bit different to the finite-dimensional Hamel bases you might be more familiar with). Then any re-arrangement of the terms of the sequence can be represented by the linear map which rearranges the basis vectors, and can be shown to be an isometry from this fact. By the definition of it being an isometry, then, the value of the sum of the sequence after its terms are rearranged is the same as the value of the sum of the sequence before they are re-arranged. There are more isometries than just these ones, so the re-arrangement theorem is more general than it might at first seem!

Similarly, in the space of integrable functions $L^1$ (and in particular, $L^1[a,b]$ of functions integrable on that interval), the norm $||\cdot||_{L^1}$ is just the integral of (the absolute value of) the function (on the interval). So an isometry here is definitionally any linear map such that application of it to a function does not change the value of the function’s integral, which would seem to be pretty much exactly what we’re after here. Any integrable function will preserve its value after the application of an isometry (on $L^1$), which seems to be the exact parallel of the sequence rearrangement theorem you inquired about.

Thanks for this question, I think you just helped me understand this content at a deeper level as well.

Interestingly enough, because of the infinite-dimensional nature of these particular Banach spaces the “rearrangement” need not even be bijective/a unitary operator, it need only be an isometry! In $\ell^1$ one such isometry is just the map that takes the value of each basis vector to the “next” one and leaves a $0$ in the first position (basically just prepending a $0$ to the start of every sequence); the sum of this sequence is exactly the same except with a $0+\dots$ appended to the start.

So, any function with finite integral can be “rearranged” and maintain its value if and only if the “rearrangement” is an isometry.

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    $\begingroup$ I think there a couple of problems in this answer: $\ell^1$ and $L^1$ are Banach spaces but not Hilbert spaces, as their norm is not given by an inner product. Additionally, the standard basis elements form a Schauder basis but not a Hamel basis, which I believe to be an important distinction to draw. $\endgroup$
    – eepperly16
    May 28, 2019 at 5:06
  • $\begingroup$ @eepperly16 Ah yes, my bad -- I am rusty on my analysis. I'll fix in edit, thanks! $\endgroup$ May 28, 2019 at 5:19

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