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As we all know in a first course in measure theory we define a symbol $\infty$ to satisfy $\infty \cdot 0=0$, but there are more two possible choices for a convention as someone has shown to me; one is $\infty\cdot 0 =\infty$ and the other $\infty \cdot 0 =-\infty$.

Can someone please provide references for those measure theories with a different convention than $\infty \cdot 0 = 0$?

Thanks in advance!

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  • $\begingroup$ Arguably $\infty \cdot 0 = \infty$ is consistent with usual measure theory: the (usual Lebesgue) measure of a singleton subset of $\mathbb{R}$ is 0 but the measure of all of $\mathbb{R}$ is infinity. $\endgroup$
    – Sam Hopkins
    Commented May 12, 2020 at 23:51
  • $\begingroup$ @SamHopkins $\mathbb{R}$ is not a countable union of singletons, though. $\endgroup$
    – user76284
    Commented May 12, 2020 at 23:55
  • $\begingroup$ I feel like any measure theory where you get rid of countable additively and insist all infinite sets have infinite measure is gonna be a very trivial theory... $\endgroup$
    – Sam Hopkins
    Commented May 13, 2020 at 0:00
  • $\begingroup$ @SamHopkins either way I'd like to see those theories. Some say all of maths when done correctly is trivial. cheers! $\endgroup$
    – Alan
    Commented May 13, 2020 at 0:03
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    $\begingroup$ $\infty\cdot0 = \infty$ I can imagine, but $\infty\cdot0 = -\infty$? Then $-\infty = \infty\cdot(0\cdot0) = (\infty\cdot0)\cdot0 = (-\infty)\cdot0 = -(\infty\cdot0) = \infty$, if multiplication is still associative (and, if not, then why not, say, $\infty\cdot0 = 1$?). $\endgroup$
    – LSpice
    Commented May 13, 2020 at 2:46

1 Answer 1

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It's not a convention, it's a theorem. Let's say I have a measure space $X$ and a function $f: X \to \overline{\mathbb{R}}$ which is identically zero off of a null set $N$, and constantly $+\infty$ on that null set. You can say that $\int f = 0$ ``by convention'', but you can also say that $\int f$ equals the area under the graph of $f$, i.e., the measure of the set $N\times [0,\infty)$. And that area has to be zero by countable additivity because the measure of $N\times [n, n+1)$ is $0\cdot 1 = 0$ for all $n$. The sum of infinitely many zeros has to be zero because that is what the partial sums converge to.

An instructive special case is the line $\{0\}\times [0,\infty) \subset \mathbb{R}^2$. Its measure is $0\cdot\infty$, right? Now for any $\epsilon$ find a sequence of rectangles which covers the line and whose areas sum to $\epsilon$. So this $0\cdot\infty$ has to be zero.

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    $\begingroup$ Another way to say it is that the integral of the given $f$ is zero no matter how you define $0 \cdot \infty$. Adopting that convention just means that you can write a more compact expression for the integral of $f = c \cdot 1_A$, one which doesn't require $c=\infty$ or $m(A)=\infty$ to be special cases. But measure theory doesn't change in any way if you use a different convention. $\endgroup$
    – Nate Eldredge
    Commented May 13, 2020 at 2:51
  • $\begingroup$ @NateEldredge right, exactly. $\endgroup$
    – Nik Weaver
    Commented May 13, 2020 at 3:02
  • $\begingroup$ @NikWeaver so we cannot have a measure theory with $\infty \cdot 0 = \infty$? I don't see how do you preclude it? $\endgroup$
    – Alan
    Commented May 23, 2020 at 21:46

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