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In the preface of Terry Tao's notes on measure theory he states that in the extended real number setting we adopt the convention that $\infty \cdot 0 = 0 \cdot \infty = 0.$

He explains that it's a useful convention which makes it natural to define integration from below (e.g. integrals as supremums of nonnegative simple functions). He seems to be saying, "Let's define it this way, because it makes our lives simpler." Is more justification not needed?

This 'fact' helped me during my midterm today in showing that $\mu(E \times \mathbb{R})=0$ for any measure-zero set $E$, but I felt sleazy using it.

Any thoughts (philosophical or otherwise) on why or how we can do this and still have self respect as mathematicians?

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  • $\begingroup$ Large values of a function on a set with small measure don't contribute much to the integral ($\infty\cdot0=0$); small values on a large set don't contribute much to the integral ($0\cdot\infty=0$). $\endgroup$ – egreg Oct 16 '14 at 8:55
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    $\begingroup$ A convention does not in itself prove a theorem; the fact that $\mu(E\times \mathbb R)=0$ for any measure-zero set $E$ is one of the justifications for using this convention, not a consequence of it. Sounds like your solution on your midterm was wrong. $\endgroup$ – Jonas Meyer Oct 16 '14 at 13:36
  • $\begingroup$ @JonasMeyer: Let $F = E \times \mathbb{R}$ so that $\mu(F) = \int_{\mathbb{R}^2} \chi_F = \int_{\mathbb{R}}\chi_E \cdot \int_{\mathbb{R}}\chi_{\mathbb{R}} = 0 \cdot \infty = 0$. Does that work? $\endgroup$ – David D. Oct 16 '14 at 18:30
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    $\begingroup$ @DavidD.: I'm not sure if you've already read my answer, but my response may be redundant. It is appropriate to use when it has been proven correct in the given situation. E.g., $\mu_1\times\mu_2(E_1\times E_2)=\mu_1(E_1)\times\mu_2(E_2)$ is a theorem if we use this convention, but the proof that it works even in the $0\times \infty$ case is not a result of the convention (but rather another reason for using the convention to condense statements). It can be proven for instance along the lines indicated at the end of my answer. $\endgroup$ – Jonas Meyer Oct 16 '14 at 19:27
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    $\begingroup$ Tao explained it very well in the book: if $x_n\geq0$ increasingly converges to $x$, $y_n\geq0$ increasingly converges to $y$, then we'd like to have $x_n\cdot y_n$ increasingly converge to $x\cdot y$. In the case of $x_n\equiv0$ and $y_n\to\infty$, we see that $x_n\cdot y_n\equiv0$ and hence $\to0$. THis leads to $0\cdot\infty=0$. $\endgroup$ – Troy Woo Jul 21 '15 at 8:02
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You need to distinguish between facts and convention. A convention is useful if it helps to organize a number of facts. In the context of measure theory, there are several facts that have the appearance of "$0\times \infty =0$" that are proven from the definitions, and thus the convention allows for an easier way to state facts.

A couple of examples:

  1. If $\mu(X)=0$ and $f(x)=+\infty$ for all $x\in X$, then the definition of $\int_X f\,d\mu$ leads to a value of $0$. On the other hand, this could be thought of as "$\mu(X)\times \infty=0\times \infty$", integrating a constant function.
  2. If $f(x)=0$ for all $x\in X$ and $\mu(X)=\infty$, then the definition of $\int_X f\,d\mu$ leads to a value of $0$. On the other hand, this could be thought of as "$\mu(X)\times 0=\infty\times 0$", integrating a constant function.

As a consequence of 1. and 2. above, adopting the convention $0\times \infty=0$ (along with the more intuitive $a\times \infty=\infty$ if $a>0$) allows the following result to be stated:

$$\int_{X} C\,d\mu=\mu(X)\times C,$$

for all constants $C\in[0,\infty]$ regardless of the size of $\mu(X)$. The convention is used only in stating and using the result in condensed form, not in proving it.

Another example that you highlight: If $E$ has measure zero, so does $E\times \mathbb R$. This is not proved using the convention. I would think of it as $$\mu(E\times \mathbb R)=\mu\left(\bigcup_n E\times [-n,n]\right)\leq\sum_n\mu(E\times[-n,n])=\sum_n 0=0,$$ where the equation $\mu(E\times[-n,n])=0$ requires justification, but nothing with infinite measure is involved.

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  • $\begingroup$ Is 2x2=4 a convention or fact? $\endgroup$ – Anixx Nov 27 '16 at 19:16
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    $\begingroup$ @Anixx: Fact or definition I would say. It could be true by definition of $4$ making it a trivial "fact" (i.e., just a definition), but if $2$, $4$, and multiplication have definitions a priori, then this fact would have to be proved. $\endgroup$ – Jonas Meyer May 10 '17 at 6:31
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Another good motive: do you find reasonable the next equalities? $$0=\int_X 0\,d\mu=0\,\mu(X).$$

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    $\begingroup$ I like this motivation although, somewhat analogously, one might question whether $\int_{(0,n)} \infty d\mu$ converges to $0$ as n goes to $0$. $\endgroup$ – David D. Oct 16 '14 at 9:11
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    $\begingroup$ @DavidD, obviously, $\lim\int=\infty$, but $\int_{\{0\}}\infty =$ area of a vertical half-line = ? $\endgroup$ – Martín-Blas Pérez Pinilla Oct 16 '14 at 9:50
  • $\begingroup$ It feels ambiguous, though! Considering the dirac delta function, $\int_{0} \infty$ would be $1$... $\endgroup$ – David D. Oct 16 '14 at 19:58
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    $\begingroup$ @DavidD.: The Dirac delta function is not a function. The definition of the integral of the function whose value at $0$ is $+\infty$ and whose value elsewhere is $0$, leads to a value of $0$. The delta "function" integral is $1$, so it is something else. $\endgroup$ – Jonas Meyer Oct 17 '14 at 3:31
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Somehow it feels like the way to be "in the same side" of Continuum Hypotesis: http://en.m.wikipedia.org/wiki/Continuum_hypothesis

The sum of countable many sets is also countable... so $\infty \cdot 0$ should be 0.

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