The problem below has been asked recently already but, as a naive user, I got burned (well singed perhaps) because I asked the question in the wrong place. So if this looks like a redundant question ... excuse me (as Steve Martin would say)!

Problem: Show that the product of two metric outer measures (also known as Borel measures on a metric space) is again a metric outer measure.

Does anyone know if the problem is valid?

The hints given already on this forum work only for the separable case, but I don't have a solution to the general case.

I found the problem originally in the text by Munroe (Measure and Integration 1953). It seemed that the obvious approach would work so I added it without further thought to our book (Real Analysis, Bruckner/Bruckner/Thomson 1996). We received some valuable feedback from many sources, mostly from R. B. Burckel who pointed out among many other things that this problem was not straightforward and may be wrong. For the second 2008 edition I left the problem (since it is interesting) but included a footnote to indicate that there was some doubt.

Anybody have a definitive answer?

The problem appears as Exercise 6:1.5. in our 2nd edition. You can get a free PDF from the website classicalrealanalysis.com.

Hope to hear from someone (not, I hope, from someone assigned this problem because it appeared in our first edition--if so blame Munroe -- please).


  • $\begingroup$ Brian, Let me start saying that I took a look at your book (Real Analysis, Bruckner/Bruckner/Thomson 1996). It is a very nice book. About Exercise 6.1.5, as a second thought, I think this exercise seems "natural" only in the separable case. $\endgroup$ – Ramiro Oct 17 '15 at 17:26
  • $\begingroup$ I checked the defns you use and posted a consistent counterexample. $\endgroup$ – hot_queen Oct 30 '15 at 22:05

Let $\kappa$ be real valued measurable with a witnessing probability measure $\mu$. Put discrete metric on $\kappa$. So $(\kappa, \mu)$ is a metric outer measure (being a total measure). Form the product $(\kappa^2 = \kappa \times \kappa, \nu = \mu \times \mu)$ in the sense of Defn 6.1 of OP. Let $A = \{(\alpha, \beta) : \alpha \leq \beta < \kappa\}$. Then for every $\mu \otimes \mu$-measurable set $W$, if $W \supseteq A$, then $(\mu \otimes \mu)(\kappa^2 \setminus W) = 0$ (Fubini's theorem). So $\nu(A) = 1$. Similarly, $\nu(\kappa^2 \setminus A) = 1$. So $(\kappa^2, \nu)$ is not a metric outer measure.

I will edit when I have a ZFC example.

Edit: Here's a sketch for the necessity of real valued measurable cardinals (rvm). Assume there is no rvm. Given $(X, \mu)$ where $X$ is metric and $\mu$ is a Borel probability measure on it, consider the subset $Y = \{x \in X : (\forall r > 0)(\mu(B(x, r)) > 0)\}$ (here, $B(X, r)$ is the open ball centered at $x$ of radius $r$). $Y$ is clearly separable. Also, $X \setminus Y$ is $\mu$-null (since otherwise there is an rvm). So $\mu$ is supported on a separable subspace of $X$ and we can argue as in the case of metric outer measures on separable metric spaces.

  • $\begingroup$ Let me add that, if not folklore, this type of reasoning should be in one of the volumes of Fremlin's measure theory. $\endgroup$ – hot_queen Oct 30 '15 at 23:13
  • $\begingroup$ Thanks for working on this. It has been a generation since I last taught or wrote about this material so it will take a while to absorb the details. I would suggest you write up a note and publish it in the REAL ANALYSIS EXCHANGE where it will reach the right audience. $\endgroup$ – B. S. Thomson Oct 31 '15 at 14:37

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