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In the introduction to Chapter 3 (Lebesgue Measure) of H. L. Royden's Real Analysis, he motivates the definition of a measure by saying that we would like to construct a set function $m$ on a collection $\mathscr{M}$ of subsets of $\mathbb{R}$ that assigns a non-negative extended real number to each set $E$ in $\mathscr{M}$ in the following way.

Ideally, we should like $m$ to have the following properties:

  1. $mE$ is defined for each set $E$ of real numbers; that is, $\mathscr{M} = \mathscr{P}(\mathbb{R})$.
  2. For an interval $I$, $mI = l(I)$, where $l(I)$ denotes the length of the interval.
  3. If $\{ E_n \}$ is a sequence of disjoint sets (for which $m$ is defined), $m(\cup_n E_n) = \sum_n E_n$.
  4. $m$ is translation invariant; that is, if $E$ is a set for which $m$ is defined and if $E + y$ is the set $\{ x + y : x \in E \}$, obtained by replacing each point $x$ in $E$ by the point $x + y$, then $$m(E+y)=mE.$$

He goes on to say that

Unfortunately, . . . it is not known whether there is a set function satisfying the first three properties.

Does this mean that this is an open problem in measure theory? Has there been any progress on this problem since the textbook was published?

Royden does mention in a footnote that assuming the Continuum Hypothesis, one can show that no such function exists. This question asks for details regarding this case.

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  • $\begingroup$ It it is actually well known that there is no such function $m$ that satisfies your (1)-(4) properties. The problem is that you cant have a function defined in the whole $\mathcal{P}(\mathbb{R})$. $\endgroup$ – dem0nakos Apr 9 '18 at 7:10
  • $\begingroup$ @dem0nakos Yes, Royden covers the construction of a non-measurable set later in the textbook. $\endgroup$ – Brahadeesh Apr 9 '18 at 7:11
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    $\begingroup$ If you would agree to omit the axiom of choice, then the answer is "yes, assuming that inaccessible cardinals are consistent" (which as far as axioms go, is a far weaker axiom than that needed for the real-valued measurable cardinals to exist). $\endgroup$ – Asaf Karagila Apr 9 '18 at 12:18
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This is in fact undecidable in standard set theory ZFC:

The existence of such an $m$ is equivalent to the existence of a real-valued measurable cardinal $\le \mathfrak{c}$ (see wikipedia e.g.) and this is known to be undecidable in ZFC; this is related to Gödel's incompleteness theorem. If ZF is consistent, ZFC without such a measure is consistent too. We could probably assume such a measure exists (this is a "leap of faith") and have a consistent theory too. But this is indeed unknown (and a strictly stronger assumption than the consistency of ZF). Probably Royden wanted to avoid these foundational discussions. It is known (and he mentions this) that if such a measure existed, we know that CH does not hold (in fact, the continuum $\mathfrak{c}$ is then a very large cardinal).

For a very thorough treatment of these and related problems, see Fremlin's measure theory volume 5. It does assume quite a bit of background.

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    $\begingroup$ "If ZF is consistent, ZFC with such a measure is consistent too." Not quite. ZFC plus the existence of real-valued measurable cardinals is of strictly stronger consistency strength than ZF (or ZFC). It is in this sense that "it is not known whether" such a measure exists, as its consistency cannot be verified in ZFC alone. $\endgroup$ – Andrés E. Caicedo Apr 9 '18 at 11:35
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    $\begingroup$ @AndrésE.Caicedo Thx. I editd the answer accordingly. $\endgroup$ – Henno Brandsma Apr 9 '18 at 15:37

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