I have read that probability measure cannot be defined on set of all subsets of unit interval namely (0,1]. Proof uses construction of the Vitali set etc. Specifically, it is well known that probability measure with the following properties does not exist on power set of (0,1].

  1. Translational invariance
  2. If $0 \leq a \leq b \leq 1$, then $\mathbb{P}((a,b]) = b-a$

This is actually Lebesgue measure on (0,1].

This might be a dumb question but I cannot find an answer to it. The domain of probability measure is a Borel Sigma algebra while that of a Lebesgue measure is a Lebesgue Sigma algebra. It is well known that Lebesgue sigma algebra has cardinality of $2^{\mathbb{R}}$ while that of Borel sigma algebra is $2^{\mathbb{N}}$.

So even though Lebesgue measure on (0,1] $ \textbf{IS}$ a probability measure satisfying 1. and 2. above, its domain has a cardinality strictly greater than that of Borel Sigma Algebra.

Where am I going wrong? Even though Lebesgue measure on (0,1] is in fact a uniform probability measure, why is there a difference in the domains? If not, please clarify.

  • $\begingroup$ A Lebesgue measurable set can be written as a union of a Borel set and a set of measure zero (which may not be Borel measurable). $\endgroup$ Commented Feb 3, 2019 at 21:45
  • $\begingroup$ Statement A: Uniform probability measure with translation invariance cannot be defined on power set of (0,1]. Statement B: Borel measure is same as Lebesgue measure on (0,1]. Statement C: Lebesgue Sigma Algebra has a cardinality strictly greater than that of Borel Sigma Algebra. Now statement B must imply that domains of Borel and Lebesgue measures are same in (0,1]. But then, statement C is implying that they have to be different. Where am I going wrong? $\endgroup$ Commented Feb 3, 2019 at 21:50
  • $\begingroup$ What do you mean by "uniform probability measure"? B is wrong and C is correct. $\endgroup$ Commented Feb 3, 2019 at 22:09
  • $\begingroup$ Uniform probability measure is defined as one which has the translational invariance property. Why is B wrong? Basically, the measure is b-a on sets (a,b] which are in (0,1] and they are both sigma algebras containing all the intervals in (0,1]. They are exactly the same right? They will differ when you define them on all of $\mathbb{R}$. $\endgroup$ Commented Feb 3, 2019 at 22:13
  • $\begingroup$ @TryingHardToBecomeAGoodPrSlvr Why does "same on intervals" imply "same on every subset"? $\endgroup$
    – Clement C.
    Commented Feb 3, 2019 at 22:16

1 Answer 1


We have the standard Lebesgue measure $\mu$, that is defined on all Borel sets, and obeys translation invariance and interval-consistency (my name for $\mu((a,b])=b-a$ for all relevant intervals; you call it uniform in the comments) . Indeed there are $\mathfrak{c}$ many Borel sets on which this $\mu$ is then defined.

Independently of that we can define null sets as:

$A \in \mathscr{N}$ iff for every $\varepsilon >0$ we can find at most countably many open intervals $(a_n, b_n)$ such that $A \subseteq \bigcup (a_n, b_n)$ and $\sum_n |b_n - a_n| < \varepsilon$ and check that such subsets are closed under subsets and countable unions (they form a $\sigma$-ideal). There are $2^\mathfrak{c}$ many null sets (e.g. as the standard Cantor set is one and so are all of its subsets.)

It turns out we can extend $\mu$ to a measure on more subsets, namely take the collection of all sets of the form $B \cup N$ where $B$ is Borel, and $N$ is a null set and just define $\mu'(B \cup N) = \mu(B)$. One can prove that this is well-defined (e.g. if a Borel set happens to be a null set, $\mu$ was $0$ on it anyway, justifying the name) and greatly extends the domain of $\mu$, while keeping translation invariance.

The existence of a Vitali set (if we assume AC) shows that we cannot extend $\mu'$ even further to $\mathscr{P}(\mathbb{R})$ while keeping translation invariance intact. $\mu'$ is called the completion of $\mu$ and it's also obtained when we apply the Carathéodory theorem on the Lebesgue measure on the half-open interval algebra; we get (IIRC) the same class of Lebesgue measurable subsets.

It depends on your application whether you want to work with only Borel sets or all Lebesgue measurable sets. Mostly in analysis the latter is done; taking the larger domain.

If we give up translation invariance and so "neutralise" the Vitali set examples (whose proof of non-measurability hinges on this translation invariance property) there are still obstacles: under CH ($\mathfrak{c}=\aleph_1$) there can be no finite measure on $(0,1]$ that measures all subsets and that gives all singletons measure $0$ (much weaker than being uniform), as was shown by Ulam. So you could never define such a measure on all subsets without additional set-theoretic assumptions. See also real valued measurable cardinals etc. It gets complicated.

  • $\begingroup$ Thanks for the nice answer. I am still relatively new to measure theory so I could not understand some parts of your answer. The part that caught my attention was the statement "and greatly extends the domain of 𝜇, while keeping translation invariance." Does that imply that Lebesgue measure, with the domain being the Lebesgue $\sigma - algebra$, is a probability measure with the property of translational invariance? $\endgroup$ Commented Feb 3, 2019 at 23:36
  • $\begingroup$ @TryingHardToBecomeAGoodPrSlvr Yes, this is clear from the definition I gave as well. Null sets are also translation invariant. $\endgroup$ Commented Feb 3, 2019 at 23:37
  • $\begingroup$ I thought that it was not possible to have translational invariant probability measures defined on sets with cardinality $2^\mathbb{R}$ whose proof can be based off of something like that given in math.stackexchange.com/questions/1544458/…. Not sure where I am going wrong here. $\endgroup$ Commented Feb 3, 2019 at 23:38
  • $\begingroup$ @TryingHardToBecomeAGoodPrSlvr You cannot define it on all subsets of $\Bbb R$, but the Vitali set is not Lebesgue measurable. The size of the $\sigma$-algebra has no role here, there are as many Lebesgue measurable (even null) sets as there are subsets of the reals, but that doesn't allow us to extend a measure in a sensible way.. $\endgroup$ Commented Feb 3, 2019 at 23:41
  • $\begingroup$ Awesome! So it is possible to have a translationally invariant uniform probability measure on set of all subsets of (0,1] but not on all subsets of $\mathbb{R}$ right? $\endgroup$ Commented Feb 3, 2019 at 23:45

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