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].
- Translational invariance
- 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.
Best Answer
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.