Extension of measure to non measurable sets

Let $(X, \Sigma, \lambda)$ and $A \notin \Sigma$. Can we extend measure $\lambda$ to $\lambda'$ measure on smallest sigma algebra $\Sigma'$ containing A and given sigma algebra?

Solution 1:

Surprisingly, it is always possible to extend a measure to a non-measurable set!

Throughout this answer, I will stick with the usual convention that measures are $\sigma$-additive and take values in $[0,+\infty]$. Given a measure space $(\Omega,\mathscr{A},\mu)$, let $\mu_*,\mu^* : \mathcal P(\Omega) \to [0,+\infty]$ denote the corresponding inner and outer measures, defined by \begin{align*} \mu_*(S) &= \sup\{\mu(A) \, \mid \, A \in \mathscr{A}, \ A \subseteq S\};\\[1ex] \mu^*(S) &= \inf\{\mu(A) \, \mid \, A \in \mathscr{A}, \ S \subseteq A\}. \end{align*}

Proposition. Let $(\Omega,\mathscr{A},\mu)$ be a measure space, and let $S \subseteq \Omega$.

1. There exist measures $\underline{\nu}$ and $\overline{\nu}$ on $\sigma(\mathscr{A} \cup \{S\})$ extending $\mu$ such that $\underline{\nu}(S) = \mu_*(S)$ and $\overline{\nu}(S) = \mu^*(S)$.
2. If $\mu^*(S) < +\infty$, then for every $\xi \in [\mu_*(S),\mu^*(S)]$ there exists a measure $\nu_\xi$ on $\sigma(\mathscr{A} \cup \{S\})$ extending $\mu$ such that $\nu_\xi(S) = \xi$.

Proof sketch. Note that every set in $\sigma(\mathscr{A} \cup \{S\})$ can be written as $(A \cap S) \cup (B \cap \overline{S})$ for some $A,B \in \mathscr{A}$. Define $\underline{\nu}(A) = \mu_*(A \cap S) + \mu^*(A \cap \overline{S})$, $\overline{\nu}(A) = \mu^*(A \cap S) + \mu_*(A \cap \overline{S})$, and $\nu_\xi = \frac{\mu^*(S) - \xi}{\mu^*(S) - \mu_*(S)}\underline{\nu} + \frac{\xi - \mu_*(S)}{\mu^*(S) - \mu_*(S)}\overline{\nu}$. It is readily verified that these are indeed well-defined measures on $\sigma(\mathscr{A} \cup \{S\})$, which extend $\mu$ and meet the required properties.

For further details, see [LM49], or Theorem 1.12.14 in [Bog07], or Exercise 1.5.12 in [Coh13].

This Proposition has the following interesting consequence:

Corollary. Let $\mu$ be a finite measure on a measurable space $(\Omega,\mathscr{A})$, and let $S \subseteq \Omega$. Then there is an extension of $\mu$ to $\sigma(\mathscr{A} \cup \{S\})$, and it is unique if and only if $S$ belongs to the completion $\mathscr{A}_\mu$ of $\mathscr{A}$ with respect to $\mu$.

Proof sketch. It is well-known that $\mu$ has a unique extension to the completion $\mathscr{A}_\mu$. Conversely, if $S$ does not belong to the completion, then $\mu_*(S) < \mu^*(S)$ (here we use that $\mu$ is finite), so it follows from part 1 of the Proposition that there are two different extensions of $\mu$ to $\sigma(\mathscr{A} \cup \{S\})$. $\quad\Box$

Closing remarks:

• There is no analogue of part 2 of the Proposition in the case $\mu^*(S) = +\infty$. It might happen that every extension $\nu$ of $\mu$ satisfies either $\nu(S) = \mu_*(S)$ or $\nu(S) = \mu^*(S)$. See Theorem 5 in [ŁM49].

• The extensions in the Proposition are not unique — there might be multiple extensions with the same prescribed value at $S$. However, if $\mu$ is finite, then the extensions in part 1 of the Proposition are unique. For a detailed discussion, see Section 7 in [ŁM49]. This non-uniqueness also has an interesting interpretation in terms of Radon–Nikodym derivatives; see Exercise 3.10.37 in [Bog07].

• The Corollary is not true for infinite measures: it could happen that $S$ does not belong to the completion but the extension is nonetheless unique. For example, let $\Omega$ be an uncountable set, let $\mathscr{A}$ be the $\sigma$-algebra of all countable or co-countable subsets of $\Omega$, and let $\mu(A)$ be the counting measure. Then $\mathscr{A}$ is already complete with respect to $\mu$. But every extension of $\mu$ to a larger $\sigma$-algebra $\mathscr{A}'$ must also be the counting measure: if $T \in \mathscr{A}' \setminus \mathscr{A}$, then $T$ is necessarily infinite, so by choosing an increasing sequence of finite subsets $A_1 \subseteq A_2 \subseteq \cdots \subseteq T$ we can show that $\mu(T) = +\infty$ must hold.

• By repeatedly applying the Proposition, we can add any finite number of additional measurable sets to a measure space. One might wonder if we can also extend $\mathscr{A}$ by infinitely many additional sets. If $\mu$ is a probability measure, then adding an arbitrary collection of pairwise disjoint sets can be done constructively, but beyond that there are non-trivial set-theoretic problems. (For example, when trying to add countably many non-disjoint subsets to $[0,1]$ with Lebesgue measure, the answer depends on the continuum hypothesis.) For more details, see Theorem 1.12.15 and the remarks thereafter in [Bog07].

• A common objection to extension questions for measure spaces is the well-known fact that the Lebesgue measure cannot be extended to $\mathcal P(\mathbb{R})$. However, it is important to note that an extension of the Lebesgue measure is required to be translation-invariant, which we do not require here. So objections inspired by classical non-measurability results do not apply. (In fact, being able to extend the Lebesgue measure to all of $\mathcal P(\mathbb{R})$ is closely related to the existence of measurable cardinals, which is undecidable in ZFC.)

References.

[ŁM49] J. Łoś and E. Marczewski, Extensions of measure, Fundamenta Mathematicae, volume 36 (1949), pp. 267–276. http://matwbn.icm.edu.pl/ksiazki/fm/fm36/fm36127.pdf

[Bog07] V.I. Bogachev, Measure Theory, Volume I, Springer, 2007.

[Coh13] Donald L. Cohn, Measure Theory, Second Edition, Birkhäuser Advanced Texts Basler Lehrbücher, Springer, 2013.