If we are looking over subsets of $\mathbb R$ and considering the outer measure defined exactly as
$$\mu^*(A) = \inf\left\{ \sum_{k=1}^\infty \ell(I_k) \text{ where the $I_k$ are open intervals such that } A\subset \bigcup_{k=1}^\infty I_k \right\}$$
and inner measure to be
$$\mu_*(A) = \sup\left\{\mu^*(F) \text{ where the $F$ is a closed subset of $\mathbb R$ such that } F\subset A \right\}$$
then measurability in the sense of Lebesgue:
- $A$ is measurable if $\color{red}{\mu^*(A)=\mu_*(A)}$
is equivalent to the Caratheodory criterion:
- $A$ is measurable if $\color{red}{\mu^*(T) = \mu^*(T\cap A) + \mu^*(T\cap A^C)}$ for all test sets $T\subset \mathbb R$
See the proof from page 41-45 of this pdf: http://www.math.harvard.edu/~shlomo/212a/11.pdf; definitions and preceding theorems are found pages 1-40).
First question $\color{red}{\text{STILL OPEN}}$ (perhaps refer to Nate’s comment below)?:
If these two definitions are in fact equivalent, has anyone tried proving the Caratheodory Extension Theorem without the Caratheodory criterion, i.e. from scratch using outer and inner measure, or at least the equivalent definition of $\color{red}{\mu(G\setminus F)<\epsilon \text{ for some } G\supset A\supset F}$ (the "inner-outer" characterization from the Harvard pdf)? If so, could you provide a rundown of the proof (or a link to some article)?
If it doesn't work generally, how about in the specific case where outer measure is defined the way I defined it above (e.g. as in the proof presented here, page 28 of the pdf: https://www.stat.washington.edu/jaw/COURSES/520s/521/bk521reJaw2012.pdf)?
And second question $\color{red}{\text{(resolved I believe by the comment by Daniel Fischer)}}$:
Consider a weaker criterion than Caratheodory's: I just want that $\mu^*(X) = \mu^*(X\cap A) + \mu^*(X\cap A^C)$ for just the "biggest set" $X$ (I'm talking about the set for which $A^C=X\setminus A$ for all $A$, i.e. the universal set; assume also that $\mu(X)$ is not infinite). We are still using outer measure as I defined it above.
Is this an equivalent condition to the notions of measurability I described before? If not, is there an explicit counterexample (one that I can use pedagogically to motivate Caratheodory's stronger criterion)?
Related questions:
- Intuition behind the definition of a measurable set
- Proving the *Carathedory Criterion* for *Lebesgue Measurability*
- Measurability criterion of Caratheodory
- https://mathoverflow.net/questions/34007/demystifying-the-caratheodory-approach-to-measurability
- Proving equivalent of definitions of measurability
- Is the Carathéodory definition of Measurable set the most general?
$\color{red}{\text{Thanks to Daniel Fischer to a comment for what I believe to be a resolution to Question 2:}}$
Best Answer
Concerning the first question, it can be done for the Lebesgue measure, but the proof is long. I taught it like that once, but I was not happy with it. I am not taking the intervals to be open (it is equivalent).
$$ \mu^{\ast}(E):=\inf\left\{ \sum_{i=1}^{\infty}\ell(R_{i}):~R_{i}\text{ intervals, }\bigcup_{i=1}^{\infty}R_{i}\supseteq E\right\} . $$
Given a set $E\subseteq\mathbb{R}$, we say that $E$ is Lebesgue measurable if for every $\varepsilon>0$ there exists an open set $U\supseteq E$ such that $$ \mu^{\ast}(U\setminus E)\leq\varepsilon. $$
Proposition 1 The following properties hold.
(i) If $E\subseteq F\subseteq\mathbb{R}$, then $\mu^{\ast}(E)\leq \mu^{\ast}(F)$.
(ii) If $E\subseteq\bigcup_{n=1}^{\infty}E_{n}$, then $\mu^{\ast }(E)\leq\sum_{n=1}^{\infty}\mu^{\ast}(E_{n})$.
(iii) If $E\subseteq\mathbb{R}$, then $$ \mu^{\ast}(E)=\inf\{\mu^{\ast}(U):~U\text{ open, }U\supseteq E\}. $$
(iv) If $E,F\subseteq\mathbb{R}$ and $\operatorname*{dist}(E,F)>0$, then $\mu^{\ast}(E\cup F)=\mu^{\ast}(E)+\mu^{\ast}(F)$.
Proof: (i) If $R_{i}$ are intervals such that $\bigcup_{i=1}^{\infty} R_{i}\supseteq F$, then $\bigcup_{i=1}^{\infty}R_{i}\supseteq E$, and so $$ \mu^{\ast}(E)\leq\sum_{i=1}^{\infty}\ell(R_{i}). $$ Taking the infimum over all sequences of intervals covering $F$ gives $\mu^{\ast}(E)\leq\mu^{\ast}(F)$.
(ii) If $\sum_{n=1}^{\infty}\mu^{\ast}(E_{n})=\infty$, there is nothing to prove. Thus, assume that $\sum_{n=1}^{\infty}\mu^{\ast}(E_{n})<\infty$. Given $\varepsilon>0$, for each $n$ find a sequence $\{R_{i}^{\left( n\right) }\}_{i}$ of intervals such that $\bigcup_{i=1}^{\infty}R_{i}^{(n)}\supseteq E_{n}$ and such that $$ \sum\limits_{i=1}^{\infty}\ell(R_{i}^{(n)})\leq\mu^{\ast}(E_{n})+\frac {\varepsilon}{2^{n}}. $$ Since $\mathbb{N}\times\mathbb{N}$ is countable, we may write $\{R_{i} ^{(n)}\}_{i,n\in\mathbb{N}}=\left\{ R_{j}\right\} _{j\in\mathbb{N}}$. Note that $$ \bigcup_{n=1}^{\infty}E_{n}\subset\bigcup_{j=1}^{\infty}R_{j}=\bigcup \limits_{n=1}^{\infty}\bigcup\limits_{i=1}^{\infty}R_{i}^{\left( n\right) }, $$ and so by part (i) and (by properties of nonnegative double series) $$ \mu^{\ast}(E)\leq\mu^{\ast}\left( \bigcup\limits_{n=1}^{\infty}E_{n}\right) \leq\sum\limits_{j=1}^{\infty}\ell(R_{j})=\sum\limits_{n=1}^{\infty} \sum\limits_{i=1}^{\infty}\ell(R_{i}^{\left( n\right) })\leq\sum \limits_{n=1}^{\infty}\mu^{\ast}(E_{n})+\varepsilon. $$ By letting $\varepsilon\rightarrow0^{+}$ we conclude the proof.
(iii) If $U$ is open and $U\supseteq E$, then by part (i), $\mu^{\ast} (E)\leq\mu^{\ast}\left( U\right) $. Taking the infimum over all open sets containing $E$ gives $$ \mu^{\ast}(E)\leq\inf\left\{ \mu^{\ast}(U):~U\text{ open, }U\supseteq E\right\} . $$ On the other hand, if $R_{i}$ are intervals such that $\bigcup_{i=1}^{\infty }R_{i}\supseteq E$, given $\varepsilon>0$ for every $i$ consider an open interval $T_{i}\supseteq R_{i}$ such that $\operatorname*{meas}T_{i} \leq\operatorname*{meas}R_{i}+\frac{\varepsilon}{2^{i}}$ and let $V:=\bigcup_{i=1}^{\infty}T_{i}$. Then $V$ is open and contains $E$ and so $$ \inf\left\{ \mu^{\ast}(U):~U\text{ open, }U\supseteq E\right\} \leq\mu ^{\ast}(V)\leq\sum_{i=1}^{\infty}\ell(T_{i})\leq\sum_{i=1}^{\infty}\ell (R_{i})+\varepsilon. $$ Letting $\varepsilon\rightarrow0^{+}$ gives $$ \inf\left\{ \mu^{\ast}(U):~U\text{ open, }U\supseteq E\right\} \leq \sum_{i=1}^{\infty}\ell(R_{i}). $$ Taking the infimum over all sequences of intervals covering $E$ gives $$ \inf\left\{ \mu^{\ast}(U):~U\text{ open, }U\supseteq E\right\} \leq\mu ^{\ast}(E). $$
(iv) Let $d:=\operatorname*{dist}(E,F)$. If $R_{i}$ are intervals such that $\bigcup_{i=1}^{\infty}R_{i}\supseteq E\cup F$, by subdividing each interval $R_{i}$ into subintervals, without loss of generality, we may assume that each interval $R_{i}$ has diameter less than $d$. Hence, if $R_{i}$ intersects $E$ it cannot intersect $F$ and viceversa. Write $\{R_{i}:~R_{i}\cap E\neq\emptyset\}=\{S_{i}\}$ and $\{R_{i}:~R_{i}\cap F\neq\emptyset \}=\{T_{i}\}$. Then $\bigcup_{i}S_{i}\supseteq E$, $\bigcup_{i}T_{i}\supseteq F$ and so $$ \mu^{\ast}(E)+\mu^{\ast}(F)\leq\sum_{i}\ell(S_{i})+\sum_{i}\ell(T_{i} )=\sum_{i=1}^{\infty}\ell(E_{i}). $$ Taking the infimum over all sequences of intervals covering $E\cup F$ gives $$ \mu^{\ast}(E)+\mu^{\ast}(F)\leq\mu^{\ast}(E\cup F). $$ The other inequality follows from part (ii).
Proposition 2 The following properties hold.
(i) Open sets are Lebesgue measurable.
(ii) If $E\subseteq\mathbb{R}$ has Lebesgue outer measure zero, then $E$ and its subsets are Lebesgue measurable.
(iii) If $E=\bigcup_{n=1}^{\infty}E_{n}$, and each $E_{n}$ is Lebesgue measurable, then $E$ is Lebesgue measurable.
(iv) Compact sets are Lebesgue measurable.
(v) Closed sets are Lebesgue measurable.
(vi) If $E\subseteq\mathbb{R}$ is Lebesgue measurable, then $\mathbb{R}\setminus E$ is Lebesgue measurable.
(vii) If $E=\bigcap_{n=1}^{\infty}E_{n}$, and each $E_{n}$ is Lebesgue measurable, then $E$ is Lebesgue measurable.
Proof (ii) Let $E\subseteq\mathbb{R}$ be such that $\mu^{\ast}(E)=0$. By Proposition 1(iii), $$ 0=\mu^{\ast}(E)=\inf\left\{ \mu^{\ast}(U):~U\text{ open, }U\supseteq E\right\} . $$ Hence, for every $\varepsilon>0$ there exists $U$ open such that $U\supseteq E$ and $\mu^{\ast}(U)\leq\varepsilon$. By Proposition 1(i), $\mu^{\ast}(U\setminus E)\leq \mu^{\ast}(U)\leq\varepsilon$, and thus $E$ is Lebesgue measurable.
Again by Proposition 1(i), if $F\subseteq E$, then $0\leq\mu^{\ast}(F)\leq\mu^{\ast}(E)=0$, and so $F$ also has Lebesgue outer measure zero, and thus is Lebesgue measurable.
(iii) Let $E=\bigcup_{n=1}^{\infty}E_{n}$, where each $E_{n}$ is Lebesgue measurable. Then for every $\varepsilon>0$ there exists an open set $U_{n}\supseteq E_{n}$ such that $$ \mu^{\ast}(U_{n}\setminus E_{n})\leq\frac{\varepsilon}{2^{n}}. $$ Define $U:=\bigcup_{n=1}^{\infty}U_{n}$. Then $U$ is open, contains $E$ and $U\setminus E=\bigcup_{n=1}^{\infty}(U_{n}\setminus E)\subseteq\bigcup _{n=1}^{\infty}(U_{n}\setminus E_{n})$. Hence, by Proposition 1(i) and (ii), $$ \mu^{\ast}(U\setminus E)\leq\sum\limits_{n=1}^{\infty}\mu^{\ast}% (U_{n}\setminus E_{n})\leq\varepsilon. $$ (iv) Let $K\subset\mathbb{R}$ be a compact set. Then $K$ is bounded and so $\mu^{\ast}(K)<\infty$. Then by Proposition 1(iii) for every $\varepsilon>0$ there exists an open set $U\supseteq K$ such that $$ \mu^{\ast}(U)\leq\mu^{\ast}(K)+\varepsilon. $$ The set $U\setminus K$ is open and so we can write $$ U\setminus K=\bigcup_{n=1}^{\infty}Q_{n}, $$ where $Q_{n}$ are closed bounded intervals with disjoint interior. Hence, for every $m\in\mathbb{N}$ the set $$ C_{m}:=\bigcup_{n=1}^{m}Q_{n} $$ is compact and does not intersect $K$. Hence, $\operatorname*{dist}% (C_{m},K)>0$. Since $K\cup C_{m}\subseteq U$, by Proposition 1(i), (iv), \begin{align*} \mu^{\ast}(U) & \geq\mu^{\ast}(K\cup C_{m})=\mu^{\ast}(K)+\mu^{\ast}(C_{m})\\ & =\mu^{\ast}(K)+\sum_{n=1}^{m}\ell(Q_{n}) \end{align*} Hence, $$ \sum_{n=1}^{m}\ell(Q_{n})\leq\mu^{\ast}(U)-\mu^{\ast}(K)\leq\varepsilon $$ for all $m$. Letting $m\rightarrow\infty$ gives $$ \mu^{\ast}(U\setminus K)\leq\sum_{n=1}^{\infty}\ell(Q_{n})\leq\mu^{\ast }(U)-\mu^{\ast}(K)\leq\varepsilon, $$ which shows that $K$ is measurable.
(v) If $C\subseteq\mathbb{R}$ is closed, then $C=\bigcup_{n=1}^{\infty} (C\cap\lbrack-n,n])$ and since each $C\cap\lbrack-n,n]$ is compact, it follows from parts (iii) and (iv) that $C$ is measurable.
(vi) If $E\subseteq\mathbb{R}$ is Lebesgue measurable, then for every $n\in\mathbb{N}$ there exists an open set $U_{n}\supseteq E$ such that $$ \mu^{\ast}(U_{n}\setminus E)\leq\frac{1}{n}. $$ By part (v) the set $C_{n}:=\mathbb{R}\setminus U_{n}$ is closed and so Lebesgue measurable. Define $F:=\bigcup_{n=1}^{\infty}C_{n}$. Then $F$ is Lebesgue measurable by part (iii). Since $U_{n}\supseteq E$, we have that $\mathbb{R}\setminus E\supseteq\mathbb{R}\setminus U_{n}=C_{n}$ for every $n$, and so $\mathbb{R}\setminus E\supseteq F$. Moreover, $$ (\mathbb{R}\setminus E)\setminus F=(\mathbb{R}\setminus E)\setminus \bigcup_{n=1}^{\infty}C_{n}\subseteq U_{n}\setminus E $$ for every $n$ and so $$ \mu^{\ast}((\mathbb{R}\setminus E)\setminus F)\leq\mu^{\ast}(U_{n}\setminus E)\leq\frac{1}{n}. $$ Letting $n\rightarrow\infty$ shows that $\mu^{\ast}((\mathbb{R}\setminus E)\setminus F)=0$ and so $(\mathbb{R}\setminus E)\setminus F$ is measurable by part (ii). Since $F$ is measurable, we have that $\mathbb{R}\setminus E$ is Lebesgue measurable since it is the union of $F$ and $(\mathbb{R}\setminus E)\setminus F$.
(vii) This follows from part (vi) and De Morgan's laws.
Remark In the proof of part (iv) we used the fact that $$ \mu^{\ast}(C_{m})=\sum_{n=1}^{m}\ell(Q_{n}). $$ To see this, if $Q_{n}$ has length $\ell_{n}$, consider the intervals with the same center $S_{n}$ of length $(1-\varepsilon)\ell_{n}$. Since the intervals $Q_{n}$ have pairwise disjoint interior, it follows that $\operatorname*{dist} (S_{n},S_{k})>0$ for $n\neq k$ and so by Propositions 1 (iv) \begin{align*} \mu^{\ast}(C_{m}) & \geq\mu^{\ast}\Big(\bigcup_{n=1}^{m}S_{n}\Big)=\sum _{n=1}^{m}\mu^{\ast}(S_{n})\\ & =\sum_{n=1}^{m}\ell(S_{n})=\sum_{n=1}^{m}(1-\varepsilon)^{N}\ell(Q_{n}). \end{align*} Letting $\varepsilon\rightarrow0^{+}$ gives $$ \mu^{\ast}(C_{m})\geq\sum_{n=1}^{m}\ell(Q_{n}). $$ The other inequality follows from Proposition 1(ii).