The definition of Lebesgue measurable set in my real analysis textbook is:
$E \subseteq \mathbb{R}^d$ is called lebesgue measurable if for all $\varepsilon > 0$ there is an open set $E \subseteq O$ of $\mathbb{R}^d$ such that $m_{*} (O / E) \leq \varepsilon$ (and from there it's kinda easy to prove that the family of all the measurable sets is a sigma algebra containing all the borel sets).
BUT my abstract measure theory book uses the Carathéodory criterion: $E \subseteq \mathbb{R}^d$ is lebesgue measurable if for all $A \subseteq \mathbb{R}^d$ we have $m_{*} (A) = m_{*} (A \cap E) + m_{*} (A \cap E^c)$. How can i prove that these two definitions are equivalent?
Equivalence of measurability definitions
lebesgue-measuremeasure-theory
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Yes, a $ \sigma- $ algebra has properties ! Let
$ \mathcal{L}= \{E \subseteq \mathbb R: m^*(A)=m^*(A\cap E)+m^*(A\cap E^c) \quad \forall A \subseteq \mathbb R\}.$
$ \mathcal{L}$ has the follwing properties (try a proof):
$ \mathbb R \in \mathcal{L}$;
$E\in \mathcal{L}$ implies $\mathbb R \setminus E \in \mathcal{L}$;
if $(E_j)$ is a sequence in $ \mathcal{L}$, then $\bigcup E_j \in \mathcal{L}.$
This shows that $ \mathcal{L}$ is a $ \sigma- $ algebra
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).
Best Answer
Let $O_n$ be the sequence such that $E \subset O_n$ and $m^*(O_n-E) \to 0$. (I am going to use $m^*$ for the outer measure).
By subadditivity, we have that
$$m^*(A) \le m^*(A \cap E) + m^*(A \cap E^c).$$
For the other inequality, since open sets are Lebesgue measurable (by either definition - you should check this), we have that for any $A \subset \mathbb{R}^d$, we have that
$$m^*(A) = m^*(A \cap O_n) + m^*(A \cap O_n^c).$$
We note that $E \subset O_n$, so $m^*(A \cap O_n) \ge m^*(A \cap E)$ and $$m^*(E^c \cap A) \le m^*(O_n^c \cap A) + m^*((O_n - E) \cap A) \Rightarrow m^*(O_n^c \cap A) \ge m^*(E^c \cap A) - m^*((O_n - E) \cap A).$$
Thus, we get that $$m^*(A) \ge m^*(A \cap E) + m^*(A \cap E^c).$$
For the reverse direction, assume that $m^*(A) = m^*(A \cap E) + m^*(A \cap E^c)$. By the definition of the outer measure, we have that
$$m^*(E) = \inf \left\{ \sum m^*(R_i) : E \subset \bigcup R_i, R_i \text{ is a rectangle}\right\}.$$
Let $(\{R^n_i\})_n$ be a sequences of covers of $E$ such that
$$m^*\left(\bigcup_i R_i^n\right) \le \sum_i m^*(R^n_i) \le m^*(E) + \frac{1}{n}.$$
Using this define $O_n = \bigcup_i R_i^n$ and apply the Caratheodory condition with $O_n$ to get
$$m^*(E) + \frac{1}{n} \ge m^*(O_n) = m^*(O_n \cap E) + m^*(O_n \cap E^c) = m^*(E) + m^*(O_n \cap E^c). $$
This implies that
$$ \frac{1}{n} \ge m^*(O_n \cap E^c) = m^*(O_n - E). $$
Thus, we have the needed sequence of sets. Thus, we are done.