Exercise 1.4.19 (Generating set of the Lebesgue $\sigma$-algebra)

Question

Show that the Lebesgue $\sigma$-algebra on $\mathbb{R}^d$ is generated by the union of the Borel $\sigma$-algebra and the null $\sigma$-algebra.

Elu Thingol · Accepted Answer

We want to demonstrate that $\sigma( \mathcal{B} \cup \mathcal{N}) = \mathcal{L}$, i.e.,
$$\bigcap \left\{\mathcal{F}\subseteq\mathcal{P}(\mathbb{R}^d): \begin{array}{ll} \mathcal{F} 	ext{ is a $\sigma$-algebra}\ \mathcal{B}\cup\mathcal{N}\subseteq \mathcal{F} \end{array} ight\} = \left\{ E \subseteq \mathbb{R}^d: \quad \forall \epsilon > 0 \quad \exists U \in \mathcal{O}(\mathbb{R}^d),E \subseteq U: \quad m^*(U/E)\leq \epsilon ight\}$$
\begin{itemize}
\item[$\subseteq$] It is easy to verify that $\mathcal{B}\subseteq \mathcal{L}$ and $\mathcal{N}\subseteq \mathcal{L}$. Null sets are Lebesgue measurable by Lemma 1.2.13 (iv). Open sets are Lebesgue measurable by Lemma 1.2.13 (i), thus a $\sigma$-algebra generated by them is contained in $\mathcal{L}$, because  $\mathcal{L}$ is already a $\sigma$-algebra itself by \href{./exercise_1-4-11}{Exercise 1.4.11}. Thus, it is obvious that $\mathcal{B} \cup \mathcal{N} \subseteq \mathcal{L}$. Again, $\mathcal{L}$ is a $\sigma$-algebra itself, thus $\sigma(\mathcal{B} \cup \mathcal{N}) \subseteq \mathcal{L}$.
\item[$\supseteq$] Pick an arbitrary $E \in \mathcal{L}$, and let $(U_n)_{n\in\mathbb{N}}$ be a sequence of open sets containing $E$ such that $m^*(U/E)\leq 1/n$. Denote $V:= \bigcap_{n=1}^{\infty} U_n$. We then still have $E \subseteq V$ and $m(V/E) = 0$. Thus, $V/E \in \mathcal{N}$, and $V \in \mathcal{B}$ as a countable intersection. Therefore $V, V/E \in \sigma(\mathcal{B}\cup\mathcal{N})$. Also, by the closure under the complements, we must have $V/(V/E) = E \in \sigma(\mathcal{B}\cup\mathcal{N})$, as desired.
\end{itemize}

Exercise 1.4.19 (Generating set of the Lebesgue $\sigma$-algebra)

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Exercise 1.4.19 (Generating set of the Lebesgue $\sigma$-algebra)

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