Exercise 1.7.9 (Approximation by an algebra II)

Question

Let $\mu_0: {\mathcal B}_0 \rightarrow [0,+\infty]$ be a pre-measure which is $\sigma$-finite (thus $X$ is the countable union of sets in ${\mathcal B_0}$ of finite $\mu_0$-measure), and let $\mu: {\mathcal B} \rightarrow [0,+\infty]$ be the Hahn-Kolmogorov extension of $\mu_0$.
\begin{enumerate}[label=(\roman*)]
\item Show that if $E \in {\mathcal B}$, then there exists $F \in \langle {\mathcal B}_0 \rangle$ containing $E$ such that $\mu(F \backslash E) = 0$ (thus $F$ consists of the union of $E$ and a null set). Furthermore, show that $F$ can be chosen to be a countable intersection $F = \bigcap_{n=1}^\infty F_n$ of sets $F_n$, each of which is a countable union $F_n = \bigcup_{m=1}^\infty F_{n,m}$ of sets $F_{n,m}$ in ${\mathcal B}_0$.
\item If $E \in {\mathcal B}$ has finite measure (i.e. $\mu(E) < \infty$), and $\varepsilon > 0$, show that there exists $F \in {\mathcal B}_0$ such that $\mu(E \Delta F) \leq \varepsilon$.
\item Conversely, if $E$ is a set such that for every $\varepsilon > 0$ there exists $F \in {\mathcal B}_0$ such that $\mu^*(E \Delta F) \leq \varepsilon$, show that $E \in {\mathcal B}$.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=( oman*)] \item Let $E\in\mathcal{B}$. First, suppose additionally that $\mu(E)<\infty^*$. Let $\frac{1}{n}>0, n\in\mathbb{N}$ be arbitrary. By definition of the outer measure, we can find a sequence $(E_m)_{m\in\mathbb{N}}$ in $\mathcal{B}_0$ such that $$\mu\left(E ight) + \frac{1}{n} \geq \sum_{m=1}^\infty \mu_0(E_m).$$ Using the axiom of choice, for each $n\in\mathbb{N}$ pick a sequence $\left(E_m^{(n)} ight)_{m\in\mathbb{N}}$ in $\mathcal{B}_0$ with the above property. Denoting $F_n := \bigcup_{m=1}^{\infty} E_m^{(n)} \in \langle \mathcal{B}_0 angle$ we immediate that $$\forall n \in \mathbb{N}: \quad \mu\left( \bigcap_{n=1}^{\infty} F_n \setminus E ight) \leq \mu\left( F_n \setminus E ight) \stackrel{*}{\leq} \sum_{m=1}^{\infty}\mu\left(E_m^{(n)} ight) - \mu(E) \leq \frac{1}{n}.$$ Thus, by setting $$F := \bigcap_{n=1}^{\infty} \bigcup_{m=1}^{\infty} E_m^{(n)}$$ we obtain the desired result. ewline In the case when $\mu(E)=\infty$, we can call our right to find the sequence $(X_m)_{m\in\mathbf{N}}$ in $\mathcal{B}_0$ with $X=\bigcup X_m$ and $\mu_0(X_m)<\infty$ and consider covers $E\cap \left(\bigcup_{m=1}^{M}X_m ight)$ for each $M\in\mathbb{M}$. Then, apply the previous part to each of these sets separately and take countable unions over $M$. \item Fix an $\epsilon > 0$. Using a similar strategy to those we have employed in the first part of this proof, find a sequence $(E_m)_{m\in\mathbb{N}}$ in $\mathcal{B}_0$ such that by the definition of $\mu^{*}$ we have $$\mu(E) + \frac{\epsilon}{2} < \sum_{m=1}^{\infty} \mu(E_m).$$ Unfortunately, $\bigcup_{m\in\mathbf{N}} E_m$ is not necessarily contained in $\mathcal{B}_0$, but $\bigcup_{m\leq M} E_m$ is, of which we make use. Set $F_M := \bigcup_{m=1}^{M}E_m$. We need to demonstrate that the following quantity can get arbitrarily small. \begin{align*} \mu\left(F_M riangle E ight) &= \mu(F\setminus E) + \mu(E\setminus F)\ &=\mu\left(\bigcup_{m=1}^{M}E_m \setminus E ight) + \mu\left(E \setminus \bigcup_{m=1}^{M}E_m ight) \end{align*} We immediately observe two symmetric facts: \begin{itemize} \item On one hand, $\bigcup_{m=1}^{1} E_m \setminus E \subseteq \bigcup_{m=1}^{M}E_m \setminus E \subseteq \ldots$ is an increasing sequence of $\mathcal{B}_0$-measurable sets. This allows us to apply the \href{./exercise_1-4-23}{upwards monotone convergence} and conclude that $$\lim_{M o\infty} \mu\left(\bigcup_{m=1}^{M}E_m \setminus E ight) = \mu\left(\bigcup_{m=1}^{\infty} E_m \setminus E ight) = \sum_{m=1}^{\infty} \mu(E_m) - \mu(E) < \frac{\epsilon}{2}.$$ Thus, we can choose $M_1\in\mathbf{N}$ large enough so that $\mu\left(\bigcup_{m=1}^{M_1} E_m \setminus E ight) \leq \epsilon/2$. \item On the other hand, $E\setminus \bigcup_{m=1}^{1} E_m \supseteq E\setminus\bigcup_{m=1}^{M}E_m \supseteq \ldots$ is a decreasing sequence of $\mathcal{B}_0$-measurable sets. This allows us to apply the \href{./exercise_1-4-23}{downwards monotone convergence} and conclude that $$\lim_{M o\infty} \mu\left(E \setminus \bigcup_{m=1}^{M}E_m ight) = \mu\left(E \setminus \bigcup_{m=1}^{\infty} E_m ight) = 0.$$ Once again, we choose $M_2\in\mathbf{N}$ large enough so that $\mu\left(E \setminus \bigcup_{m=1}^{M}E_m ight) \leq \epsilon/2$. \end{itemize} Finally, combining these two results by setting $M:=\max\{M_1,M_2\}$ gives us the desired result $\mu(F_M riangle E) \leq \epsilon$. \end{enumerate}

Exercise 1.7.9 (Approximation by an algebra II)

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Exercise 1.7.9 (Approximation by an algebra II)

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