If $A\in\mathfrak{M}(\mu), \epsilon>0$, then there exist $F$ closed, $G$ open such that $F\subset A\subset G$ and $\mu(G-A), \mu(A-F)<\epsilon$.

Question

Source: The problem arises from Walter Rudin's Principles of Mathematical Analysis:

My Background: Rudin's PMA Chapter 1-7.

A Flawed Attempt (by Kit-Wing Yu): We know from the definition that $A\in\mathfrak{M}(\mu)$ implies that $A=\bigcup_{n=1}^{\infty}A_n$, where each $A_n\in\mathfrak{M}_F(\mu)$. From the definition, for each $n\in\mathbb{N}$ there exists a sequence $\{A_{n_k}\}\subset\mathscr{E}$ such that $A_{n_k}\to A_n$ as $k\to\infty$. Since $\mu^*$ is regular on $\mathscr{E}$, there exists $F_{n_k}\in\mathscr{E}$ closed and $G_{n_k}\in\mathscr{E}$ open, such that $F_{n_k}\subset A_{n_k}\subset G_{n_k}$ and $\mu^*(G_{n_k})-2^{-n}2^{-k}2^{-1}\varepsilon\leq\mu^*(A_{n_k})\leq \mu^*(F_{n_k})+2^{-n}2^{-k}2^{-1}\varepsilon$; in particular, $\mu^*(G_{n_k})-\mu^*(A_{n_k})\leq2^{-n}2^{-k}2^{-1}\varepsilon$.

Note $G_n=\bigcup_{k=1}^{\infty}G_{n_k}$ and $G=\bigcup_{k=1}^{\infty}G_n$ are open. Also, we know that $G_{n_k}, G_n, G\in\mathfrak{M}(\mu)$.

IF we have $\mathbf{A_n=\bigcup_{k=1}^{\infty}A_{n_k}}$, then we'd have $A_n\subset G_n$, $A\subset G$, $G_n\backslash A_n\subset\bigcup_{k=1}^{\infty}(G_{n_k}\backslash A_{n_k})$ and $G\backslash A\subset\bigcup_{n=1}^{\infty}(G_n\backslash A_n)$. Thus, $\mu^*(G\backslash A)\leq\mu^*(\bigcup_{n=1}^{\infty}(G_n\backslash A_n))\leq\sum_{n=1}^{\infty}\mu^*(G_n\backslash A_n)\leq\sum_{n=1}^{\infty}\mu^*(\bigcup_{k=1}^{\infty}(G_{n_k}\backslash A_{n_k}))\leq\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\mu^*(G_{n_k}\backslash A_{n_k})\leq\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}[\mu^*(G_{n_k})-\mu^*(A_{n_k})]\leq\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}2^{-n}2^{-k}2^{-1}\varepsilon<\varepsilon$. The other half of the proposition can be similarly obtained. That IF, however, does not seem justified. Even if we let $A_{n_k}'=A_n\cap A_{n_k}$ and therefore have $A_{n_k}'\to A_n$ and $A_n\supset\bigcup_{k=1}^{\infty}A_{n_k}'$, we still don't know if each $A_{n_k}'$ is indeed elementary, and if $A_n\subset\bigcup_{k=1}^{\infty}A_{n_k}'$.

Any help would be greatly appreciated.

Answer 1

If you are looking for countable unions of sets in $\mathbb{R^d}$, then the rationals are very useful. Consider how you could measure $A$ if you had $A \cap \mathbb{Q^d}$.