Help understanding where my proof is wrong (proving that a certain set is a Dynkin system)

Question

So I was reading Measures, Integrals and Martingales (second edition), by Schilling, and upon reaching chapter 5 (the one on uniqueness of measures), I came to exercise 5.12 which states:

A Boolean algebra $\mathscr{C}$ on a set $\mathit{X}$ is a subset of $\mathscr{P(\mathit{X})}$ (the power set of $\mathit{X}$) such that $\mathit{X}$ is in $\mathscr{C}$, and $\mathscr{C}$ is closed under the fomation of finite unions and intersections and complements. If $\mathscr{A}= \sigma(\mathscr{C})$ (the sigma algebra generated by the collection $\mathscr{C}$), and $\mu$ is a finite measure on the measurable space ($\mathit{X}$, $\mathscr{A}$), then prove that:

$\forall$$\mathit{A}$$\in$$\mathscr{A}$, $\forall$$\varepsilon$$\gt$$0$ $\exists$$\mathit{C}$$\in$$\mathscr{C}$ such that $\mu(\mathit{A}\bigtriangleup\mathit{C} )$ $\lt$ $\varepsilon$, where $\bigtriangleup$ is the symmetric difference of two sets, that is $\mathit{A}\bigtriangleup\mathit{C}$=$(\mathit{A}\setminus\mathit{C})$$\sqcup$$(\mathit{C}\setminus\mathit{A})$

The exercise itself has two more requests, but I managed to do them just fine. To show that this statement is true, as the author suggests, I tried to prove that the set:

$$\mathscr{D}=\{\, \mathit{A}\in\mathscr{A}\, \mid\, (\forall\varepsilon\gt0 \, \exists\mathit{C}\in\mathscr{C}\, \mid \, \mu(\mathit{A}\bigtriangleup\mathit{C})\lt\varepsilon)\, \}$$

is a Dynkin system containing $\mathscr{C}$, so that we may use the fact that by minimality, the smallest Dynkin system which contains $\mathscr{C}$ (which is also the sigma algebra generated by $\mathscr{C}$ since this set is closed under finite intersections) is contained in $\mathscr{D}$. My main concern is, upon attempting to prove that the disjoint union of countable sets from $\mathscr{D}$ is in $\mathscr{D}$ I fail to realize where my proof needs the fact that this union is actually disjoint. Moreover I think I don't even need the fact that the measure is finite. So I thought it'd be better posting my attempt at proving this on here to see if anyone can spot the error(s). That being said my proof is as follows:

Let $(\mathit{A_n})$$\subset$$\mathscr{D}$ be a countable sequence and $\varepsilon$$\gt$$0$ be arbitrary. By definition of $\mathscr{D}$, we may then find another countable sequence of sets $(\mathit{C_n})$$\subset$$\mathscr{C}$ such that $\mu(\mathit{A_n}\bigtriangleup\mathit{C_n})$$\lt$$(\frac{\varepsilon}{4})$$\frac{1}{2^n}$$\\,$$\forall$$n$$\in$$\mathbb{N}$. Write $\mathit{A}$=$\bigcup{\mathit{A_n}}$, and define $\mathit{G_n}$=$\bigcup_1^n{C_j}$. My claim is that $\lim_{n\to \infty}$$\mu(\mathit{A}\bigtriangleup\mathit{G_n})$$\leqslant$$\frac{\varepsilon}{2}$. To prove this, note first that $\mathit{G_n}$$\uparrow$$\mathit{C}$$=$$\bigcup{C_j}$, which in turn implies both that ($\mathit{G_n}$$\setminus$$\mathit{A}$)$\uparrow$($\mathit{C}$$\setminus$$\mathit{A}$), and ($\mathit{A}$$\setminus$$\mathit{G_n}$)$\downarrow$($\mathit{A}$$\setminus$$\mathit{C}$).

Next note that by the definition of the symmetric difference we have: $$\mu(\mathit{A}\bigtriangleup\mathit{G_n})=\mu(\mathit{A}\setminus\mathit{G_n})+\mu(\mathit{G_n}\setminus\mathit{A})$$ And by properties of limits and measures we then have $$\lim_{n\to \infty}\mu(\mathit{A}\bigtriangleup\mathit{G_n})=\lim_{n\to \infty}\mu(\mathit{A}\setminus\mathit{G_n})+\lim_{n\to \infty}\mu(\mathit{G_n}\setminus\mathit{A})=\mu(\mathit{A}\setminus\mathit{C})+\mu(\mathit{C}\setminus\mathit{A})$$ If we now write $\mathit{A}\setminus\mathit{C}$$\,$ as $\,$$\bigcup{\mathit{A_k}}\setminus\bigcup{\mathit{C_k}}$, we note that $\mathit{A}\setminus\mathit{C}$$\,$$\subseteq$$\,$$\bigcup({\mathit{A_k}\setminus\mathit{C_k}})$. Similarly one has that $\mathit{C}\setminus\mathit{A}$$\,$$\subseteq$$\,$$\bigcup{(\mathit{C_k}\setminus\mathit{A_k})}$. From this we know that $$\mu(\mathit{A}\setminus\mathit{C})+\mu(\mathit{C}\setminus\mathit{A})\,\le\,\mu(\bigcup({\mathit{A_k}\setminus\mathit{C_k}})+\mu(\bigcup{(\mathit{C_k}\setminus\mathit{A_k})})$$ And by properties of measures we may write a further inequality, which is: $$\mu(\bigcup({\mathit{A_k}\setminus\mathit{C_k}})+\mu(\bigcup{(\mathit{C_k}\setminus\mathit{A_k})})\,\le\,\sum{\mu(\mathit{A_k}\setminus\mathit{C_k})}+\sum{\mu(\mathit{C_k}\setminus\mathit{A_k})}$$ If we now recall how the sequence ($\mathit{C_n}$) was chosen, we note that both the arguments in the infinte sums are bounud to be less than or equal to $(\frac{\varepsilon}{4})$$\frac{1}{2^k}$, hence by a simple use of a geometric series we arrive at the inequality $$\mathit{l}=\lim_{n\to \infty}\mu(\mathit{A}\bigtriangleup\mathit{G_n})\,\le \, \frac{\varepsilon}{2}$$ Now we use the definition of a limit to find ourselves an number $\mathit{N}\in\mathbb{N}\,$ such that if $n\gt\mathit{N}\,$ then $$| \mu(\mathit{A}\bigtriangleup\mathit{G_n})-\mathit{l} |\,\lt \,\frac{\varepsilon}{2}$$ Then by properties of absolute values and the fact that our limit $\mathit{l}$ is less than or equal to $\frac{\varepsilon}{2}$, we easily get that $\forall n\gt\mathit{N}$ $$\mu(\mathit{A}\bigtriangleup\mathit{G_n})\, \lt \, \varepsilon$$ And since our set $\mathscr{C}$, is closed under finite unions, we managed to show that $\mathit{A}=\bigcup{A_n}$, is in $\mathscr{D}$, which in the end implies that this set is a sigma algebra.

Answer 1

I agree that you did not need the assumption that the $A_n$ are disjoint. So indeed, you actually proved directly that $\mathscr{D}$ is a $\sigma$-algebra (complements are easy) and so there is no need to use the Dynkin lemma. This is how the traditional proof goes; see for instance Approximating a $\sigma$-algebra by a generating algebra. Maybe the author is just so accustomed to using the Dynkin lemma (it is a very powerful tool) that they didn't notice it is unnecessary here.

But you did use the assumption that $\mu$ is finite. When you claimed that $$\lim_{n\to \infty}\mu(\mathit{A}\bigtriangleup\mathit{G_n})=\lim_{n\to \infty}\mu(\mathit{A}\setminus\mathit{G_n})+\lim_{n\to \infty}\mu(\mathit{G_n}\setminus\mathit{A})=\mu(\mathit{A}\setminus\mathit{C})+\mu(\mathit{C}\setminus\mathit{A})$$ you had $A \setminus G_n \, \downarrow\, A \setminus C$ and wanted to conclude that $\lim \mu(A \setminus G_n) = \mu(A \setminus C)$, the so-called "continuity from above" property. But that is only valid under the assumption that $\mu(A \setminus G_n) <\infty$ for all (sufficiently large) $n$.

Other than that, I think your proof is fine.