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The following inequality appears in the proof of this:

$$ P(\Gamma_\epsilon \triangle \Lambda) \leq P(\Lambda - G) + P(G\triangle \Gamma_\epsilon). $$

Why is it true? I know it uses subadditivity, but I'm not sure that the last equality in $\Lambda \triangle \Gamma_\epsilon = (\Lambda -G \cup G)\triangle \Gamma_\epsilon =(\Lambda- G) \cup (G\triangle \Gamma_\epsilon)$ is valid.

EDIT:

The details are as follow. There is a theorem that states: "Let $\mathcal{F}_0 \subset \mathcal{F}$ be a field and $\Lambda\in \sigma(\mathcal{F}_0)$. For any $\epsilon>0$, there exists $\Gamma_\epsilon \in\mathcal{F}_0$ such that $P(\Lambda \triangle \Gamma_\epsilon)< \epsilon$." The proof goes to defining the set $\mathcal{G} = \lbrace \Lambda\in\mathcal{F} : {\rm for\;all} \;\epsilon>0\; {\rm there\;is}\;\Gamma_\epsilon\in \mathcal{F}_0 \;{\rm such\;that}\; P(\Lambda\triangle \Gamma_\epsilon < \epsilon) \rbrace$. Then the (part of) proof, let $\epsilon>0$ and let $\Lambda_1\subset\Lambda_2\subset\dots \in \mathcal{G}$. Choose $n$ such that $P(\Lambda - G)< \epsilon /2$, with $G=\cup_{j=1}^n \Lambda_j$. For $j=1,\dots,n$ choose $\Gamma_j \in\mathcal{F}_0$ such that $P(\Lambda_j \triangle \Gamma_j)<\epsilon/2^{2j+1}$. Then set $\Gamma_\epsilon = \cup_{j=1}^n \Gamma_j$. Note that $\Gamma_\epsilon \in\mathcal{F}_0$ since $\mathcal{F}_0$ is a field, and $P(\Gamma_\epsilon \triangle \Lambda) \leq P(\Lambda - G) + P(G\triangle \Gamma_\epsilon)$ (...)

user2820579
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  • What is $G {}$? – William M. Apr 25 '22 at 14:33
  • I'm voting to close this question as you do not know what you are trying to ask. You change the question after receiving a valid answer to the original. If you have a doubt about how the proof of the theorem (that you wrote in the edit), you should ask that doubt in a separate question. – William M. Apr 25 '22 at 16:53
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    There is a very short proof of your result using the $\pi$-$\lambda$ theorem. – Mason Apr 26 '22 at 01:40
  • I'm very curious, have some reference you may recommend? – user2820579 Apr 26 '22 at 18:30
  • Notice that $P[A\triangle B]=E[\mathbb{1}_A-\mathbb{1}_B]=\int|\mathbb{1}_A-\mathbb{1}_B|,dP$ and that if $A\subset B$, $P[A\triangle B]=P[B\setminus A]$ since $|\mathbb{1}_A-\mathbb{1}_B|=\mathbb{1}_B-\mathbb{1}_A$ in this case. – Mittens Apr 27 '22 at 16:53

1 Answers1

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I am now assuming that $\Delta = \bigcup\limits_{n = 1}^\infty \Delta_n,$ but this is not stated (you need to learn to properly ask self-contained questions). In fact, the proof below only assumes that $G \subset \Delta$ and $\Gamma$ is another set.

By definition, $\Gamma \triangle \Delta$ is the set of points that belong to one but not both of $\Gamma$ or $\Delta.$ Clearly then $\Gamma \triangle \Delta \subset (\Gamma \triangle G) \cup (\Delta \setminus G)$ for $$ \begin{align*} \Gamma \triangle \Delta &= (\Gamma \setminus \Delta) \cup (\Delta \setminus \Gamma) \\ &= (\Gamma \setminus \Delta) \cup (G \setminus \Gamma) \cup (\Delta \setminus (G \cup \Gamma)) \\ &\subset (\Gamma \setminus G) \cup (G \setminus \Gamma) \cup (\Delta \setminus (G \cup \Gamma)) \\ &\subset (\Gamma \triangle G) \cup (\Delta \setminus G). \end{align*} $$

William M.
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  • Something that often causes confusion is that $A \setminus B = A \setminus (A \cap B)$ (you only take away those points of $B$ that are in $A$). Therefore, $(A \setminus B) \setminus C = A \setminus (B \cup C).$ – William M. Apr 25 '22 at 15:23
  • I think it's ok, but some things are now the other way around. See my edit... – user2820579 Apr 25 '22 at 16:18
  • Your edit does not connect very well with the original question except that $G \subset \Delta.$ What I wrote should clarify every doubt of the original question, if not, that means you do not really know what you are trying to ask (this is an unfortunate recurrent occurrence here in MSE). – William M. Apr 25 '22 at 16:32
  • Chill out... this is as it comes in my book. You can check the reference, it's Walsh Knowing the odds, AMS 2010, p6. Part of the question is precisely this, I don't understand what the book is assuming! – user2820579 Apr 25 '22 at 16:34