Equivalence relation of atoms

Question

If $(X,\mathcal{S},\mu)$ is a measure space, a set $A\in\mathcal{S}$ is called an atom of $\mu$ iff $0<\mu(A)<\infty$ and for every $C\subset A$ with $C\in\mathcal{S}$, either $\mu(C)=0$ or $\mu(C)=\mu(A)$.

For two atoms $A$ and $B$, define a relation $A\approx B$ iff $\mu(A\cap B)=\mu(A)$.

For $\mu$ finite, show that $\approx$ is an equivalence relation.

My efforts:

$A\approx A$ since $\mu(A\cap A)=\mu(A)$.

Suppose $A\approx B$. Since $A$ is an atom, $\mu(A)>0$. Since $A\approx B$, $\mu(A\cap B)=\mu(A)>0$. Since $B$ is an atom and $A\cap B\subset B$, either $\mu(A\cap B)=0$ or $\mu(A\cap B)=\mu(B)$. Since $\mu(A\cap B)>0$, we must have $\mu(A\cap B)=\mu(B)$. So $B\approx A$.

Suppose $A\approx B$ and $B\approx C$. We need to show $A\approx C$. Since $C$ is an atom and $A\cap C\subset C$, either $\mu(A\cap C)=0$ or $\mu(A\cap C)=\mu(C)$. We only need to show $\mu(A\cap C)>0$. Assume $\mu(A\cap C)=0$. Then $0<\mu(A)=\mu(A\cap C)+\mu(A\setminus C)=\mu(A\setminus C)$. Since $A\setminus C\subset A\setminus(B\cap C)$, we have $\mu(A\setminus C)\leq\mu(A\setminus (B\cap C))=\mu(A)-\mu(A\cap B\cap C)$.

Then I don't know how to proceed.

Answer 1

The idea is that $A\approx B$ iff the difference between $A$ and $B$ is negligible (in measure). We do just that: Suppose $A\approx B$ and $B\approx C$.

Notice that \begin{align*} A\setminus C &\subseteq ((A\setminus B)\cup B)\setminus C\\ &=\left[(A\setminus B)\setminus C\right]\cup \left[B\setminus C\right]\\ &\subseteq (A\setminus B)\cup (B\setminus C) \end{align*} so $\mu(A\setminus C)\leq\mu(A\setminus B)+\mu(B\setminus C)$.

Now we have the decomposition of $A$ as a disjoint union $A=(A\setminus B)\sqcup (A\cap B)$, so from $A\approx B$ we get $$\mu(A)=\mu(A\setminus B)+\mu(A\cap B)=\mu(A\setminus B)+\mu(A),$$ so $\mu(A\setminus B)=0$. Similarly, from $B\approx C$ you get $\mu(B\setminus C)=0$.

Then $$0\leq\mu(A\setminus C)\leq\mu(A\setminus B)+\mu(B\setminus C)=0$$ Therefore $\mu(A\setminus C)=0$. Now we decompose $A=(A\setminus C)\sqcup(A\cap C)$ to obtain $$\mu(A)=\mu(A\setminus C)+\mu(A\cap C)=0+\mu(A\cap C)=\mu(A\cap C),$$ which means that $A\approx C$.

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So in the end, what I'm doing is using the following lemma (which follows by decomposing $A=(A\setminus B)\cup (A\cap B)$) a few times, which rewrites $\approx$ in terms of differences and makes some computations easier:

For atoms $A,B$, the following are equivalent:

• $A\approx B$;
• $\mu(A\setminus B)=0$.

Answer 2

Use $\mu((A \cap B) \cup (B \cap C)) = \mu(A \cap B) + \mu(B \cap C) - \mu(A \cap B \cap C)$. $A \cap B \subseteq B$ and $B \cap C \subseteq B$ so $(A \cap B) \cup (B \cap C) \subseteq B$. By monotonicity, \begin{equation*} 0 < \mu(A \cap B) \leq \mu((A \cap B) \cup (B \cap C)) \end{equation*} so $\mu((A \cap B) \cup (B \cup C)) = \mu(A) = \mu(B) = \mu(C)$. From this, we conclude that $\mu(A \cap B \cap C) = \mu(A) + \mu(A) - \mu(A) = \mu(A)$ and it follows that $\mu(A \cap C) = \mu(A) = \mu(C)$.

Answer 3

I'd do it in a different way (each step is justified by the previous one):

1. $\mu(B\cup C)=\mu(B)+\mu(C)-\mu(B\cap C)=\mu(B\cap C)+\mu(C)-\mu(B\cap C)=\mu(C)$

2. $\mu((A\cap B)\cup C)=\mu((A\cup C)\cap(B\cup C))\le\mu(B\cup C)=\mu(C)$

3. $\mu(A\cap B\cap C)=\mu(A\cap B)+\mu(C)-\mu((A\cap B)\cup C)\ge\mu(A\cap B)+\mu(C)-\mu(C)=\mu(A\cap B)=\mu(A)$

4. $\mu(A)\ge\mu(A\cap C)\ge\mu(A\cap B\cap C)\ge\mu(A)\Rightarrow\mu(A\cap C)=\mu(A)$