I want to prove that the Lebesgue measure is nonatomic. Here's the definition of nonatomic that I use:
Let $(X, \mathscr A, \mu)$ be a measure space. An element $A \in \mathscr A$ is said to be a $\mu$ atom if $\mu (A) > 0$ and for all $B\in \mathscr A$, we have that $\mu (A \cap B) =0$ or $\mu (A \cap B ^{c}) =0$. The measure space $(A, \mathscr A, \mu)$ is called nonatomic if there are no $\mu$-atoms.
Here's my attempt:
Suppose that $A \in \mathcal B (\mathbb R )$ be an atom. So, $m (A) >0$. Then by definition of atom, for each $x \in \mathbb R$, we have that $m (A \cap (-\infty , x]) = 0$ or $m (A \cap (x, \infty)) = 0$ ($\star$).
Define $L= \{ x\in \mathbb R : m (A \cap (-\infty , x]) = 0 \}$ and $R = \{ x \in \mathbb R : m (A \cap (x, \infty)) = 0 \} $. We claim that $\mathbb R = L \sqcup R$. First of all from $(\star)$, we have that $\mathbb R = L \cup R$. To see that $L \cap R = \emptyset$, suppose that $x \in L \cap R$. Then we have that $m (A \cap (-\infty , x]) = 0$ and $m (A \cap (x, \infty)) = 0$. Hence, $m (A) = m (A \cap (-\infty, x]) + m (A \cap (x, \infty)) =0$ which is contrary to our assumption.
We claim that $x \le y$ for all $x \in L$ and $y \in R$ $(\star \star)$. Suppose not; then $y < x$ for some $y \in R$ and some $x \in L$. Hence, $A = (A \cap (-\infty , x]) \cup (A \cap (y, \infty))$. Thus, $m (A) \le m (A \cap (-\infty , x]) + m (A \cap (y, \infty)) = 0$.
We claim that both $L$ and $R$ are nonempty. If $R =\emptyset$ then $L=\mathbb R$. Since $A \cap (-\infty , n]$ is a sequence of increasing sets increasing to $A$ all of whose measure is $0$, we have that $m (A) = 0$ which is contrary to our assumption.
Thus, we can let $\alpha = \sup L$ and $\beta = \inf A$. It is evident from $(\star \star)$ that $\alpha \le \beta$. We claim that $\alpha = \beta$. If not, we can select $x \in (\alpha , \beta)$. By the fact that $\mathbb R = L \sqcup R$, we must have that $x \in L$ or $x \in R$ but if $x \in L$ then $x \le \alpha$ which cannot happen because $\alpha < x$ and similar argument applies for $y \in R$. Thus, $\alpha = \beta$.
Now, we can select a nondecreasing sequence $\{ x_n \}$ in $L$ converging to $\alpha$. Then $\bigcup_{n} (-\infty , x_n] = (-\infty , \alpha]$. Thus, $\bigcup_{n} \left( (-\infty , x_n] \cap A \right)= (-\infty , \alpha] \cap A$. Since $\left( (-\infty , x_n] \cap A \right)$ is a sequence of increasing sets with measure $0$, we have that $m((-\infty , \alpha] \cap A)=0$. Likewise we can select a nonincreasing sequence $\{ y_n \}$ in $R$ converging to $\beta = \alpha$ and from this, we can conclude that $m((\alpha, \infty ) \cap A)=0$. Thus, $m (A) = 0$ which is a contradiction and completes the proof.
Is my attempt correct? I am also looking for other ways to prove it. So, alternative proofs are welcome. $\smile$
