Show that $\mu(a)=1$ for some $a$ if $\mu(f^{-1}([a_n, b_n)))=1$ and $\lim_{n\to\infty}(b_n-a_n)=0$.

Question

The problem arises from a solution to an exercise in Walter Rudin's Real & Complex Analysis.

The Problem: Let $X$ be an uncountable set, let $\mathfrak{M}$ be the collection of all sets $E\subset X$ such that either $E$ or $E^c$ is at most countable. Define $\mu(E)=0$ in the first case, $\mu(E)=1$ in the second. It can be shown that $\mathfrak{M}$ is a $\sigma$-algebra and $\mu$ is a measure on $\mathfrak{M}$.

Now, let $f: X\to\mathbb{R}$ be a measurable function. Since $X$ is uncountable, we have $\mu(X)=1$. For every $n\in\mathbb{Z}$, we know that $f^{-1}([n, n+1))=f^{-1}(\mathbb{R})-[f^{-1}((-\infty, n))\cup f^{-1}([n+1, +\infty))]$, thus $f^{-1}([n, n+1))\in\mathfrak{M}$. Let $E_n=f^{-1}([n, n+1))$, then either $\mu(E_n)=0$ or $\mu(E_n)=1$. Note $X=\bigcup_{n=-\infty}^{+\infty}E_n$, and $\{E_n\}$ is pairwise disjoint, thus $\mu(X)=\mu(\bigcup_{n=-\infty}^{+\infty}E_n)=\sum_{n=-\infty}^{+\infty}\mu(E_n)$. Then $\mu(f^{-1}([n, n+1)))=1$ for some $n\in\mathbb{N}$; WLOG, suppose $n=0$, i.e., $\mu(f^{-1}([0, 1)))=1$. If we write $[0, 1)=[0, \frac{1}{2})\cup[\frac{1}{2}, 1)$, then it is clear that either $\mu(f^{-1}([0, \frac{1}{2})))=1$ or $\mu(f^{-1}([\frac{1}{2}, 1)))=1$. It follows that there is a sequence of intervals $\{[a_n, b_n)\}$ such that $\mu(f^{-1}([a_n, b_n)))=1, 0\leq b_n-a_n\leq\frac{1}{2^n}$. In other words, it means that $\mu(f^{-1}(a))=1$ and $\mu(f^{-1}(b))=0$ for some $a\in\mathbb{R}$ and all other real numbers $b\neq a$.

My Question: Why do we have "In other words, it means that $\mu(f^{-1}(a))=1$ and $\mu(f^{-1}(b))=0$ for some $a\in\mathbb{R}$ and all other real numbers $b\neq a$"? Indeed, $\underset{n\to\infty}{\lim}b_n-a_n=0$; but how exactly do we find the $a\in\mathbb{R}$ such that $\mu(f^{-1}(a))=1$?

Answer 1

• $A_n:=[a_n,b_n)$. $B_n:=[a_n,b_n]$, $A'_n=f^{-1}(A_n)$, $B'_n=f^{-1}(B_n)$.
• We know that $1=\mu(A'_n)\leq \mu(B'_n)\leq \mu(X) = 1$, so $\mu(B'_n)$=1.
• $B_1\supset B_2\supset B_3\ldots$ and $\mathrm{diam}(B_n)\to 0$, so from Cantor's intersection theorem we get $\bigcap_{n=1}^\infty B_n=\{a\}$.
• $B'_1\supset B'_2\supset B'_3\ldots$ and $\mu(B'_n)=1$, so from continuity of measure we get $$\mu\left(f^{-1}(\{a\})\right)=\mu\left(\bigcap_{n=1}^\infty B'_n\right)=\lim_{n\to\infty}\mu(B_n') =1.$$
• If $b\neq a$ then $$1+\mu\left(f^{-1}(\{b\})\right)=\mu\left(f^{-1}(\{a\})\right)+\mu\left(f^{-1}(\{b\})\right)\leq 1.$$ Therefore $\mu\left(f^{-1}(\{b\})\right)=0$.