Show that $\mu(\bigcap\limits_{n=1}^{\infty}F_{n})=\lim\mu(F_{n})$ can fail if $\mu(F_1) = +\infty$

Question

Show that lemma 3.4 (b) in The Elements of Integration by Bartle may fail if the finiteness condition $\mu(F_{1})<+\infty$ is dropped.

Lemma 3.4(b): Let $\mu$ be a measure defined on a $\sigma$-algebra $X$. If $(F_{n})$ is decreasing sequence in $X$ and if $\mu(F_{1})<+\infty$, then $$\mu \Bigg(\bigcap\limits_{n=1}^{\infty}F_{n}\Bigg)=\lim\mu(F_{n})$$

Answer 1

You can see that in the proof of Lema 3.4(b) when he defines the sequence of sets $E_n=F_1 - F_n $ , you can't say that $\mu (E_n) = \mu (F_1) - \mu (F_n)$ if $\mu(F_1)$ is not finite.

This could get you in trouble because expression like $\infty - \infty$ aren't well defined in this context.

For a counter-example, you could use $X=\mathbb{R}$ as the set; the power set $\mathcal{P}(\mathbb{R})$ as the $\sigma$-álgebra and the following measure:

$$\mu (E)=\begin{cases} \vert E \vert \mbox{, if $E$ is finite}\\ +\infty\mbox{, otherwise} \end{cases}$$

Then use the sequence $F_n=(-\frac{1}{n},\frac{1}{n})$.

$\mu(\bigcap\limits _{n=1}^{\infty}F_n)=\mu (\{0\})=1 $, but $\lim\mu(F_n)=+\infty$.

Answer 2

As Robson said, please include your thoughts on the question. You'll get better responses if we know what you are thinking already.

However, I thought the question was interesting, so:

Hint: Can any of the sets have finite measure if the lemma is to fail?