7
$\begingroup$

If $E_1 \supset E_2 \supset ...$ and $\mu(E_1)<\infty$ then $\mu(\bigcap E_j)=\lim \mu(E_j)$. But why need $\mu(E_1)<\infty$? Is $(-\infty,-n)$ an counter example?

$\endgroup$
0
6
$\begingroup$

Assume you have a family of sets $E_n = [n, +\infty)$ and a Lebesgue measure $\mu$. Then $\mu( \bigcap E_n) = \mu(\emptyset) = 0$ on the other hand for each $n$ $\mu(E_n) = \infty$ so $\lim_{n \to \infty} \mu(E_n) = \infty$

This fact works because proof (as I know it ) of the theorem in question relies on the fact that $$\mu(E_1) - \mu(\bigcap E_n) =\lim_{n \to \infty} \mu(E_1 \setminus E_n) = \mu(E_1) - \lim_{n \to \infty} \mu(E_n)$$ which cannot be correctly processed if $\mu(E_1) = \infty$ by definition of algebra for extended real numbers. To be more general we will have equality $$ \mu(\bigcap E_n) - \lim_{n \to \infty} \mu(E_n) = \infty - \infty $$ which is indeterminant.

The straight implication of this fact is that many theorems of probability won't work in case of general measures e. g. Egoroff theorem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.