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?


1 Answer 1


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.


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