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Free [DOWNLOAD] Now.. Buy Intuit TurboTax All Editions for Mac now for $99.99. It is available at the lowest price of $99.99 and saves you $10.00 over the list price.Q:

probability theory question about proof

Is there a better way to prove the theorem?

$P(\Omega\backslash A_n) \to 0$ as $n \to \infty$.

Assume $\Omega \subset \mathbb R^k$, $\forall n \in \mathbb N \; A_n \subset \Omega \;$ and $\Omega$ is open.

Let $A_n$ have the property that if $x\in \Omega \;$ then $x \in A_n$ for all $n\in \mathbb N \;$

then

$\Omega\backslash A_n \subset \Omega\backslash A_{n+1}$

Proof:

Let $x\in \Omega\backslash A_n$ then $x\in \Omega \;$ but $x

otin A_{n+1}$

then $x \in \Omega \backslash A_{n+1}$

let $n \to \infty \;$

since $A_n \subset \Omega \; \forall n\in \mathbb N$

we have $\Omega \backslash A_n \subset \Omega \backslash A_{n+1}$

now

$\Omega \backslash A_n \subset \Omega \backslash A_{n+1} \implies \Omega \backslash A_n \subset \Omega \backslash A_{\infty}$

which implies $P(\Omega \backslash A_n) \to 0 \implies P(\Omega \backslash A_{\infty})=0 \implies A_{\infty} = \emptyset$

Now suppose $A_{\infty}

eq \emptyset$

then $A_{\infty} \subset \Omega \;$

since $\Omega$

 

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