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Let $P$ be a probability on $R$. Prove that for every $\epsilon > 0$ there is a compact set $K$ such that $P(K) > 1 - \epsilon$.

First poster here, I read the H.W. FAQ, first course in non-measure based probability theory so only the most basic definitions are known.

Here are my thought thus far:

Let $F_p$ be the distribution function of $P$ such that: $F_p$ = $P((\infty, t])$

Because $F_p$ is increasing and right continuous, an interval $(a,b]$ must exist such that $P((a,b]) > 1- \epsilon$. (I am at a loss as to how to show this)

Then there is the issue of the interval being a compact set $K$. I have never learned about compact sets, from my understanding the interval $(a,b]$ is not a compact set, while the interval $[a,b]$ is a compact set.(Not sure about this either...)

Thanks for all your input!

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    Satis$f$ied with an answer below?2012-09-19

2 Answers 2

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Hint: Let $K_N=[-N,N]$. Then $P(K_N)\to P(\mathbb R)=1$ when $N\to+\infty$.

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Hint: You know, I hope, that $\lim_{x \to +\infty} F_p(x) = 1$ and $\lim_{x \to -\infty} F_p(x) = 0$ (if you don't know those, prove them using countable additivity). Take the compact set to be $[a,b]$ where $b$ is a large enough positive number that $F_p(b) > \ldots$ and $a$ is a large enough negative number that $F_p(a) < \ldots$.