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This is somewhat a basic question, but I'm having difficulty proceeding with a certain part of the proof. I was reading Billingsley "Convergence of Probability Measures", and I encountered the following question:

Question: Given a metric space $X$ and $\mathscr{B}$ the Borel sigma algebra, prove that a probability measure $P$ on $X$ is tight iff $$\forall A \in \mathscr{B},\quad \sup \{P(K): K\subset A, K \mbox{ compact}\} = P(A)$$

My Proof:

($\Rightarrow$) Put $A=X$. Then we have $$1 = P(X) = \sup \{P(K): K\subset X, K \mbox{ compact}\}$$ By definition of sup, we get $\forall \epsilon > 0$, $\exists K$ compact such that $$P(K) \geq 1-\epsilon$$ Thus P is tight.

($\Leftarrow$)

Billingsley suggests using the following theorem:

Probability measures on metric spaces are regular. That is $$\forall A \in \mathscr{B}, \forall \epsilon > 0, \exists F \mbox{ closed}, G \mbox{ open such that}$$ $$F \subset A \subset G, \quad P(G \setminus F) < \epsilon$$

I do not know how to use compactness here. I reasoned that if I could prove that $\forall \epsilon > 0$, $\exists K \subset A$ compact such that $P(A\setminus K) < \epsilon$, I would be done. Since the compact sets in "tightness", needn't be subsets of A, I don't know how to use it. I was able to show the easy inequality which is:

$$\sup \{P(K): K\subset A, K \mbox{ compact}\} \leq P(A)$$ (since $K\subset A$)

I would appreciate any ideas, hints and tips (if not answers) to this. References are also welcome. I searched for "Tight Regular Measures" and "Tight Probability" but to no useful results.

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First, we will see how Billingsley's claim helps to conclude. For each $n$, consider $K_n$ compact such that $\mu(K_n)>1-n^{-1}$. We can assume the sequence $(K_n,n\geqslant 1)$ to be increasing. Take $B\in\mathcal B(X)$ and $\varepsilon>0$. By the claim, there is $F\subset B$ such that $\mu(B\setminus F)<\varepsilon$. Fix $n$ such that $\varepsilon>n^{-1}$. Then $$\mu(B\cap K_n)=\mu(B)-\mu(B\setminus (B\cap K_n))=\mu(B)-\mu(B\setminus K_n)\\\geqslant \mu(B)-\mu(K_n^c)\geqslant \mu(B)-n^{-1}\geqslant \mu(B)-\varepsilon.$$ We thus get $\mu(F\cap K_n)>\mu(B)-2\varepsilon$, and $F\cap K_n$ is a compact subset of $B$.

Now we shall see the claim is true. First, note that in a metric space, an closed set is a countable intersection of open sets, and using the fact that the measure is finite, we deduce that the assertion is true for $B$ open.

Then the remaining task is to show that the collection of the $B$ for which the property holds is a $\sigma$-algebra.

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  • $\begingroup$ Thanks Davide. So it doesn't directly follow... Also I wasn't aware that a closed set was a countable intersection of open sets. Are you sure this doesn't require separability of the space?, since Billingsley quotes it for the general metric space. $\endgroup$ – Gautam Shenoy May 8 '13 at 9:20
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    $\begingroup$ If $F$ is closed, then $F:=\bigcap_{n\geqslant 1}O_n$, where $O_n:=\{x,d(x,O)< n^{-1}\}$, and there is no need of separability. $\endgroup$ – Davide Giraudo May 8 '13 at 9:58
  • $\begingroup$ No wait. I got it. Your statement "We thus get ... compact subset of B" completed the proof. $\endgroup$ – Gautam Shenoy May 8 '13 at 10:09
  • $\begingroup$ Hi Davide, did you just prove that any probability measure on Polish space is tight? For example here in this note, they have proved it but I am not understanding from the line $\because P$ is a probability measure then there is a $k$ for some $n_k$....., could you explain me ''why'' for this particular line is true? Here Thm 2.49 stat.columbia.edu/~porbanz/teaching/G6106S16/… $\endgroup$ – Marso Jul 3 at 10:14
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Thanks to Davide Giraudo for this:

The remaining steps are as follows. Let $K_n$, be a collection of compact sets such that $\forall n \geq 1$ $$P(K_n) > 1-\frac{1}{2n}$$

Now as per the provided theorem, given $A \in \mathscr{B}$, and $n \geq 1$ we have $F\subset A$ closed such that $$P(A) - P(F) = P(A \setminus F) < \frac{1}{2n}$$

Now as Davide Giraudo showed, $P(A \cap K_n) \geq P(A) - \frac{1}{2n}$ Add the previous two statements, and use the fact that $F \subset A$ to get $$P(A) - \frac{1}{n} \leq P(A \cap K_n) - P(A) + P(F)$$ $$ = P(F) - P(A \cap K_n^c) = P(F \cap K_n) + P(F \cap K_n^c)- P(A \cap K_n)$$ $$\leq P(F \cap K_n)$$

Now $F \cap K_n$ is a closed subset of a compact set in a metric space and hence is compact. Additionally $F \cap K_n \subset A$ !!.

Therefore we have $$P(A) \leq P(F \cap K_n) + \frac{1}{n} \leq \sup\{P(K): K \subset A, \mbox{K compact}\} + \frac{1}{n}$$ $$\Rightarrow P(A) \leq \sup\{P(K): K\subset A, \mbox{K compact}\} + \frac{1}{n}$$ Take $n \to \infty$ to get the desired inequality.

QED

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An alternative approach to prove $P(A) \leq \sup\{P(K): K \subset A, K \hspace{1mm} compact\}$.

For any given $n$, we can find $F_n$ closed such that $P(A) \leq P(F_n) + \frac{1}{n}$. (By regularity of the measure P)

Also, we can find $K_n$ compact such that $P(K_n) > 1 - \frac{1}{n}$ (since we assume P is tight). Now $F_n \cap K_n$ is compact with $P(F_n \cap K_n) = P(F_n) + P(K_n) - P(F_n \cup K_n)$.

Hence we have, $P(F_n) = P(F_n \cap K_n) - P(K_n) + P(F_n \cup K_n) \leq P(F_n \cap K_n) - 1 + \frac{1}{n} + 1$.

We thus have $P(A) \leq P(F_n \cap K_n) + \frac{1}{2n}$.

Therefore, for any $n$, we have $J_n := F_n \cap K_n \subset A$ and $J_n$ compact such that $P(A) \leq P(J_n) + \frac{1}{2n}$.

Since the above is true for all $n$, we have $P(A) \leq \sup\{P(J): J \subset A, J \hspace{1mm} compact\}$.

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Let $\epsilon>0$ be given. There are closed F (By Theorem. 1.1 of billingsly) and compact $K$ such that $$P(K)>1-\frac{\epsilon}{2},$$ i.e. $$P(K')<\epsilon/2,$$ and $$P(A-F)<\epsilon/2,$$ Then we have $P(A-(F\cap K ))<\epsilon$, which proposes $F\cap K$ as desired compact set.

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