2
$\begingroup$

Suppose $Q$ is a probability measure on a Euclidean space $\mathbf{E}$. Suppose $\mathcal{K}$ is the collection of all compact sets on $\mathbf{E}$.

Consider the function $f: \mathcal{K} \rightarrow [0,\infty)$ defined as $$f(K) = -\log(1 - Q(K)) $$ for every compact set $K$.

$f$ is a superadditive set function in the sense that, given two disjoint $K_1, K_2 \in \mathcal{K}$, $$ f(K_1 \cup K_2) = - \log(1 - Q(K_1) -Q(K_2)) \geq -\log(1 - Q(K_1) - Q(K_2) + Q(K_1)Q(K_2)) = -\log((1-Q(K_1))(1-Q(K_2)) = f(K_1) + f(K_2). $$

My question is whether there exists a locally finite measure $\Lambda$ (not a probability measure) on $\mathbf{E}$ such that

$$ \Lambda(K) \geq f(K)$$

for all compact $K$.

If such a $\Lambda$ exists, then it must satisfy:

(1) $\Lambda(K) \geq Q(K)$ for all $K$ since $f(K) \geq Q(K)$.

(2) $\Lambda(\mathbf{E}) = +\infty$ since $f(K) \rightarrow +\infty$ as $K \nearrow \mathbf{E}$.

Yet, the tricky part is that $\Lambda$ is additive but $f$ is superadditive for disjoint sets.

I played around the case where $\mathbf{E} = \mathbb{R}$ and $Q$ being a standard Gaussian, but have not found a construction for such $\Lambda$.

$\endgroup$
3
  • $\begingroup$ Please be more precise --- what measures do you allow for $\Lambda$? Possibly infinite on compact sets but determined by their finite values? $\endgroup$
    – user95282
    Jul 2, 2019 at 1:32
  • $\begingroup$ In what sense is there such a measure $\Lambda$ when $Q$ is supported on a singleton? $\endgroup$
    – user95282
    Jul 2, 2019 at 1:39
  • $\begingroup$ @user95282 Sorry for the confusion. Let us assume that $\Lambda$ is locally finite and $Q$ is supported on $\mathbf{E}$. $\endgroup$ Jul 2, 2019 at 4:29

1 Answer 1

1
$\begingroup$

(Edited once more, to eliminate extraneous symbols.)

There is such a measure, assuming $Q(K)<1$ for every compact set. The assumption holds if the support of $Q$ is the whole $E$.

Define $B_0=\emptyset$ and for $j=1,2,\dots$ let $B_j$ be the closed ball with center at the origin and radius $j$. By the assumption $Q(B_j)<1$, hence $f(B_j)<\infty$, for every $j$.

Define constants $c_j$ for $j=1,2,\dots$ so that $$ c_{j} Q(B_j \setminus B_{j-1}) = f(B_j)- f(B_{j-1}) $$ (if $Q(B_j \setminus B_{j-1}) = f(B_j)- f(B_{j-1}) =0$ then choose $c_j$ arbitrarily, for example $0$ or $1$). The function $x\mapsto -\log(1-x)$ is convex, hence $$ c_{j} Q(K\cap(B_j \setminus B_{j-1})) \geq f(K\cap B_j)- f(K\cap B_{j-1}) $$ for every compact set $K$. By summing that from $1$ to $n$ we get $$ \sum_{j=1}^n c_{j} Q(K\cap(B_j \setminus B_{j-1})) \geq f(K\cap B_n). $$

Define the measure $\Lambda$ by $\Lambda(A) = \sum_{j=1}^\infty c_j Q(A\cap (B_j \setminus B_{j-1}))$ for Borel sets $A$. For every compact set $K\subseteq E$ there is $n$ such that $K\subseteq B_n $, hence $\Lambda(K)$ is finite and $$ \Lambda(K) = \sum_{j=1}^n c_j Q (K\cap (B_j \setminus B_{j-1})) \geq f(K \cap B_n) = f(K). $$

$\endgroup$
2
  • $\begingroup$ Thanks! I am not sure I understand these steps. (i) How does the existence such $\Lambda_j$ follow from the supermodularity of $f$? (ii) How did you get the inequality from convexity? By definition and superadditivity of $f$, $c_{j+1} \geq 1$. Are you using convexity with weights $c_{j+1} / (1 + c_{j+1})$ and $1 / (1 + c_{j+1})$? I cannot get what you have there with these weights. (iii) Why is $\Lambda = \sum_{j=1}^{\infty} c_j \Lambda_j$ locally finite? $\endgroup$ Jul 2, 2019 at 22:38
  • $\begingroup$ @RichardGuo You were right about problems with the previous answer. They are corrected now, I hope. $\endgroup$
    – user95282
    Jul 3, 2019 at 2:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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