Can a superadditive set function be upper-bounded by a measure?

Question

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 $\Lambda$.

Answer 1

(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 $$ \Lambda(K) = \sum_{j=1}^n c_j Q (K\cap (B_j \setminus B_{j-1})) \geq f(K \cap B_n) = f(K). $$