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For $r>0$, write $m_r=v_d{r}^d$, $a_r={\int }_{B(0,r)}{e}^{-f(x)}dx$, and $L=\mathbb{E}{e}^{-M(f)}$. The count-weighted Laplace functional of a Poisson random measure and the Poisson distribution giveTo compute the other numerator, on $\{N_r=0\}$ the inside contribution to $M(f)$ vanishes. The restrictions of the Poisson random measure to the ball and its complement are independent; this follows first for their simple-function integrals from independent counts, then by approximation. HenceSubtract from $L$ and divide by $\mathbb{P}(N_r\ge 1)$ to findContinuity of $f$ at zero givesAlso $(1-e^{-z})/z\to 1$ as $z\downarrow 0$. Applying these facts to the two exact expressions proves the small-ball conditioning for a Poisson random measure:One can also see equality of the limits without the exact first formula: since $0\le e^{-M(f)}\le 1$, the difference of the two quantities lies between zero and $[m_r-(1-e^{-m_r})]/(1-e^{-m_r})\to 0$. The conditioning is used only at positive $r$, where its event has positive probability; it is not conditioning directly on a point at the origin.

First derive the Laplace functional of a Poisson random measure. For a simple nonnegative $g={\sum }_j{t}_j{1}_{A_j}$ on disjoint finite-intensity sets, independent Poisson random variables giveIncreasing simple approximations, monotone convergence for the integrals, and bounded convergence theorem for the random exponentials prove the same identity for every nonnegative measurable $g$.

Here the intensity is Lebesgue measure. Put $A=B(0,r)$ andThe integral is finite because $f$ has compact support. Apply the Laplace functional of a Poisson random measure to $f+t{1}_A$, $t\ge 0$:Differentiate from the right at $t=0$. The difference quotient on the random side is dominated by $N_r$, which is integrable with mean $v_d{r}^d$; the intensity integral is over a finite-volume ball. We obtain the count-weighted Laplace functional of a Poisson random measure:At $r=0$, the ball is empty and both sides are zero.

For a Poisson random measure on ${\mathbb{R}}^d$ with Lebesgue measure intensity and a nonnegative compactly supported continuous function $f$, put $m_r=\mu (B(0,r))$ and $a_r={\int }_{B(0,r)}{e}^{-f}d\mu$. Continuity gives $a_r/m_r\to e^{-f(0)}$. With $L=\mathbb{E}{e}^{-M(f)}$, the count-weighted Laplace functional of a Poisson random measure and independent inside/outside counts give respectivelyBoth converge to the displayed value. The conditioning is meaningful for every $r>0$; it is not conditioning on a positive-probability point at the origin.