Compact Exhaustions of Noncompact Parameter Spaces

Why Truncate a Prior on a Noncompact Space?

In admissibility arguments, an integrated risk under a prior can turn a pointwise comparison of decision rules into a useful global quantity. On a noncompact parameter space, however, the prior may have infinite total mass, or an integral over the entire space may be difficult to handle directly. A natural approach is to first integrate over compact regions and then let those regions increase. The key question is not just whether the regions cover the space, but whether the integrals over them converge in the way the argument requires.

A compact exhaustion gives a systematic sequence of such regions. For nonnegative risks, increasing the regions guarantees convergence of the truncated integrals to the full integral, though that full integral may be infinite. Signed risk differences require more care: cancellations can make truncated integrals misleading unless there is adequate control of their tails.

Compact Exhaustions and Truncated Integrals

Definition: A compact exhaustion of a topological space \(\Theta\) is a sequence of compact sets \(K_1\subseteq K_2\subseteq\cdots\) such that \(\bigcup_{m=1}^{\infty}K_m=\Theta\). Given a measure \(\pi\) on the measurable subsets of \(\Theta\), its truncation to \(K_m\) is the restricted measure \(\pi_m(A)=\pi(A\cap K_m)\). For a measurable function \(R:\Theta\to[0,\infty]\), the truncated integral is \(I_m=\int_{K_m}R(\theta)\,\pi(d\theta)\).

A compact exhaustion need not exist for every topological space. A common sufficient setting is a locally compact, second-countable space; in particular, \(\mathbb{R}^d\) has the exhaustion \(K_m=[-m,m]^d\). For the truncation to be useful as a prior on a compact set, it is also common to assume that \(\pi(K_m)<\infty\). A measure with this property along the exhaustion is locally finite on these sets. A probability prior automatically has finite mass on each \(K_m\).

The definition does not require the sets to be strictly increasing, and an exhaustion is about covering the whole parameter space, not about assigning equal prior mass to its pieces. In applications, \(\pi\) may be a proper prior or a locally finite measure with infinite total mass. For either kind of measure, the unnormalized truncated integral is precisely the integral over \(K_m\).

Worked Example: Exhausting an Open Parameter Interval

Take \(\Theta=(0,1)\), with its usual topology. For integers \(m\geq 3\), define \(K_m=[1/m,1-1/m]\).

Compactness and nesting. Each \(K_m\) is closed and bounded in \(\mathbb{R}\), so it is compact. Since \(1/(m+1)<1/m\) and \(1-1/(m+1)>1-1/m\), we have \(K_m\subset K_{m+1}\).

Coverage. For any \(\theta\in(0,1)\), choose an integer \(m\geq3\) large enough that \(1/m\leq\theta\) and \(1/m\leq1-\theta\). Then \(\theta\in K_m\). Thus the sets exhaust \((0,1)\), even though neither endpoint belongs to the parameter space.

For Lebesgue measure and the constant risk \(R(\theta)=1\), the truncated integral is \(\int_{K_m}1\,d\theta=1-2/m\), which tends to \(1=\int_{(0,1)}1\,d\theta\). This example also shows why an exhaustion of an open parameter space need not contain its boundary points.

Nonnegative Risks: Monotone Convergence

Theorem: Let \((\Theta,\mathcal{A},\pi)\) be a measure space, let \(K_m\in\mathcal{A}\) increase to \(\Theta\), and let \(R:\Theta\to[0,\infty]\) be measurable. Then \(\int_{K_m}R\,d\pi\) increases to \(\int_{\Theta}R\,d\pi\), with the equality understood in \([0,\infty]\).

This is the monotone convergence theorem applied to the functions \(R_m=R\mathbf{1}_{K_m}\). These functions are measurable, nonnegative, and increase pointwise to \(R\): if \(\theta\in K_j\), then \(\theta\in K_m\) for all \(m\geq j\), so \(R_m(\theta)=R(\theta)\) from then on. Consequently,

\lim_{m\to\infty}\int_{K_m}R(\theta)\,\pi(d\theta) = \lim_{m\to\infty}\int_{\Theta}R(\theta)\mathbf{1}_{K_m}(\theta)\,\pi(d\theta) = \int_{\Theta}R(\theta)\,\pi(d\theta).

In particular, if the full integral is finite, the truncated integrals approximate that finite value. If the full integral is infinite, the truncated integrals still increase to it; one must not describe them as converging to a finite integrated risk.

Worked Example: Truncating a Finite Integrated Risk

Let \(\Theta=\mathbb{R}\), take \(K_m=[-m,m]\), let \(\pi\) be Lebesgue measure, and set \(R(\theta)=1/(1+\theta^2)\). This risk is measurable and nonnegative, so the theorem applies.

An antiderivative is \(\arctan(\theta)\). Therefore,

I_m=\int_{-m}^{m}\frac{1}{1+\theta^2}\,d\theta =\arctan(m)-\arctan(-m) =2\arctan(m).

As \(m\to\infty\), \(\arctan(m)\to\pi/2\), and hence \(I_m\to\pi\). Directly, the full integral is \(\int_{-\infty}^{\infty}(1+\theta^2)^{-1}\,d\theta=\pi\), in agreement with monotone convergence. Each individual truncation has finite mass, since the Lebesgue measure of \(K_m\) is \(2m\), even though Lebesgue measure on all of \(\mathbb{R}\) is infinite.

Signed Risk Differences Need Tail Control

In an admissibility comparison, the function of interest may be a difference \(g(\theta)=R(\theta,\delta)-R(\theta,\delta_0)\) between the risks of two rules. Unlike a risk itself, \(g\) can be positive in some regions and negative in others. The monotone convergence theorem does not apply to \(g\) as though it were nonnegative. A useful sufficient condition is absolute integrability.

Theorem: Let \((\Theta,\mathcal{A},\pi)\) be a measure space, let \(K_m\in\mathcal{A}\) increase to \(\Theta\), and let \(g:\Theta\to\mathbb{R}\) be measurable with \(\int_{\Theta}|g|\,d\pi<\infty\). Then \(\int_{K_m}g\,d\pi\to\int_{\Theta}g\,d\pi\).

Proof. The functions \(g\mathbf{1}_{K_m}\) are measurable and converge pointwise to \(g\), since every point eventually belongs to \(K_m\). Also, \(|g\mathbf{1}_{K_m}|\leq |g|\), and the dominating function \(|g|\) is integrable by hypothesis. The dominated convergence theorem therefore gives

\lim_{m\to\infty}\int_{K_m}g\,d\pi = \lim_{m\to\infty}\int_{\Theta}g(\theta)\mathbf{1}_{K_m}(\theta)\,\pi(d\theta) = \int_{\Theta}g(\theta)\,\pi(d\theta).

Thus the signed comparison converges, and the absolute-integrability condition prevents the limit from relying on uncontrolled cancellation between tails. \(\square\)

For risk differences, a simple sufficient condition is that both risks have finite integrals: if \(\int R(\theta,\delta)\,d\pi<\infty\) and \(\int R(\theta,\delta_0)\,d\pi<\infty\), then \(|g|\leq R(\theta,\delta)+R(\theta,\delta_0)\), so \(g\) is absolutely integrable. In some applications, one can instead establish absolute integrability of the difference directly, even when the separate risk integrals are infinite. What matters for this theorem is the integrability of \(|g|\), not finiteness of the two integrals separately.

Worked Example: A Signed Difference with Controlled Tails

Let \(\Theta=\mathbb{R}\), \(K_m=[-m,m]\), and let \(\pi\) be Lebesgue measure. Consider \(g(\theta)=e^{-|\theta|}\cos(\theta)\). Its absolute value is bounded by \(e^{-|\theta|}\), whose integral over \(\mathbb{R}\) is \(2\). Thus \(g\) is absolutely integrable.

By symmetry,

\int_{-m}^{m} e^{-|\theta|}\cos(\theta)\,d\theta =2\int_0^m e^{-t}\cos(t)\,dt =1+e^{-m}(-\cos(m)+\sin(m)).

For the last equality, an antiderivative of \(e^{-t}\cos(t)\) is \(\tfrac12 e^{-t}(-\cos(t)+\sin(t))\); evaluating at \(m\) and \(0\) and multiplying by \(2\) gives the displayed expression. Since \(|-\cos(m)+\sin(m)|\leq2\), the final term has absolute value at most \(2e^{-m}\), which tends to zero. The truncated integrals therefore converge to \(1\), the full integral. This is a signed-integral calculation justified by absolute integrability, not by monotonicity.

Proper Priors and Normalized Truncations

If \(\pi\) is a probability measure, a truncated prior is often normalized to remain a probability measure. When \(\pi(K_m)>0\), define \(\pi^{(m)}(A)=\pi(A\cap K_m)/\pi(K_m)\). For a nonnegative measurable risk \(R\), its risk integrated under this normalized prior is \[ \int R\,d\pi^{(m)} = \frac{\int_{K_m}R\,d\pi}{\pi(K_m)}. \] The denominator matters: convergence of the unnormalized numerator alone does not establish convergence of the ratio. Since \(K_m\) increases to \(\Theta\), continuity from below gives \(\pi(K_m)\to1\). If \(\int R\,d\pi<\infty\), the numerator tends to \(\int R\,d\pi\) by monotone convergence, so the ratio tends to that same integral. If the full risk integral is infinite, the normalized integrals increase to infinity because the denominators tend to \(1\) while the numerators tend to infinity.

For a prior measure with infinite total mass, dividing the restriction by \(\pi(K_m)\) creates a proper prior whenever \(0<\pi(K_m)<\infty\). But these normalized priors are a different object from the unnormalized truncated integrals used in many integrated-risk comparisons. Their normalizing constants can grow without bound, so a limit for unnormalized integrals does not by itself give a limit for normalized risks.

Worked Example: Correct Risk at a Single Parameter Value

Let \(X\sim N(\theta,1)\), with \(\theta\in\mathbb{R}\). Consider the rule \(\delta(X)=1\) if \(X>0\), and \(\delta(X)=-1\) otherwise, under the loss \(L(\theta,a)=\mathbf{1}\{a\theta<0\}\). We compute its risk \(R(\theta,\delta)=E_\theta[L(\theta,\delta(X))]\), then integrate it over the compact exhaustion \(K_m=[-m,m]\) using Lebesgue measure.

At \(\theta=0\), the event in the loss is \(a\theta<0\), which becomes \(0<0\) for either action and is false. Therefore \(L(0,a)=0\) for either action and \(R(0,\delta)=0\). For \(\theta>0\), the loss is one exactly when the rule chooses \(-1\), which occurs when \(X\leq0\); the normal distribution is continuous, so \(P_\theta(X\leq0)=\Phi(-\theta)\). For \(\theta<0\), the loss is one when the rule chooses \(1\), which occurs when \(X>0\); this has probability \(1-\Phi(-\theta)=\Phi(\theta)=\Phi(-|\theta|)\). Thus

R(\theta,\delta)= \begin{cases} \Phi(-|\theta|),&\theta\ne0,\\ 0,&\theta=0. \end{cases}

Because the exceptional value at \(0\) is on a set of Lebesgue measure zero, it does not change the integral. Symmetry and the standard normal density \(\phi\) give

\int_{-m}^{m}R(\theta,\delta)\,d\theta =2\int_0^m\Phi(-t)\,dt =2\bigl(\phi(0)-\phi(m)+m\Phi(-m)\bigr).

The last equality follows by differentiating \(t\Phi(-t)-\phi(t)\), whose derivative is \(\Phi(-t)\), and evaluating at \(0\) and \(m\). As \(m\to\infty\), \(\phi(m)\to0\), and \(m\Phi(-m)\to0\); for example, \(0\leq m\Phi(-m)\leq\phi(m)\), by the normal-tail bound \(\Phi(-m)\leq\phi(m)/m\) for \(m>0\). The integral tends to \(2\phi(0)=\sqrt{2/\pi}\). The risk at zero is exactly zero; the integral calculation is unchanged because a singleton has Lebesgue measure zero.

Common Pitfall: Coverage Is Not Enough for Signed Limits

A frequent error is to argue that because \(K_m\) increases to all of \(\Theta\), every truncated integral must converge to the full integral. Coverage alone proves no such statement for an arbitrary signed function. For a nonnegative measurable integrand, monotone convergence supplies the result. For a signed integrand, a condition such as absolute integrability supplies a valid route through dominated convergence.

Incorrect reasoning: “The compact sets eventually include every parameter value, so positive and negative contributions must settle to their full total.”
Correction: Pointwise coverage does not control the size of contributions in distant tails or justify cancellation. Establish nonnegativity for monotone convergence, or verify an integrable dominating function (for example, \(|g|\)) before passing to the limit.
Another check: If using normalized restrictions of a prior, track the factor \(\pi(K_m)\) as well as the truncated numerator.

In an admissibility argument, this discipline keeps a local calculation from being mistaken for a global integrated-risk comparison. First specify the measure, the exhaustion, and whether the integrand is a risk or a signed difference. Then check the relevant convergence hypothesis before using the limiting integral.

Check Your Understanding

For each question, identify the convergence claim and the condition that supports it.

For a nonnegative measurable risk, what does monotone convergence say about its integrals over an increasing compact exhaustion if the full integral is infinite?
Let \(\pi\) be a probability measure and \(K_m\uparrow\Theta\), with \(\pi(K_m)>0\). Write the integrated risk under the normalized restriction \(\pi^{(m)}\) in terms of the unnormalized truncated integral.
Why is absolute integrability of a signed risk difference a sufficient condition for the truncated integrals to converge?
In the normal sign-decision example, what is the risk at \(\theta=0\), and why does that value not alter the Lebesgue integral?
Give one reason that convergence of unnormalized truncated integrals does not automatically imply convergence of integrals under normalized truncations of an infinite-mass prior.