Null Sets, Integrated Risk, and Pointwise Dominance

Why Pointwise and Integrated Comparisons Differ

Admissibility is defined through comparisons of risk at every parameter value: a procedure is inadmissible if another procedure has no greater risk anywhere and strictly smaller risk somewhere. Integrating risk against a probability measure on the parameter space gives a different comparison. It summarizes performance relative to that measure, so it can overlook parameter values assigned probability zero and can conceal differences when the integrated risks are infinite.

The distinction matters whenever integrated-risk minimization is used as evidence for admissibility. A lower integrated risk does not by itself establish pointwise domination; conversely, equal integrated risks do not rule out domination. To draw a pointwise conclusion from an integral, we must specify the measure and check the conditions that make the integral sensitive to the risk difference.

Definition: Let $R(\theta,\delta)$ be the risk of a decision rule $\delta$ at parameter $\theta$. Given a measure $\mu$ on the parameter space $\Theta$, the integrated risk of $\delta$ is $$ \mathcal{R}_{\mu}(\delta)=\int_{\Theta} R(\theta,\delta)\,\mu(d\theta), $$ when this integral is defined, possibly with value $+\infty$. We say $\delta'$ pointwise dominates $\delta$ if $R(\theta,\delta')\leq R(\theta,\delta)$ for every $\theta\in\Theta$, with strict inequality for at least one $\theta$. A risk inequality holds $\mu$-almost everywhere if it fails only on a set of $\mu$-measure zero.

The usual definition of dominance is pointwise and does not depend on a chosen measure. Integrated risk does depend on $\mu$. In Bayesian settings $\mu$ may be a prior distribution, but the same mathematical distinction applies to any measure used to average risk. In what follows, risks and their differences are assumed measurable when integrated. Risks are nonnegative, as they are under nonnegative loss.

What an Integral Can and Cannot Detect

If two rules have finite integrated risks, the difference of their integrated risks is the integral of their risk difference. Thus, if one rule has no greater risk $\mu$-almost everywhere and has strictly smaller risk on a set of positive $\mu$-measure, it has strictly smaller integrated risk. A strict improvement at a single parameter is different: unless that point has positive measure, the integral need not detect it.

Worked Example: A Dominator Hidden at a Null Point

Let $\Theta=[0,1]$, take no observation, and let the action set contain $a$ and $b$. Define the losses by $L(\theta,a)=1$ for every $\theta$, and $L(0,b)=0$, $L(\theta,b)=1$ for $0<\theta\leq 1$. Let $\delta_a$ and $\delta_b$ be the rules that always choose the named actions. Their risks equal their losses because there is no observation.

At $\theta=0$, $R(0,\delta_b)=0<1=R(0,\delta_a)$. At every $\theta>0$, the risks are equal to 1. Hence $\delta_b$ pointwise dominates $\delta_a$.

Now use the uniform probability measure on $[0,1]$, which assigns measure zero to $\{0\}$. The integrated risks are $$ \mathcal{R}_{\mu}(\delta_a)=\int_0^1 1\,d\theta=1, \qquad \mathcal{R}_{\mu}(\delta_b)=\int_0^1 R(\theta,\delta_b)\,d\theta =\int_0^1 1\,d\theta=1. $$ The value at $\theta=0$ does not change the integral. In the class consisting of these two rules, $\delta_a$ is an integrated-risk minimizer, yet it is dominated. If instead $\mu$ puts probability one on $\theta=0$, the integrated risks are 1 and 0, so this particular measure detects the improvement.

An almost-everywhere comparison can also fail to give dominance in the other direction: a rule can be better almost everywhere yet worse at an exceptional parameter. That single exception is enough to defeat pointwise dominance.

Worked Example: Better Almost Everywhere, but Not a Dominator

Again take $\Theta=[0,1]$ with no observation and two actions. Let action $a$ have loss 1 at every parameter. Let action $b$ have loss 2 at $\theta=0$ and loss 0 for $0<\theta\leq1$. Under the uniform measure, the integrated risks are $$ \mathcal{R}_{\mu}(\delta_a)=\int_0^1 1\,d\theta=1, \qquad \mathcal{R}_{\mu}(\delta_b)=\int_0^1 R(\theta,\delta_b)\,d\theta=0. $$ Thus $\delta_b$ has strictly lower integrated risk and lower risk $\mu$-almost everywhere. But at $\theta=0$, its risk is 2 rather than 1. It therefore does not dominate $\delta_a$. A favorable average, even a strict one, cannot erase a pointwise loss increase.

When a Pointwise Dominator Must Improve the Integral

There is a useful sufficient condition for converting pointwise domination into a strict integrated-risk improvement. The strict inequality must occur on a set of positive measure, and the integral must be finite enough for a strict risk difference to remain visible. The next result states this carefully.

Theorem: Suppose $R(\theta,\delta)$ and $R(\theta,\delta')$ are measurable and nonnegative, and $\mathcal{R}_{\mu}(\delta)<\infty$. If $R(\theta,\delta')\leq R(\theta,\delta)$ for every $\theta$, and the set $$ A=\{\theta:R(\theta,\delta')<R(\theta,\delta)\} $$ has positive $\mu$-measure, then $\mathcal{R}_{\mu}(\delta')<\mathcal{R}_{\mu}(\delta)$.

Proof. Define $d(\theta)=R(\theta,\delta)-R(\theta,\delta')$. It is measurable and nonnegative. Since $0\leq R(\theta,\delta')\leq R(\theta,\delta)$ and the latter has finite integral, $\mathcal{R}_{\mu}(\delta')$ is finite as well. Thus the integral difference is well-defined and $$ \mathcal{R}_{\mu}(\delta)-\mathcal{R}_{\mu}(\delta')=\int_{\Theta}d(\theta)\,\mu(d\theta). $$ On $A$, $d(\theta)>0$. Since $$ A=\bigcup_{m=1}^{\infty}\{\theta:d(\theta)\geq 1/m\}, $$ and $\mu(A)>0$, at least one set in this union has positive measure. For that $m$, $$ \int_{\Theta}d(\theta)\,\mu(d\theta) \geq \frac{1}{m}\,\mu\{\theta:d(\theta)\geq 1/m\}>0. $$ Therefore $\mathcal{R}_{\mu}(\delta')<\mathcal{R}_{\mu}(\delta)$. $\square$

The theorem explains a valid route from integrated-risk minimization to admissibility, but its hypotheses must apply to every possible dominator under consideration. If $\delta$ minimizes integrated risk within a class, has finite integrated risk, and every pointwise dominator in that class has a strict improvement on a positive-$\mu$-measure set, then no such dominator can exist. Consequently $\delta$ is admissible within that class. The conclusion is only as broad as the class being compared.

Continuity and Full Support

A strict improvement at one point need not lie on a positive-measure set. Regularity of the risk functions and the support of the measure can bridge that gap. A probability measure $\mu$ on $\Theta\subseteq\mathbb{R}$ has full support on $\Theta$ if every nonempty relatively open subset of $\Theta$ has positive $\mu$-measure.

Corollary: Suppose $\mathcal{R}_{\mu}(\delta)<\infty$, $\mu$ has full support on $\Theta$, and both risk functions are continuous on $\Theta$. If $\delta'$ pointwise dominates $\delta$ and is strictly better at some $\theta_0\in\Theta$, then $\mathcal{R}_{\mu}(\delta')<\mathcal{R}_{\mu}(\delta)$.

Proof. The difference $d(\theta)=R(\theta,\delta)-R(\theta,\delta')$ is continuous. At $\theta_0$, it is strictly positive. By continuity, there is a relatively open neighborhood $U$ of $\theta_0$ and an $\varepsilon>0$ such that $d(\theta)>\varepsilon$ throughout $U$. Full support gives $\mu(U)>0$, so the strict-improvement set has positive measure. The theorem applies and gives the claimed strict inequality. $\square$

Worked Example: A Finite Integrated-Risk Comparison

Let $\Theta=\mathbb{R}$, let $\mu=N(0,1)$, and consider two rules with continuous risks $R(\theta,\delta)=1+\theta^2$ and $R(\theta,\delta')=\theta^2$. These risks can, for example, be represented by action-specific losses in a no-observation decision problem.

For every $\theta$, $R(\theta,\delta')=R(\theta,\delta)-1$, so $\delta'$ pointwise dominates $\delta$. Under $N(0,1)$, $E[\theta^2]=1$. Therefore $$ \mathcal{R}_{\mu}(\delta)=E[1+\theta^2]=1+1=2, \qquad \mathcal{R}_{\mu}(\delta')=E[\theta^2]=1. $$ The normal measure has full support, the risks are continuous, and $\delta$ has finite integrated risk. The strict integrated-risk improvement agrees with the pointwise comparison, as the corollary requires.

Why Finiteness Cannot Be Omitted

The finiteness condition is not a technical decoration. If both integrated risks are infinite, a pointwise strict improvement can still leave the two extended-real integrals tied at $+\infty$. An integrated-risk minimizer may then fail to exclude a dominator.

Worked Example: Strict Domination with Two Infinite Integrals

Let $\Theta=\mathbb{R}$, use no observation, and let $\mu=N(0,1)$. Consider actions $a$ and $b$ with losses $$ L(\theta,a)=e^{\theta^2/2}, \qquad L(\theta,b)=e^{\theta^2/2}-1. $$ Both losses are nonnegative, and action $b$ has strictly smaller loss at every parameter. Thus the rule $\delta_b$ dominates $\delta_a$.

The density of $\mu$ is $\phi(\theta)=(2\pi)^{-1/2}e^{-\theta^2/2}$. For action $a$, the integrated risk has integrand $$ L(\theta,a)\phi(\theta)=\frac{1}{\sqrt{2\pi}}, $$ so its integral over $\mathbb{R}$ is infinite. For action $b$, integration over $[-M,M]$ gives $$ \int_{-M}^{M} L(\theta,b)\phi(\theta)\,d\theta =\frac{2M}{\sqrt{2\pi}}-P(|Z|\leq M), \qquad Z\sim N(0,1). $$ As $M\to\infty$, the first term diverges and the probability term is at most 1. Hence the integrated risk of $b$ is also infinite. In a class containing only these two rules, both attain the minimum extended integrated risk $+\infty$, even though $a$ is dominated. Minimizing an extended-real integral when every candidate has value $+\infty$ gives no useful exclusion of dominators.

Common Mistake and Practical Check

Wrong reasoning: “A rule minimizes integrated risk, so no other rule can dominate it. If another rule were better somewhere and nowhere worse, its integral would have to be smaller.”

Correction: “Better somewhere” may mean better only on a $\mu$-null set, leaving the integral unchanged. Even if the improvement occurs on a positive-measure set, both integrated risks may be infinite. To use integrated-risk minimization to rule out a pointwise dominator, check that the candidate’s integrated risk is finite and that every strict improvement allowed by pointwise domination is detected by the measure—for example, because the risks are continuous and $\mu$ has full support.

State the comparison.
Is the claim pointwise domination, an almost-everywhere inequality, or simply a lower integrated risk?

Inspect the exceptional set.
If a strict improvement occurs only on a $\mu$-null set, the integrated comparison may not detect it. If a rule is worse on a null set, it still does not pointwise dominate.

Check finiteness and regularity.
For an admissibility argument based on integrated-risk minimization, verify finite integrated risk and a condition ensuring that a strict pointwise improvement has positive $\mu$-measure.

Key takeaway: Pointwise dominance is a measure-free, every-parameter comparison. Integrated risk is measure-dependent and can ignore null sets or tie at infinity. Integrated-risk minimization rules out pointwise dominators only when finiteness and the measure’s ability to detect strict improvement are established.

Check Your Understanding

Use the definitions and conditions in this tutorial to answer the following questions.

Can a rule be strictly better $\mu$-almost everywhere than another rule without dominating it? Describe how.
In the null-point example, why do the two integrated risks agree under the uniform measure but differ under a point mass at zero?
State the finiteness and positive-measure conditions that ensure a pointwise dominator has strictly lower integrated risk.
Why does continuity of both risks, together with full support of $\mu$, make a strict improvement at one parameter detectable by integration?
In the example with losses $e^{\theta^2/2}$ and $e^{\theta^2/2}-1$, why does strict pointwise improvement not give a strict integrated-risk inequality?