Limit Inferior Methods for Prior-Sequence Arguments

Why Limit-Inferior Bounds Matter

In an admissibility argument, a risk comparison is often integrated against a sequence of priors. The prior may change with the sequence index, and the risk difference being integrated may change as well. It is therefore not enough to know that each integral exists: one needs a justified way to relate the sequence of integrals to a limiting risk difference and prior.

A useful one-sided tool is Fatou’s lemma. It gives a lower bound on the limit inferior of integrals of nonnegative functions. In this setting, that direction can be exactly what is needed: if a positive limiting risk improvement remains under priors that retain enough mass where the improvement occurs, then the integrated improvement cannot vanish along the sequence. The nonnegativity and measurability conditions matter; without them, the conclusion need not hold.

Integrated Risk Differences and a Fatou Bound

Definition: Let $\Theta$ be a parameter space with a sigma-finite measure $\mu$. For each $n$, suppose the prior $\pi_n$ has density $p_n$ with respect to $\mu$, and let $d_n(\theta)$ be a measurable risk difference. Its integrated risk difference is $$ \Delta_n=\int_{\Theta} d_n(\theta)\,\pi_n(d\theta) =\int_{\Theta} p_n(\theta)d_n(\theta)\,\mu(d\theta), $$ when the integral is defined. Here $d_n(\theta)=R(\theta,\delta_n)-R(\theta,\tilde{\delta}_n)$; thus $d_n\geq 0$ means $\tilde{\delta}_n$ has no greater risk than $\delta_n$ at that parameter.

The following form makes the necessary comparison explicit. The measures must be represented using the same dominating measure, and the products of prior densities and risk differences must be nonnegative and measurable.

Theorem (Fatou bound for prior sequences): Suppose $p_n$ and $d_n$ are measurable and nonnegative $\mu$-almost everywhere. Let $p$ and $d$ be measurable and nonnegative, and suppose $$ \liminf_{n\to\infty}p_n(\theta)d_n(\theta)\geq p(\theta)d(\theta) $$ for $\mu$-almost every $\theta$. Then $$ \int_{\Theta}p(\theta)d(\theta)\,\mu(d\theta) \leq \liminf_{n\to\infty}\int_{\Theta}p_n(\theta)d_n(\theta)\,\mu(d\theta). $$ The integrals may be infinite.

Proof. Set $h_n=p_nd_n$. Each $h_n$ is measurable and nonnegative. Fatou’s lemma, which requires only these conditions on a measure space, gives $$ \int_{\Theta}\liminf_{n\to\infty}h_n\,d\mu \leq \liminf_{n\to\infty}\int_{\Theta}h_n\,d\mu. $$ By the assumed almost-everywhere inequality and monotonicity of the integral, $$ \int_{\Theta}pd\,d\mu \leq \int_{\Theta}\liminf_{n\to\infty}h_n\,d\mu. $$ Combining the two inequalities proves the result. $\square$

A convenient sufficient condition is $p_n(\theta)\to p(\theta)<\infty$ and $\liminf_n d_n(\theta)\geq d(\theta)$, almost everywhere. To see why, fix a point where these conditions hold. If $p(\theta)>0$, then for every $\varepsilon>0$, eventually $p_n(\theta)\geq p(\theta)-\varepsilon>0$, while $d_n$ has limit inferior at least $d(\theta)$. Letting $\varepsilon$ decrease to zero gives $\liminf_n p_nd_n\geq pd$. If $p(\theta)=0$, the same inequality follows from $p_nd_n\geq0=pd$. This reasoning also applies when $d(\theta)=+\infty$, by using arbitrarily large finite lower bounds for $d_n$ eventually.

Worked Example: Varying Densities and Risk Gaps

Take $\Theta=[0,1]$ with Lebesgue measure. For each positive integer $n$, define the prior density and nonnegative risk difference by $$ p_n(x)=1+\frac{2x-1}{n}, \qquad d_n(x)=x+\frac{1}{n}. $$ The density integrates to one because $\int_0^1(2x-1)\,dx=0$. It is nonnegative: its minimum on $[0,1]$ is $1-1/n$, and for $n=1$ it is $2x$, also nonnegative. The risk difference is nonnegative as well.

For every $x\in[0,1]$, $p_n(x)\to1$ and $d_n(x)\to x$. The theorem therefore gives the limiting lower bound $\int_0^1x\,dx=1/2$. We can check the sequence of integrated differences directly: $$ \begin{aligned} \Delta_n &=\int_0^1\left(1+\frac{2x-1}{n}\right)\left(x+\frac{1}{n}\right)\,dx\\ &=\int_0^1x\,dx+\frac{1}{n}\int_0^1 1\,dx +\frac{1}{n}\int_0^1(2x-1)x\,dx +\frac{1}{n^2}\int_0^1(2x-1)\,dx\\ &=\frac12+\frac1n+\frac{1}{n}\left(\frac23-\frac12\right)+0\\ &=\frac12+\frac{7}{6n}. \end{aligned} $$ Thus $\lim_n\Delta_n=1/2$, agreeing with the bound. The calculation also shows that the theorem supplies a lower bound; it does not claim equality at each finite $n$.

Lower Semicontinuity Gives Local Bounds

Fatou’s lemma uses a pointwise limit-inferior condition on the integrands. Lower semicontinuity is one way to establish useful local lower bounds on a risk difference. A function $g:\Theta\to[0,\infty]$ is lower semicontinuous if, for every real $a$, the strict superlevel set $\{\theta:g(\theta)>a\}$ is open. On a metric space, this is equivalent to $g(\theta)\leq\liminf_n g(\theta_n)$ whenever $\theta_n\to\theta$.

The local consequence is direct. If $g(\theta_0)>c$, where $c$ is a real number, then $\theta_0$ belongs to the open set $\{g>c\}$. Hence there is a neighborhood $U$ of $\theta_0$ on which $g>c$. Suppose, in addition, that for all $\theta\in U$, $$ d_n(\theta)\geq g(\theta)-\varepsilon_n $$ for a sequence $\varepsilon_n\to0$. If $c>\varepsilon_n$, then $d_n>c-\varepsilon_n$ on $U$. For any prior $\pi_n$, nonnegativity outside $U$ gives $$ \int_\Theta d_n\,d\pi_n \geq (c-\varepsilon_n)\pi_n(U). $$ In particular, if $\liminf_n\pi_n(U)\geq q>0$, then $$ \liminf_n\int_\Theta d_n\,d\pi_n\geq cq. $$ For the last step, use $\pi_n(U)\geq0$, $c-\varepsilon_n\to c>0$, and $\liminf_n\pi_n(U)\geq q$. This bound needs local prior mass; lower semicontinuity alone says nothing about how much mass the priors place in $U$.

Worked Example: A Local Gap and Priors with Mass Nearby

Let $\Theta=[0,1]$, and define $$ g(x)=\max\{0,\,1-4|x-\tfrac12|\}. $$ This is continuous, hence lower semicontinuous. At $x_0=1/2$, $g(x_0)=1$. Choose $c=1/2$ and $U=(3/8,5/8)$. For $x\in U$, $|x-1/2|<1/8$, so $g(x)>1/2$.

For $n\geq 9$, let $\pi_n$ put unit mass at $x_n=1/2+1/n$, and let $d_n(x)=g(x)-1/n$ on $U$, with $d_n(x)=0$ outside $U$. These are nonnegative: on $U$, $g>1/2$ and $1/n\leq1/9$. Also $x_n\in U$ for $n\geq9$, so $\pi_n(U)=1$, and $$ \int_\Theta d_n(x)\,\pi_n(dx) =d_n(x_n) =1-\frac4n-\frac1n =1-\frac5n. $$ Consequently, the integrated difference tends to $1$. The local argument alone, with $c=1/2$, gives the weaker but valid lower bound $1/2$. Here the prior mass condition is satisfied exactly, and the direct calculation confirms that the bound does not depend on treating a pointwise risk comparison as an integral limit.

Checking a Lower-Semicontinuity Claim

Consider $g:[0,1]\to[0,1]$ defined by $g(0)=0$ and $g(x)=1$ for $x>0$. Its strict superlevel sets are $$ \{x:g(x)>a\}= \begin{cases} [0,1], & a<0,\\ (0,1], & 0\leq a<1,\\ \varnothing, & a\geq1. \end{cases} $$ All three sets are open in the relative topology on $[0,1]$. Therefore $g$ is lower semicontinuous. This example illustrates why the topology must be stated: $(0,1]$ is open relative to $[0,1]$, even though it is not open as a subset of the real line.

Worked Example: A Moving Spike Does Not Permit a Reverse Bound

On $[0,1]$ with Lebesgue measure, let $p_n(x)=1$ and $$ d_n(x)=n\,\mathbf{1}_{(0,1/n)}(x). $$ These are measurable nonnegative functions. At every fixed $x>0$, $d_n(x)=0$ for all sufficiently large $n$; also $d_n(0)=0$. Thus $d_n(x)\to0$ at every $x$, and Fatou’s lower bound is $0\leq\liminf_n\int_0^1d_n(x)\,dx$.

But direct integration gives $$ \int_0^1d_n(x)\,dx =n\cdot\frac1n=1 $$ for every $n$. The pointwise limit has integral zero, while the integrals remain one. Fatou’s lemma is satisfied, but its reverse inequality would be false. The example also warns that pointwise convergence alone does not justify exchanging an integral and a limit.

A Finite-State Calculation with Changing Priors

Worked Example: Computing the Bound on Two Parameter Values

Let $\Theta=\{0,1\}$, with counting measure, and define proper priors and nonnegative risk differences by $$ \begin{aligned} p_n(0)&=\frac14+\frac{1}{8n},& p_n(1)&=\frac34-\frac{1}{8n},\\ d_n(0)&=1-\frac{1}{2n},& d_n(1)&=2+\frac{1}{n}. \end{aligned} $$ The prior probabilities sum to one. Both risk differences are nonnegative for every positive integer $n$. Their pointwise limits are $p(0)=1/4,\ p(1)=3/4,\ d(0)=1,\ d(1)=2$.

Since an integral with respect to counting measure is a sum, $$ \begin{aligned} \Delta_n &=p_n(0)d_n(0)+p_n(1)d_n(1)\\ &=\left(\frac14+\frac{1}{8n}\right)\left(1-\frac{1}{2n}\right) +\left(\frac34-\frac{1}{8n}\right)\left(2+\frac{1}{n}\right)\\ &=\left(\frac14-\frac{1}{16n^2}\right) +\left(\frac32+\frac{1}{2n}-\frac{1}{8n^2}\right)\\ &=\frac74+\frac{1}{2n}-\frac{3}{16n^2}. \end{aligned} $$ Hence $\lim_n\Delta_n=7/4$. The limiting integrated difference is also $$ p(0)d(0)+p(1)d(1)=\frac14+\frac32=\frac74. $$ This discrete calculation is the same Fatou comparison in a setting where each contribution and each prior probability can be checked individually.

Common Mistakes and Practical Use

The main value of a limit-inferior argument is its reliability under limited information. It can establish that integrated risk differences cannot fall below a positive threshold without requiring the integrals themselves to converge. In applications, write the sequence of integrands explicitly, identify a common measure when using densities, and verify the pointwise lower bound before invoking Fatou’s lemma.

Reversing Fatou’s inequality. Fatou gives $\int\liminf h_n\leq\liminf\int h_n$ for nonnegative measurable $h_n$. It does not give an upper bound on the limit inferior from the limiting integrand.
Applying the result to signed risk differences. A risk difference can be negative, even when its integral exists. Fatou’s lemma in the stated form requires nonnegative integrands. Do not silently apply it to signed differences.
Ignoring prior variation. When $p_n$ changes, convergence of $d_n$ by itself does not establish convergence of $p_nd_n$. Check the product or verify the sufficient pointwise conditions on both factors.
Assuming lower semicontinuity supplies prior mass. It supplies a neighborhood with a lower bound on the function. A separate bound on $\pi_n(U)$ is needed to obtain a positive integrated lower bound.
Confusing a lower bound with convergence. A positive Fatou lower bound does not show that the sequence of integrated differences converges, nor does it identify its limit.

Key takeaway: To control changing integrated risk differences, form the nonnegative products of prior densities and risk gaps and establish their pointwise limit-inferior bound. Lower semicontinuity can provide a local gap, but a nonvanishing amount of prior mass in that region is an additional requirement.

Check Your Understanding

Check both the hypotheses of each bound and the direction of its conclusion.

State the measurability and sign conditions needed to apply Fatou’s lemma to $p_nd_n$.
If $p_n(\theta)\to p(\theta)$ and $\liminf_n d_n(\theta)\geq d(\theta)$, what additional sign and finiteness conditions make the product bound applicable?
Why does lower semicontinuity at a point with a positive gap not, by itself, give a positive lower bound on an integrated risk difference?
For $d_n(x)=n\mathbf{1}_{(0,1/n)}(x)$, calculate the pointwise limit and the integral of $d_n$. Which proposed reverse limit inequality does this refute?
In the two-state example, calculate the limiting prior-weighted risk difference directly from the limiting prior probabilities and gaps.