Why a Conditional Version of Dominated Convergence Matters
Conditional expectation averages a random variable while retaining the information in a specified sigma-algebra. When a sequence of random variables converges, it is natural to ask whether those conditional averages converge as well. The answer is yes under a useful domination assumption: almost-sure convergence together with a single integrable bound.
The conclusion is stronger than convergence of ordinary expectations. The conditional expectations converge almost surely as random variables, and they converge in \(L^1\). A key part of the proof is to show that the conditional expectation of the absolute error itself tends to zero almost surely.
Statement of Conditional Dominated Convergence
The domination hypothesis is the same kind of integrable bound used in the ordinary dominated convergence theorem. The difference is that the sequence of interest is now passed through conditional expectation.
The conditional expectations here may initially be chosen as separate versions. Since there are only countably many \(n\), the conclusions hold outside one common null set; changing any version on a null set does not affect either conclusion.
First note that \(|X|\le Y\) almost surely. Indeed, on the probability-one event where \(X_n\to X\) and all the bounds \(|X_n|\le Y\) hold, taking limits gives \(|X|\le Y\). Thus \(X\) is integrable. Also, \(|X_n-X|\le 2Y\) almost surely, so the ordinary dominated convergence theorem gives \(\mathbb E|X_n-X|\to0\).
Proof of Almost-Sure Convergence
We use the conditional Fatou lemma: for nonnegative random variables \(V_n\),
Apply it to the nonnegative variables \(V_n=2Y-|X_n-X|\). These variables converge almost surely to \(2Y\). The conditional Fatou inequality then bounds the limiting conditional absolute error from above by zero.
Set \(A_n=\mathbb E[|X_n-X|\mid\mathcal G]\) and \(C=2\mathbb E[Y\mid\mathcal G]\). By linearity, a common choice of versions gives \(\mathbb E[V_n\mid\mathcal G]=C-A_n\). Conditional Fatou therefore yields
The last equality holds pointwise wherever \(C\) is finite; it is finite almost surely because \(Y\) is integrable. Subtracting \(C=2\mathbb E[Y\mid\mathcal G]\) gives \(\limsup_n A_n\le0\). Since every \(A_n\) is nonnegative, \(A_n\to0\) almost surely.
The conditional absolute-value bound gives $$ \left|\mathbb E[X_n\mid\mathcal G]-\mathbb E[X\mid\mathcal G]\right| \le \mathbb E[|X_n-X|\mid\mathcal G]=A_n \quad\text{almost surely.} $$ Consequently the conditional expectations converge almost surely. \(\square\)
Why the Convergence Also Holds in \(L^1\)
Almost-sure convergence does not, by itself, imply \(L^1\) convergence. Here the same domination hypothesis supplies the needed control. The conditional expectation is an \(L^1\) contraction: for every integrable \(U\), \(\mathbb E|\mathbb E[U\mid\mathcal G]|\le\mathbb E|U|\). Apply this to \(U=X_n-X\).
By linearity and the contraction inequality,
Since \(X_n\to X\) almost surely and \(|X_n-X|\le2Y\), the ordinary dominated convergence theorem gives \(\mathbb E|X_n-X|\to0\). The left-hand side therefore tends to zero, which is exactly convergence in \(L^1\). \(\square\)
The conditional absolute-error conclusion also gives a direct view of what the theorem says: given the information in \(\mathcal G\), the expected size of the error tends to zero almost surely. Taking expectations of that conditional error recovers the same \(L^1\) bound.
Worked Example: A Finite Partition
Let \(\Omega=\{1,2,3,4\}\), with each point having probability \(1/4\), and let \(\mathcal G\) be generated by the partition \(\{\{1,2\},\{3,4\}\}\). Define $$ X_n(1)=1+\frac1n,\qquad X_n(2)=-1+\frac2n,\qquad X_n(3)=2-\frac1n,\qquad X_n(4)=\frac1n. $$ Then \(X_n\) converges pointwise to \(X=(1,-1,2,0)\), and \(Y=2\) dominates \(|X_n|\) for every \(n\). On each of the two partition cells, conditional expectation is the average of the values on that cell. Hence $$ \mathbb E[X_n\mid\mathcal G](1)=\mathbb E[X_n\mid\mathcal G](2) =\frac{(1+1/n)+(-1+2/n)}2=\frac{3}{2n}, $$ while $$ \mathbb E[X_n\mid\mathcal G](3)=\mathbb E[X_n\mid\mathcal G](4) =\frac{(2-1/n)+1/n}{2}=1. $$ The limit \(X\) has conditional expectation \(0\) on \(\{1,2\}\) and \(1\) on \(\{3,4\}\), so these conditional expectations converge pointwise to \(\mathbb E[X\mid\mathcal G]\). Their \(L^1\) distance from that limit is $$ \frac14\left(\frac{3}{2n}+\frac{3}{2n}+0+0\right)=\frac{3}{4n}\longrightarrow0. $$ This example also shows why the limit in the theorem is \(\mathbb E[X\mid\mathcal G]\), not necessarily \(X\): the limiting values \(1\) and \(-1\) on the first cell average to zero.
Worked Example: Powers on the Unit Interval
Take \([0,1]\) with Lebesgue probability measure, and let \(\mathcal G\) be generated by the two intervals \([0,1/2]\) and \((1/2,1]\). Set \(X_n(t)=t^n\). The sequence converges to zero almost everywhere and is dominated by the integrable constant \(Y(t)=1\). A conditional expectation on either cell is the average of \(t^n\) over that cell. Thus, on the first cell, $$ \mathbb E[X_n\mid\mathcal G] =2\int_0^{1/2}t^n\,dt =\frac{2^{-n}}{n+1}, $$ and on the second cell, $$ \mathbb E[X_n\mid\mathcal G] =2\int_{1/2}^{1}t^n\,dt =\frac{2(1-2^{-(n+1)})}{n+1}. $$ Both values tend to zero. The conditional expectation of the almost-sure limit is also zero. Moreover, its \(L^1\) distance from zero is $$ \mathbb E|\mathbb E[X_n\mid\mathcal G]| =\mathbb E[X_n]=\int_0^1t^n\,dt=\frac1{n+1}\longrightarrow0. $$
Worked Example: Why Domination Cannot Be Omitted
On \([0,1]\) with Lebesgue probability measure, let \(X_n(t)=n\mathbf 1_{(0,1/n)}(t)\), and take \(\mathcal G=\{\varnothing,[0,1]\}\). For each \(t>0\), \(t\notin(0,1/n)\) for all sufficiently large \(n\), so \(X_n(t)\to0\); convergence also holds at \(t=0\). However, $$ \mathbb E[X_n\mid\mathcal G]=\mathbb E[X_n] =n\,\mathbb P((0,1/n))=1 $$ for every \(n\), whereas the conditional expectation of the limit is zero. Thus the claimed convergence fails without an integrable dominator. In fact, no integrable \(Y\) can dominate all the \(X_n\): such a \(Y\) would satisfy \(\mathbb E[Y]\ge\mathbb E[X_n]=1\), which alone is not a contradiction, but pointwise domination forces \(Y(t)\ge n\) whenever \(0<t<1/n\). The resulting lower bound grows on intervals approaching zero and has infinite integral. For example, for \(m\ge2\) and \(t\in[1/m,1/(m-1))\), choosing \(n=m-1\) gives \(t<1/(m-1)\), so \(Y(t)\ge m-1\); summing these interval contributions gives a divergent harmonic series.
Interpretation and a Common Pitfall
The theorem preserves the form of the limit under conditional expectation, but it does not claim that conditioning leaves the limit unchanged. In general, \(X\) need not be \(\mathcal G\)-measurable, so \(\mathbb E[X\mid\mathcal G]\) can differ from \(X\). The finite-partition example makes this distinction explicit.
Another important point is the role of one common integrable dominator. Individual integrability of each \(X_n\), even together with almost-sure convergence, is insufficient. The last example has integrable \(X_n\) for every \(n\), but their expectations do not approach the expectation of the limit. The bound \(|X_n|\le Y\) prevents this kind of concentration of larger and larger values on smaller sets.
In applications, it is often useful to establish the stronger intermediate statement \(\mathbb E[|X_n-X|\mid\mathcal G]\to0\) almost surely. It immediately bounds the difference of conditional expectations and identifies the conditional error that must be controlled.
Check Your Understanding
Use the hypotheses and proof of conditional dominated convergence to answer these questions.
- Why does the common bound \(|X_n|\le Y\) imply \(|X|\le Y\) almost surely?
- In the proof, why is \(2Y-|X_n-X|\) nonnegative almost surely?
- How does conditional Fatou imply that \(\mathbb E[|X_n-X|\mid\mathcal G]\) converges to zero almost surely?
- Which inequality turns convergence of \(X_n\) to \(X\) in \(L^1\) into \(L^1\) convergence of their conditional expectations?
- In the finite-partition example, why does the conditional limit on \(\{1,2\}\) differ from the values of \(X\) there?