Why Convergence in Probability Needs an Additional Condition
Convergence in probability controls how often a random variable differs substantially from its limit. It does not, by itself, control the size of the differences on the events where they occur. A sequence can therefore converge in probability while retaining a nonvanishing expectation error.
Uniform integrability supplies the missing control: it prevents the expectations from being carried by increasingly extreme values. Vitali’s convergence theorem combines these two ideas and gives a useful criterion for convergence in \(L^1\).
Definitions and the Main Statement
Definition: A sequence \((X_n)\) of integrable random variables is
uniformly integrable if the tail expectations vanish uniformly in \(n\):
$$
\lim_{K\to\infty}\sup_{n\in\mathbb N}
\mathbb{E}\!\left[|X_n|\mathbf{1}_{\{|X_n|>K\}}\right]=0.
$$
Here \(\mathbf{1}_A\) denotes the indicator of an event \(A\). The condition says that values beyond a large threshold contribute only a small amount to the expectations, uniformly over the sequence.
The sequence \((X_n)\) converges in \(L^1\) to an integrable random variable \(X\) if \(\mathbb{E}|X_n-X|\to 0\). It converges in probability to \(X\) if, for every \(\varepsilon>0\), the probability of \(|X_n-X|>\varepsilon\) tends to zero.
Theorem (Vitali’s convergence theorem): Let \((X_n)\) be a uniformly integrable sequence of integrable random variables, and let \(X\) be a random variable such that \(X_n\to X\) in probability. Then \(X\) is integrable and
$$
\mathbb{E}|X_n-X|\longrightarrow 0.
$$
Conversely, if \(X_n\to X\) in \(L^1\), then \((X_n)\) is uniformly integrable and \(X_n\to X\) in probability.
Proof of Vitali’s Theorem
We prove the forward implication first. The proof has two parts: establish that the limit is integrable, then control the difference by truncating both variables.
Proof.
Uniform integrability implies a uniform bound on the first moments. Indeed, choose \(K_0\) large enough that
\(\sup_n\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K_0\}}]\le 1\). Splitting according to whether \(|X_n|\le K_0\) gives
$$
\mathbb{E}|X_n|
\le K_0+
\mathbb{E}\!\left[|X_n|\mathbf{1}_{\{|X_n|>K_0\}}\right]
\le K_0+1.
$$
Write \(C=K_0+1\). Since \(X_n\to X\) in probability, every subsequence has a further subsequence converging to \(X\) almost surely. To see this, from any subsequence choose indices \(n_j\) such that
\(\mathbb{P}(|X_{n_j}-X|>2^{-j})<2^{-j}\). The Borel–Cantelli lemma then gives almost-sure convergence along these indices. Fatou’s lemma yields
\(\mathbb{E}|X|\le\liminf_j\mathbb{E}|X_{n_j}|\le C\). Thus \(X\) is integrable.
For \(K>0\), define the clipping function \(T_K(x)=\max(-K,\min(x,K))\). It satisfies
\(|x-T_K(x)|\le |x|\mathbf{1}_{\{|x|>K\}}\), and it is Lipschitz with constant one. Hence
\[
|X_n-X|
\le |X_n-T_K(X_n)|
+|T_K(X_n)-T_K(X)|
+|T_K(X)-X|.
\]
The expectations of the first and third terms are bounded respectively by
\(\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K\}}]\) and
\(\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]\). The first bound tends to zero uniformly in \(n\) by uniform integrability; the second tends to zero because \(X\) is integrable.
It remains to control the middle term. It is bounded by \(2K\), and clipping is Lipschitz, so \(T_K(X_n)\to T_K(X)\) in probability. For any \(\delta>0\),
\[
\mathbb{E}|T_K(X_n)-T_K(X)|
\le \delta+2K\,\mathbb{P}(|T_K(X_n)-T_K(X)|>\delta),
\]
which has limit superior at most \(\delta\). Letting \(\delta\) decrease to zero shows that this expectation tends to zero for fixed \(K\). First choose \(K\) so the two tail bounds are small, then let \(n\to\infty\) for the middle term. This proves \(\mathbb{E}|X_n-X|\to0\).
$$
\square
$$
Why the Converse Holds
The converse is a separate useful fact: \(L^1\) convergence forces both convergence in probability and uniform integrability. Markov’s inequality immediately gives the first conclusion:
\(\mathbb{P}(|X_n-X|>\varepsilon)\le \mathbb{E}|X_n-X|/\varepsilon\).
For uniform integrability, we need a bound on the tails that works for all sufficiently large \(n\), as well as control of the finitely many initial terms.
Proposition: If \(X_n\to X\) in \(L^1\), then \((X_n)\) is uniformly integrable.
Proof.
For \(K>0\), split the event \(\{|X_n|>K\}\) according to whether \(|X|>K/2\). On the part where \(|X|>K/2\), retain the term involving \(|X|\); on the remaining part, use \(|X_n-X|>K/2\). The triangle inequality and Markov’s inequality give
$$
\begin{aligned}
\mathbb{E}\!\left[|X_n|\mathbf{1}_{\{|X_n|>K\}}\right]
&\le \mathbb{E}|X_n-X|
+\mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>K/2\}}\right]\\
&\quad+\frac K2\,\mathbb{P}(|X_n-X|>K/2)\\
&\le 2\mathbb{E}|X_n-X|
+\mathbb{E}\!\left[|X|\mathbf{1}_{\{|X|>K/2\}}\right].
\end{aligned}
$$
To verify the first inequality’s last contribution, on the event
\(\{|X_n|>K,\ |X|\le K/2\}\) one has
\(|X_n-X|>K/2\), and \(|X|\le K/2\). Thus its contribution is at most
\((K/2)\mathbb{P}(|X_n-X|>K/2)\).
Given \(\eta>0\), choose \(N\) so that \(\mathbb{E}|X_n-X|<\eta\) for all \(n\ge N\). For these \(n\), the displayed bound is at most
\(2\eta+\mathbb{E}[|X|\mathbf{1}_{\{|X|>K/2\}}]\), whose second term tends to zero as \(K\to\infty\). For each of the finitely many \(n<N\), integrability of \(X_n\) implies
\(\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K\}}]\to0\). Taking the maximum over those finitely many terms and then letting \(\eta\) decrease to zero proves uniform integrability.
$$
\square
$$
Worked Examples
Worked Example: Bounded Indicators Converge in L1
Take the probability space \([0,1]\) with Lebesgue probability measure and let \(X_n=\mathbf{1}_{[0,1/n]}\), with \(X=0\). For \(0<\varepsilon<1\),
$$
\mathbb{P}(|X_n-X|>\varepsilon)=\frac1n\longrightarrow0.
$$
Thus \(X_n\to X\) in probability. Since \(|X_n|\le1\), the tail expectation is zero for every \(K\ge1\), so the sequence is uniformly integrable. Direct calculation confirms the theorem’s conclusion:
$$
\mathbb{E}|X_n-X|=\int_0^1 X_n\,dx=\frac1n\longrightarrow0.
$$
Worked Example: Convergence in Probability Without Uniform Integrability
On the same probability space, let \(X_n=n\mathbf{1}_{[0,1/n]}\) and \(X=0\). For fixed \(\varepsilon>0\), when \(n>\varepsilon\) the event \(\{|X_n|>\varepsilon\}\) is \([0,1/n]\), so its probability is \(1/n\). Therefore \(X_n\to0\) in probability. But
$$
\mathbb{E}|X_n|=n\cdot\frac1n=1
$$
for every \(n\), so there is no \(L^1\) convergence. In fact, for every \(K>0\), choose an integer \(n>K\). Then
\[
\mathbb{E}\!\left[|X_n|\mathbf{1}_{\{|X_n|>K\}}\right]=1,
\]
so the supremum of the tail expectations is at least one for every \(K\). Uniform integrability fails, exactly as the theorem permits.
Worked Example: An L1-Convergent Sequence with an Unbounded Limit
Let \(Y\) have the exponential distribution with mean one, and set \(X=Y\) and \(X_n=Y+1/n\). Then
\[
\mathbb{E}|X_n-X|=\frac1n\longrightarrow0,
\]
so the sequence converges to \(X\) in \(L^1\). Its uniform integrability can also be checked directly. For \(K>1\), put \(a=K-1\). Since \(0<1/n\le1\), the event \(\{X_n>K\}\) implies \(Y>a\), and \(X_n\le Y+1\). Consequently,
$$
\begin{aligned}
\mathbb{E}\!\left[|X_n|\mathbf{1}_{\{|X_n|>K\}}\right]
&\le \mathbb{E}[(Y+1)\mathbf{1}_{\{Y>a\}}]\\
&=(a+1)e^{-a}+e^{-a}\\
&=(a+2)e^{-a}.
\end{aligned}
$$
The exponential tail calculation uses
\(\mathbb{P}(Y>a)=e^{-a}\) and
\(\mathbb{E}[Y\mathbf{1}_{\{Y>a\}}]=(a+1)e^{-a}\). The final bound tends to zero as \(K\to\infty\) and does not depend on \(n\), proving uniform integrability even though the limit is unbounded.
Convergence of Expectations
Vitali’s theorem gives more than convergence in probability. If its hypotheses hold, then \(X_n\to X\) in \(L^1\), and the triangle inequality for integrals yields
$$
|\mathbb{E}X_n-\mathbb{E}X|
=
|\mathbb{E}(X_n-X)|
\le \mathbb{E}|X_n-X|
\longrightarrow0.
$$
Thus uniform integrability together with convergence in probability guarantees convergence of expectations. A common pitfall is to infer convergence of expectations from convergence in probability alone: the indicator-spike example shows why that inference is invalid. Large values on events of small probability can remain significant in expectation unless their tail contributions are controlled.
Takeaway: For a sequence of integrable random variables converging in probability to \(X\), uniform integrability is the condition that upgrades the convergence to \(L^1\). Conversely, \(L^1\) convergence guarantees uniform integrability, and either conclusion gives convergence of expectations.
Check Your Understanding
Answer these questions using the theorem and its proof.
- Which part of the Vitali proof uses uniform integrability, and which part uses convergence in probability?
- Why must the limit \(X\) be shown integrable before using its tail expectation in the truncation argument?
- For \(X_n=n\mathbf{1}_{[0,1/n]}\), calculate the tail expectation at a fixed threshold \(K\) by choosing \(n>K\).
- How does Markov’s inequality show that \(L^1\) convergence implies convergence in probability?
- Why do finitely many initial terms not prevent a sequence from being uniformly integrable?