From Convergence in Probability to Convergence in Mean
Convergence in probability controls the chance of a noticeable error, but it does not by itself control the size of that error when it occurs. A random variable can be very large on an event whose probability is very small. Uniform integrability rules out this kind of escaping tail contribution. Together, the two conditions give convergence in \(L^1\), meaning that the expected absolute error tends to zero.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((X_n)_{n\ge1}\) be a sequence of integrable real-valued random variables, and let \(X\) be a real-valued random variable. Convergence in probability, written \(X_n\to X\) in probability, means that for every \(\varepsilon>0\), \(\mathbb{P}(|X_n-X|>\varepsilon)\to0\). Convergence in \(L^1\) means \(\mathbb{E}|X_n-X|\to0\).
Uniform integrability is stronger than having finite expectations one variable at a time: it requires the tail expectations to become small uniformly over all indices. The key proof idea is to truncate each variable at a fixed level \(K\). The truncated variables are bounded, so convergence in probability controls their mean absolute difference. Uniform integrability then makes the truncation errors small.
Proof. We first record two standard measure-theoretic facts used in the argument. Fatou's lemma states that for nonnegative random variables \(Y_r\), \(\mathbb{E}[\liminf_r Y_r]\le\liminf_r\mathbb{E}[Y_r]\). Also, convergence in probability has an almost surely convergent subsequence. For this second fact, choose indices \(n_r\) so that \(\mathbb{P}(|X_{n_r}-X|>2^{-r})<2^{-r}\). The probability that any of these inequalities fails for some \(r\ge m\) is at most \(\sum_{r\ge m}2^{-r}\), which tends to zero as \(m\to\infty\). Thus, with probability one, only finitely many fail, and \(X_{n_r}\to X\) almost surely.
First, uniform integrability implies \(\sup_n\mathbb{E}|X_n|<\infty\). Indeed, choose \(K_0\) such that \(\sup_n\mathbb{E}[|X_n|\mathbf{1}_{\{|X_n|>K_0\}}]\le1\). Splitting according to \(|X_n|\le K_0\) and \(|X_n|>K_0\) gives \(\mathbb{E}|X_n|\le K_0+1\) for every \(n\). Along the almost surely convergent subsequence above, Fatou's lemma therefore gives \(\mathbb{E}|X|\le\liminf_r\mathbb{E}|X_{n_r}|\le K_0+1\). In particular, \(X\) is integrable.
For each fixed \(K>0\), define \(f_K(x)=|x|\mathbf{1}_{\{|x|>K\}}\). Along the same subsequence, \(f_K(X_{n_r})\) has lower limit at least \(f_K(X)\): when \(|X|>K\), eventually \(|X_{n_r}|>K\) and \(f_K(X_{n_r})\to|X|\); when \(|X|\le K\), \(f_K(X)=0\). Fatou's lemma now gives $$ \mathbb{E}\bigl[|X|\mathbf{1}_{\{|X|>K\}}\bigr] \le \liminf_r\mathbb{E}\bigl[|X_{n_r}|\mathbf{1}_{\{|X_{n_r}|>K\}}\bigr] \le \sup_n\mathbb{E}\bigl[|X_n|\mathbf{1}_{\{|X_n|>K\}}\bigr]. $$ Consequently, the tail expectation of \(X\) also tends to zero as \(K\to\infty\).
Now truncate by \(c_K(x)=\max(-K,\min(x,K))\). This function is 1-Lipschitz, so \(c_K(X_n)\to c_K(X)\) in probability, and \(|c_K(X_n)-c_K(X)|\le2K\). For any \(\varepsilon>0\), splitting according to whether the difference exceeds \(\varepsilon\) yields $$ \mathbb{E}|c_K(X_n)-c_K(X)| \le \varepsilon+2K\,\mathbb{P}(|c_K(X_n)-c_K(X)|>\varepsilon). $$ For fixed \(K\) and \(\varepsilon\), the probability on the right tends to zero. Taking the upper limit in \(n\), then letting \(\varepsilon\downarrow0\), proves that the truncated variables converge in \(L^1\).
Finally, \(|X_n-c_K(X_n)|\le |X_n|\mathbf{1}_{\{|X_n|>K\}}\), and the corresponding inequality holds for \(X\). The triangle inequality therefore gives $$ \mathbb{E}|X_n-X| \le \mathbb{E}\bigl[|X_n|\mathbf{1}_{\{|X_n|>K\}}\bigr] +\mathbb{E}|c_K(X_n)-c_K(X)| +\mathbb{E}\bigl[|X|\mathbf{1}_{\{|X|>K\}}\bigr]. $$ For fixed \(K\), the middle term tends to zero. The upper limit of the first term is bounded by the supremum of the sequence's tail expectations at that same \(K\), and the last term has the same bound by the preceding Fatou argument. Letting \(K\to\infty\) makes both bounds tend to zero. Since \(\mathbb{E}|X_n-X|\ge0\), this proves \(\mathbb{E}|X_n-X|\to0\). \(\square\)
The proof separates the two roles of the hypotheses. Convergence in probability handles what happens inside a fixed bounded range. Uniform integrability controls the part outside that range, uniformly in \(n\), and also controls the limit's tail. Neither condition is being used as a substitute for the other.
Worked Example: Indicators of Shrinking Events
Let \(\Omega=[0,1]\) with Lebesgue probability, and set \(X_n=\mathbf{1}_{(0,1/n]}\). Show that the sequence is uniformly integrable and converges to zero in \(L^1\).
For every \(\varepsilon\) with \(0<\varepsilon<1\), the event \(\{|X_n|>\varepsilon\}\) is \((0,1/n]\), so $$ \mathbb{P}(|X_n|>\varepsilon)=\frac1n\longrightarrow0. $$ Thus \(X_n\to0\) in probability. For \(K\ge1\), \(X_n\) takes only the values 0 and 1, so the strict event \(\{|X_n|>K\}\) is empty and every tail expectation is zero. This proves uniform integrability.
Directly, \(\mathbb{E}|X_n-0|=\mathbb{E}X_n=1/n\), which tends to zero. Check: the event where \(X_n=1\) has probability \(1/n\), so its contribution to the expectation is \(1\cdot(1/n)=1/n\), as computed. The general theorem gives the same conclusion from convergence in probability and uniform integrability.
Worked Example: A Rare Spike That Does Not Converge in L1
On \([0,1]\) with Lebesgue probability, define \(X_n=n\mathbf{1}_{[0,1/n]}\). Show that \(X_n\to0\) in probability but not in \(L^1\), and identify the failed hypothesis.
For any \(\varepsilon>0\), if \(n\le\varepsilon\), then \(\mathbb{P}(|X_n|>\varepsilon)=0\). If \(n>\varepsilon\), the strict event \(\{|X_n|>\varepsilon\}\) is \([0,1/n]\), of probability \(1/n\). Hence the probability tends to zero. But the expectation is $$ \mathbb{E}|X_n|=n\mathbb{P}([0,1/n])=n\cdot\frac1n=1. $$ Therefore \(\mathbb{E}|X_n-0|=1\) for every \(n\), so there is no \(L^1\) convergence.
For any finite \(K\ge0\), choose an integer \(n>K\). On \([0,1/n]\), \(|X_n|=n>K\), and thus $$ \mathbb{E}\bigl[|X_n|\mathbf{1}_{\{|X_n|>K\}}\bigr]=n\cdot\frac1n=1. $$ For every \(n\le K\), the tail event is empty; for every \(n>K\), the tail expectation is 1. Therefore the supremum over \(n\) at this same threshold \(K\) is exactly 1, for every finite \(K\). The sequence is not uniformly integrable. Check: the event probability tends to zero, but its contribution to the expectation remains \(n(1/n)=1\). This is why small event probabilities alone do not imply small mean absolute error.
Worked Example: Truncations of an Unbounded Random Variable
Let \(\Omega=\{1,2,\ldots\}\), with \(\mathbb{P}(\{k\})=2^{-k}\), and define \(X(k)=k\) and \(X_n=\min(X,n)\). Show that \((X_n)\) is uniformly integrable and converges to \(X\) in \(L^1\).
The probabilities sum to one, since \(\sum_{k=1}^{\infty}2^{-k}=1\). For every \(k\), \(X_n(k)\to k=X(k)\), so the convergence is pointwise and hence in probability. The expectation of \(X\) is $$ \mathbb{E}X=\sum_{k=1}^{\infty}\frac{k}{2^k}=2. $$ For example, the series identity follows from \(\sum_{j\ge1}r^j=r/(1-r)\): differentiating for \(0<r<1\) gives \(\sum_{j\ge1}jr^j=r/(1-r)^2\), and setting \(r=1/2\) gives 2.
For \(K\ge0\), let \(m=\lfloor K\rfloor\). Since \(X_n\le X\), the event \(X_n>K\) implies \(X>K\). Therefore $$ \mathbb{E}\bigl[X_n\mathbf{1}_{\{X_n>K\}}\bigr] \le \mathbb{E}\bigl[X\mathbf{1}_{\{X>K\}}\bigr] =\sum_{k=m+1}^{\infty}\frac{k}{2^k} =\frac{m+2}{2^m}. $$ For the last equality, write \(k=m+j\); the sum becomes \(2^{-m}(m\sum_{j\ge1}2^{-j}+\sum_{j\ge1}j2^{-j})=2^{-m}(m+2)\). This bound tends to zero as \(K\to\infty\), proving uniform integrability.
For the \(L^1\) error, \(X-X_n=0\) when \(k\le n\), and \(X-X_n=k-n\) when \(k>n\). Thus $$ \mathbb{E}|X-X_n| =\sum_{k=n+1}^{\infty}\frac{k-n}{2^k} =2^{-n}\sum_{j=1}^{\infty}\frac{j}{2^j} =2^{1-n}\longrightarrow0. $$ Check: when \(n=2\), \(\mathbb{E}X_2=\tfrac12+2\sum_{k=2}^\infty2^{-k}=\tfrac12+1=\tfrac32\), while \(\mathbb{E}X=2\); their difference is \(\tfrac12=2^{1-2}\), agreeing with the error formula.
What the Result Does—and Does Not—Say
The theorem can be applied whenever the limit is specified as a random variable and convergence in probability is known. It does not require almost-sure convergence. In the proof, almost-sure convergence was used only along a subsequence to control the limit's tails; the original hypothesis remains convergence in probability.
One direction needs no uniform-integrability assumption: if \(X_n\to X\) in \(L^1\), then \(X_n\to X\) in probability. Indeed, for \(\varepsilon>0\), $$ \varepsilon\,\mathbb{P}(|X_n-X|>\varepsilon) \le \mathbb{E}|X_n-X|, $$ because \(|X_n-X|>\varepsilon\) on the event being measured. If the right side tends to zero, so does the probability. The converse is not automatic, as the rare-spike example demonstrates.
Common mistake. It is incorrect to argue that \(X_n\to0\) in probability implies \(\mathbb{E}|X_n|\to0\) because the event where \(X_n\ne0\) becomes unlikely. In the rare-spike example, that event has probability \(1/n\), but the value on it is \(n\), so the expectation is \(n(1/n)=1\), not a quantity tending to zero. The missing check is uniform integrability at a common threshold \(K\): here the supremum tail expectation remains 1 no matter how large finite \(K\) is.
Check Your Understanding
For each question, identify which part of the convergence argument is relevant.
- State the definition of uniform integrability, including the tail event used in it.
- Why does convergence in probability alone not imply convergence in \(L^1\)? Give the expectation calculation for the rare-spike sequence.
- For the sequence \(X_n=\mathbf{1}_{(0,1/n]}\), compute \(\mathbb{E}|X_n|\) and describe its tail expectations when \(K\ge1\).
- In the truncation proof, why is \(|c_K(X_n)-c_K(X)|\le2K\) useful?
- For \(X(k)=k\) with \(\mathbb{P}(\{k\})=2^{-k}\), what is \(\mathbb{E}|X-X_n|\) when \(X_n=\min(X,n)\)?