Definitions (probability setting)
Interpreting “convergence in measure” on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), convergence in measure is the same as convergence in probability.
- Convergence in measure (in probability): \(X_n \to X\) in measure if for every \(\varepsilon>0\), \[ \mathbb{P}(|X_n-X|>\varepsilon)\to 0. \]
- \(L^1\) convergence: \(X_n \to X\) in \(L^1\) if \[ \mathbb{E}\,|X_n-X|\to 0. \]
Baseline implication (easy direction)
\(L^1\) convergence always implies convergence in measure. For any \(\varepsilon>0\), Markov’s inequality gives \[ \mathbb{P}(|X_n-X|>\varepsilon)\le \frac{\mathbb{E}|X_n-X|}{\varepsilon}\to 0. \]
Why the reverse fails without extra conditions
Convergence in measure does not control rare but very large deviations. A standard counterexample on \((0,1)\) with Lebesgue probability: \[ X_n(\omega)=n\,\mathbf{1}_{(0,\,1/n)}(\omega), \qquad X(\omega)=0. \]
- For any \(\varepsilon>0\), \(\mathbb{P}(|X_n-0|>\varepsilon)=\mathbb{P}\big((0,1/n)\big)=1/n\to 0\), so \(X_n\to 0\) in measure.
- But \(\mathbb{E}|X_n-0|=\int_0^1 n\,\mathbf{1}_{(0,1/n)}\,d\omega = 1\) for all \(n\), so \(X_n\) does not converge to \(0\) in \(L^1\).
The failure comes from “mass moving to the tails”: the event is small (\(1/n\)) but the magnitude is large (\(n\)), keeping the mean absolute deviation at \(1\).
Conditions for convergence in measure to imply convergence in L1
Condition A: Uniform integrability (Vitali-type condition)
A family \(\{X_n\}\) is uniformly integrable if \[ \lim_{M\to\infty}\ \sup_n \mathbb{E}\Big(|X_n|\mathbf{1}_{\{|X_n|>M\}}\Big)=0. \]
Vitali convergence theorem (probability form): If \(X_n \to X\) in measure and \(\{X_n\}\) is uniformly integrable, then \(X \in L^1\) and \[ \mathbb{E}|X_n-X|\to 0. \]
Condition B: Domination by an \(L^1\) function (a practical sufficient condition)
Suppose there exists an integrable random variable \(Y\) such that \[ |X_n-X|\le Y \quad \text{a.s. for all } n,\qquad \mathbb{E}|Y|<\infty, \] and \(X_n\to X\) in measure. Then \[ \mathbb{E}|X_n-X|\to 0. \]
This is a “dominated convergence in measure” principle: domination prevents the tail-mass pathology seen in the counterexample.
Condition C: An \(L^p\) bound for some \(p>1\) (de la Vallée-Poussin criterion)
If there exists \(p>1\) and a constant \(C\) such that \[ \sup_n \mathbb{E}|X_n|^p \le C, \] then \(\{X_n\}\) is uniformly integrable. Consequently, if also \(X_n \to X\) in measure, then \[ \mathbb{E}|X_n-X|\to 0. \]
Proof sketch for Condition A (uniform integrability + convergence in measure)
Set \(D_n = X_n - X\). The goal is \(\mathbb{E}|D_n|\to 0\).
Step 1: Truncate to control tails
For \(M>0\), write \[ |D_n| = |D_n|\mathbf{1}_{\{|X_n|\le M,\ |X|\le M\}} + |D_n|\mathbf{1}_{\{|X_n|>M\}} + |D_n|\mathbf{1}_{\{|X|>M\}}. \]
Using \(|D_n|\le |X_n|+|X|\), the tail terms satisfy \[ \mathbb{E}\big(|D_n|\mathbf{1}_{\{|X_n|>M\}}\big)\le \mathbb{E}\big(|X_n|\mathbf{1}_{\{|X_n|>M\}}\big) + \mathbb{E}\big(|X|\mathbf{1}_{\{|X_n|>M\}}\big), \] and similarly for \(\mathbf{1}_{\{|X|>M\}}\). Uniform integrability of \(\{X_n\}\) allows choosing \(M\) so that the \(\sup_n \mathbb{E}(|X_n|\mathbf{1}_{\{|X_n|>M\}})\) term is as small as desired; integrability of \(X\) makes \(\mathbb{E}(|X|\mathbf{1}_{\{|X|>M\}})\) small for large \(M\).
Step 2: Convergence of the bounded (truncated) part
On the event \(\{|X_n|\le M,|X|\le M\}\), the difference is bounded: \[ 0 \le |D_n|\mathbf{1}_{\{|X_n|\le M,|X|\le M\}} \le 2M. \]
Convergence in measure implies \(D_n \to 0\) in measure, hence the bounded variable \(|D_n|\mathbf{1}_{\{|X_n|\le M,|X|\le M\}}\) converges to \(0\) in measure. For bounded nonnegative \(Z_n \le 2M\) with \(Z_n\to 0\) in measure, \[ \mathbb{E}Z_n = \int_0^{2M} \mathbb{P}(Z_n>t)\,dt \to 0 \] by dominated convergence on the integral in \(t\), since \(\mathbb{P}(Z_n>t)\to 0\) for each \(t>0\) and is bounded by \(1\).
Step 3: Combine bounded part + tails
With \(M\) chosen so tails are small and \(n\) large so the bounded part has small expectation, conclude \(\mathbb{E}|D_n|\to 0\), i.e. \(X_n \to X\) in \(L^1\).
Quick checklist (what a correct answer should state)
| Goal | Given | Extra condition (sufficient) | Conclusion |
|---|---|---|---|
| \(X_n \to X\) in \(L^1\) | \(X_n \to X\) in measure | Uniform integrability of \(\{X_n\}\) | \(\mathbb{E}|X_n-X|\to 0\) |
| \(X_n \to X\) in \(L^1\) | \(X_n \to X\) in measure | \(|X_n-X|\le Y\) a.s. with \(\mathbb{E}|Y|<\infty\) | \(\mathbb{E}|X_n-X|\to 0\) |
| \(X_n \to X\) in \(L^1\) | \(X_n \to X\) in measure | \(\sup_n \mathbb{E}|X_n|^p<\infty\) for some \(p>1\) | \(\mathbb{E}|X_n-X|\to 0\) |