Absolute continuity compares two fixed measures. Contiguity compares changing sequences and asserts that asymptotically impossible events under one sequence remain impossible under the other. It is the precise condition that permits \(o_p(1)\) remainders proved under a null law to remain negligible under local alternatives.
One-sided and mutual contiguity
Definition. A sequence \(Q_n\) is contiguous with respect to \(P_n\), written \[ Q_n\triangleleft P_n, \] if, for every event sequence \(A_n\), \[ P_n(A_n)\to0 \quad\Longrightarrow\quad Q_n(A_n)\to0. \] The sequences are mutually contiguous, \(Q_n\triangleleft\triangleright P_n\), if each is contiguous with respect to the other.
Proposition. If \(Q_n\triangleleft P_n\) and \(T_n=o_{P_n}(1)\), then \(T_n=o_{Q_n}(1)\). If \(T_n=O_{P_n}(1)\), contiguity alone also implies \(T_n=O_{Q_n}(1)\).
Proof. For \(o_p\), apply the definition to \(A_n=\{|T_n|>\varepsilon\}\). For \(O_p\), choose \(M_k\uparrow\infty\) and a subsequence argument: if \(Q_n(|T_n|>M)\) failed to become uniformly small for large \(M\), one could choose \(M_n\uparrow\infty\) with \(P_n(|T_n|>M_n)\to0\) but \(Q_n(|T_n|>M_n)\not\to0\), contradicting contiguity. \(\square\)
Uniform integrability of likelihood ratios
Theorem. Suppose \(Q_n\ll P_n\) with \[ L_n=\frac{\mathrm{d}Q_n}{\mathrm{d}P_n}. \] Then \[ Q_n\triangleleft P_n \quad\Longleftrightarrow\quad \{L_n:n\geq1\} \text{ is uniformly integrable under }P_n, \] meaning \[ \lim_{M\to\infty}\sup_n E_{P_n}\{L_n1(L_n>M)\}=0. \]
Proof. If the likelihood ratios are uniformly integrable and \(P_n(A_n)\to0\), then \[ Q_n(A_n) =E_{P_n}(L_n1_{A_n}) \leq MP_n(A_n) +E_{P_n}\{L_n1(L_n>M)\}. \] Take \(\limsup_n\), then \(M\to\infty\). Conversely, if uniform integrability fails, choose \(M_k\uparrow\infty\) and indices \(n_k\) such that \[ E_{P_{n_k}}\{L_{n_k}1(L_{n_k}>M_k)\}\geq\varepsilon. \] Set \(A_{n_k}=\{L_{n_k}>M_k\}\) and \(A_n=\varnothing\) otherwise. Markov's inequality gives \(P_{n_k}(A_{n_k})\leq1/M_k\to0\), while \(Q_{n_k}(A_{n_k})\geq\varepsilon\), contradicting contiguity. \(\square\)
Tightness of \(L_n\) under \(P_n\) is not enough. A small probability can carry a very large likelihood ratio and a nonvanishing amount of \(Q_n\)-mass. The relevant condition is tail control weighted by \(L_n\), namely uniform integrability.
A moment criterion
Corollary. If for some \(\eta>0\), \[ \sup_nE_{P_n}L_n^{1+\eta}<\infty, \] then \(Q_n\triangleleft P_n\).
Proof. On \(\{L_n>M\}\), \(L_n\leq M^{-\eta}L_n^{1+\eta}\). Therefore the likelihood-ratio tails are uniformly \(O(M^{-\eta})\). \(\square\)
Local Gaussian shifts
Example. Let \[ P_n=N(0,1)^{\otimes n}, \qquad Q_n=N(h/\sqrt n,1)^{\otimes n}. \] Then \[ \log L_n = h\sqrt n\,\bar X_n-\frac12h^2 \rightsquigarrow N(-h^2/2,h^2) \quad\text{under }P_n. \] The limit \(L=\exp(hZ-h^2/2)\) is positive and has mean one. Le Cam's first lemma therefore gives mutual contiguity.
Counterexample (fixed separation). If instead \(Q_n=N(\mu,1)^{\otimes n}\) for fixed \(\mu\ne0\), the event \[ A_n=\{|\bar X_n-\mu|<|\mu|/3\} \] has \(P_n(A_n)\to0\) and \(Q_n(A_n)\to1\). The sequences are asymptotically singular, not contiguous. The root-\(n\) local scale is exactly the scale at which neither model becomes perfectly distinguishable.
Contiguity and total variation
If \(\|P_n-Q_n\|_{\mathrm{TV}}\to0\), the sequences are mutually contiguous, but the converse is false. Local normal alternatives are mutually contiguous while their total-variation distance converges to a positive constant depending on \(h\). Contiguity preserves null events; it does not say the experiments are indistinguishable.
Why local asymptotics needs contiguity
An expansion such as \[ T_n=\Delta_n+o_{P_{\theta_0}}(1) \] is initially proved under \(P_{\theta_0}^{(n)}\). If local alternatives are contiguous, the same remainder is \(o_{P_{\theta_0+h/\sqrt n}}(1)\). Le Cam's third lemma then changes the law of the leading term \(\Delta_n\). Contiguity and change of measure therefore perform separate jobs: one transfers negligibility, the other calculates the shifted limit.