Continuity Theorem

Also known as Lévy’s Continuity Theorem. It identifies weak convergence of probability measures with pointwise convergence of their characteristic functions, turning a question about distributions into a question about a single complex-valued function.

Preliminary: Continuity of characteristic functions

Before stating the theorem, we record a basic fact used in its proof.

Proposition

Statement

Let $\varphi$ be the characteristic function of a probability measure $\mu$. Then $\varphi$ is continuous on $\mathbb{R}$.

Proof

Let $X \sim \mu$, so $\varphi(t) = \mathbb{E}[e^{itX}]$. Fix $t \in \mathbb{R}$ and let $s \to t$. Then:

1. Pointwise convergence: As $s \to t$, $e^{isX} \to e^{itX}$ pointwise on $\Omega$, since the map $s \mapsto e^{isx}$ is continuous for every $x \in \mathbb{R}$.

2. Domination: $|e^{isX}| = 1$, which is integrable since $\mu$ is a probability measure.

By the dominated convergence theorem, $\varphi(s) = \mathbb{E}[e^{isX}] \to \mathbb{E}[e^{itX}] = \varphi(t)$.

Theorem

Theorem (Lévy)

Statement

Let $\{\mu_n\}_{n \ge 1}$ and $\mu$ be probability measures on $(\mathbb{R}, \mathcal{B})$ with characteristic functions $\varphi_n, \varphi$. Then

$$\mu_n \Rightarrow \mu \quad \Longleftrightarrow \quad \varphi_n(t) \to \varphi(t) \text{ for all } t \in \mathbb{R}.$$

In short: weak convergence of measures is the same as pointwise convergence of characteristic functions.

Proof

The forward direction is a one-line application of Portmanteau. The reverse direction is the substantive content: pointwise convergence of $\varphi_n$ contains enough information to force tightness, after which Helly and the inversion formula close the argument.

(⟹) Weak convergence implies pointwise convergence of ch.f.

Let $X_n \sim \mu_n$ and $X \sim \mu$. The map $x \mapsto e^{itx}$ is bounded ($|e^{itx}| = 1$) and continuous. By the Portmanteau theorem (condition 2: expectations of bounded continuous functions converge):

$$\mathbb{E}[e^{itX_n}] \to \mathbb{E}[e^{itX}],$$

which is exactly $\varphi_n(t) \to \varphi(t)$ for every $t \in \mathbb{R}$.

(Strictly speaking, $e^{itx}$ is complex-valued. Apply Portmanteau separately to the real part $\cos(tx)$ and the imaginary part $\sin(tx)$, both of which are bounded continuous real-valued functions.)

(⟸) Pointwise convergence of ch.f. implies weak convergence.

This direction has three steps: derive a tail bound from $\varphi_n$, conclude tightness, then identify every subsequential weak limit with $\mu$.

1. Tail bound from the characteristic function (Fubini trick).

   Let $X_n \sim \mu_n$. For any $u > 0$, compute

   $$\frac{1}{u} \int_{-u}^{u} \big(1 - \varphi_n(t)\big) \, dt = \frac{1}{u} \int_{-u}^{u} \mathbb{E}\big[1 - e^{itX_n}\big] \, dt.$$

   The integrand is bounded by $2$ on a bounded interval, so by Fubini we may swap:

   $$= \mathbb{E}\left[\frac{1}{u} \int_{-u}^{u} \big(1 - e^{itX_n}\big) \, dt\right] = \mathbb{E}\left[2 - \frac{2\sin(uX_n)}{uX_n}\right] = 2\,\mathbb{E}\left[1 - \frac{\sin(uX_n)}{uX_n}\right],$$

   where the inner integral $\int_{-u}^{u} e^{itx} dt = \frac{2 \sin(ux)}{x}$ was evaluated directly.

   Now $|\sin y / y| \le 1$ always, and for $|y| \ge 2$ we have $|\sin y / y| \le 1/|y| \le 1/2$. Restricting to the event $\{|X_n| \ge 2/u\}$, i.e., $\{|uX_n| \ge 2\}$:

   $$2\,\mathbb{E}\left[1 - \frac{\sin(uX_n)}{uX_n}\right] \ge 2\,\mathbb{E}\left[\left(1 - \frac{1}{|uX_n|}\right) \mathbb{1}_{\{|X_n| \ge 2/u\}}\right] \ge 2 \cdot \tfrac{1}{2} \cdot \mathbb{P}\!\left(|X_n| \ge \tfrac{2}{u}\right).$$

   The first inequality also used that the integrand is non-negative wherever $|\sin y / y| \le 1$, so dropping the complement of $\{|X_n| \ge 2/u\}$ can only decrease the expectation. We have arrived at the bound

   $$\mathbb{P}\!\left(|X_n| \ge \tfrac{2}{u}\right) \le \frac{1}{u} \int_{-u}^{u} \big(1 - \varphi_n(t)\big) \, dt. \quad\textcolor{gray}{\text{---(1.1)}}$$

   Choosing $u$ small makes $2/u$ large, so $(1.1)$ controls the probability that $X_n$ falls outside a large interval.

2. Tightness.

   Fix $\epsilon > 0$. By the preliminary proposition, $\varphi$ is continuous, and $\varphi(0) = 1$ (property 1 of CHFs). Choose $u > 0$ small enough that

   $$\frac{1}{u} \int_{-u}^{u} \big(1 - \varphi(t)\big) \, dt < \frac{\epsilon}{2}. \quad\textcolor{gray}{\text{---(1.2)}}$$

   Each $\varphi_n$ is bounded by $1$ in modulus, so $|1 - \varphi_n(t)| \le 2$. Pointwise convergence $\varphi_n \to \varphi$ together with this bound on $[-u, u]$ and the DCT gives

   $$\frac{1}{u} \int_{-u}^{u} \big(1 - \varphi_n(t)\big) \, dt \;\longrightarrow\; \frac{1}{u} \int_{-u}^{u} \big(1 - \varphi(t)\big) \, dt < \frac{\epsilon}{2}.$$

   So there exists $N$ such that for all $n \ge N$, the left side is below $\epsilon$. Combining with $(1.1)$,

   $$\mathbb{P}\!\left(|X_n| \ge \tfrac{2}{u}\right) < \epsilon \quad \text{for all } n \ge N.$$

   The remaining finitely many indices $n < N$ can each be handled by enlarging $M$, since each $\mu_n$ is a probability measure. Setting $M_\epsilon$ to be the maximum of $2/u$ and the values needed for $n < N$ gives $\mathbb{P}(|X_n| \ge M_\epsilon) < \epsilon$ for all $n$. So $\{\mu_n\}$ is tight.

3. Every subsequential weak limit equals $\mu$.

   By the tightness theorem, every subsequence of $\{\mu_n\}$ has a further subsequence $\{\mu_{n_k}\}$ converging weakly to some probability measure $\nu$. Applying the forward direction (already proved) to this sub-subsequence,

   $$\varphi_{n_k}(t) \to \widehat{\nu}(t) \quad \text{for all } t.$$

   But by assumption $\varphi_{n_k}(t) \to \varphi(t) = \widehat{\mu}(t)$. So $\widehat{\nu} = \widehat{\mu}$, and by the inversion formula characteristic functions determine the measure: $\nu = \mu$.

   Every weak subsequential limit of $\{\mu_n\}$ equals $\mu$. This forces $\mu_n \Rightarrow \mu$: if $F_n(x) \not\to F(x)$ at some continuity point $x$ of $F$, some subsequence stays bounded away from $F(x)$, but tightness lets us extract a sub-subsequence converging weakly to $\mu$, hence $F_{n_{k_j}}(x) \to F(x)$ at continuity points, contradicting the choice of subsequence.

Why this is useful

Two distributions are equal iff their characteristic functions agree (by the inversion formula); the continuity theorem extends this to the dynamic setting: $\mu_n$ converges weakly to $\mu$ iff their characteristic functions converge pointwise. Among the consequences:

1. The Central Limit Theorem reduces to showing that the characteristic function of the standardized sum converges to $e^{-t^2/2}$, the characteristic function of the standard normal.
2. Sums of independent random variables become tractable because characteristic functions of independent sums multiply, so weak convergence of sums reduces to a multiplicative computation.
3. A pointwise limit $\varphi_n \to g$ does not automatically yield weak convergence to a probability measure: it requires $g$ to be the characteristic function of a probability measure, which by the proposition above requires $g$ to be continuous at $0$. Continuity of the limit at $0$ is what guarantees no mass escapes to infinity.