---
tag: K5H0RT71A3
kind: definition
name: Weak convergence of measures
msc: [60B10, 28A33]
defines: MBW9JCGZC1
uses: [0VKD1TP7F3, 059TJN0NZ1, 0WQXY32Q8D, 0X3WGX95FE, DWJF7ABPR1, 0ACRGYNHSR]
sources:
  - url: https://en.wikipedia.org/wiki/Convergence_of_measures
---

## Definition

Let $X$ be a [[metric space|0VKD1TP7F3]] with its [[Borel σ-algebra|0WQXY32Q8D]].
Let $\mu$ and $\mu_1, \mu_2, \dots$ be [[probability measures|059TJN0NZ1]] on $X$.
The sequence $(\mu_n)$ *converges weakly* to $\mu$ if
$$\lim_{n \to \infty} \int_X f \, d\mu_n = \int_X f \, d\mu$$
for every bounded [[continuous function|0X3WGX95FE]] $f \colon X \to \R$.
The integrals are [[Lebesgue integrals|DWJF7ABPR1]].

## Remarks

Common notations are $\mu_n \Rightarrow \mu$ and $\mu_n \xrightarrow{w} \mu$.
Some authors call the same notion narrow convergence.
Weak limits are unique on a metric space.

Example: let $\delta_a$ be the [[Dirac measure|0ACRGYNHSR]] at $a$ on $\R$.
Then $\delta_{1/n}$ converges weakly to $\delta_0$.
Indeed $\int f \, d\delta_{1/n} = f(1/n)$, and $f(1/n) \to f(0)$ because $f$ is continuous.

Non-example: $\delta_n$ does not converge weakly to $\delta_0$.
Take $f(x) = 1/(1 + x^2)$, which is bounded and continuous.
Then $\int f \, d\delta_n = 1/(1 + n^2) \to 0$, but $\int f \, d\delta_0 = 1$.

Random variables converge in distribution exactly when their laws converge weakly.

## Explanations

### L1 human

Weak convergence asks that every bounded continuous average converges.
It allows mass to shift slightly, which makes it the natural notion for limit theorems in probability.