Measure and integration Signed measures and differentiation of measures Theorem

Nikodym convergence theorem

Builds on Measurable space Complex measure Countable additivity

Statement

Let $(X, \Sigma)$ be a measurable space and let $(\mu_n)_{n \ge 1}$ be a sequence of complex measures on $\Sigma$ . Assume that for every $E \in \Sigma$ the limit $\mu(E) = \lim_{n \to \infty} \mu_n(E)$ exists in $\C$ . Then:

The $\mu_n$ are uniformly countably additive: for every sequence $E_1 \supseteq E_2 \supseteq \cdots$ in $\Sigma$ with $\bigcap_k E_k = \emptyset$ , one has $\sup_n \abs{\mu_n(E_k)} \to 0$ as $k \to \infty$ .
The set function $\mu$ is countably additive, so it is a complex measure on $\Sigma$ .

Remarks

One proof reduces to the Vitali–Hahn–Saks theorem. Put $\lambda = \sum_n 2^{-n} \abs{\mu_n} / (1 + \abs{\mu_n}(X))$ , a finite measure. Each $\mu_n$ is absolutely continuous with respect to $\lambda$ , and the Vitali–Hahn–Saks theorem gives the conclusion.

Example: let $f_n$ be integrable functions on a measure space with $\int_E f_n$ convergent for every measurable $E$ . Then the limit $E \mapsto \lim_n \int_E f_n$ is again countably additive.

Going deeper

For a student

The theorem is often stated together with the Nikodym boundedness theorem, which says that a family of complex measures bounded on each set is uniformly bounded. Both are measure-theoretic forms of the uniform boundedness principle. They are basic tools in the study of weak convergence in spaces of measures and in $L^1$ .

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Tag	`FCAA02SN0E`
URI	`m2://FCAA02SN0E`
Kind	statement, theorem
Writing	W0 stub
MSC	28A33, 28A10
Status	This copy of the site has no build data. Run `m2 status FCAA02SN0E`.
About	Complex measure
Uses	Measurable space, Complex measure, Countable additivity
Changed	2 November 2026