Bourbaki

Proposition 17

Tag 08TD, permanent. Cite it as [Bourbaki, Tag 08TD].

Integration, IV, p. 50, § 4. Integrable functions and sets

In order that a nonempty set Φ\Phi of subsets of a set AA be a clan, it is necessary and sufficient that it satisfy the following condition:

(CL) For every pair of sets M,NM, N belonging to Φ\Phi, the sets M∪NM \cup N and M∩CNM \cap \mathbf{C}N belong to Φ\Phi.¹

The condition is necessary, by virtue of the relations

φM∪N=φM+φN−φMφN,φM∩CN=φM−φMφN. \varphi_{M \cup N} = \varphi_M + \varphi_N - \varphi_M \varphi_N , \quad \varphi_{M \cap \mathbf{C}N} = \varphi_M - \varphi_M \varphi_N .

To show that it is sufficient, we first observe that it implies that, for any two sets M,NM, N in Φ\Phi, M∩NM \cap N belongs to Φ\Phi since M∩N=M∩C(M∩CN)M \cap N = M \cap \mathbf{C}(M \cap \mathbf{C}N). Let E(Φ)\mathscr{E}(\Phi) be the set of linear combinations, with real coefficients, of the characteristic functions of the sets of Φ\Phi. Since φMφN=φM∩N\varphi_M \varphi_N = \varphi_{M \cap N}, E(Φ)\mathscr{E}(\Phi) is an algebra. Everything comes down to showing that if MM is a subset of AA such that φM=∑iciφMi\varphi_M = \sum_i c_i \varphi_{M_i}, where the MiM_i belong to Φ\Phi, then M∈ΦM \in \Phi. This will result from the following lemma:

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.