Corollary 3

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

Integration, IV, p. 53, § 4. Integrable functions and sets

If $\mu$ and $\nu$ are two measures on $X$ such that $\mu(K) = \nu(K)$ for every compact subset $K$ of $X$ , then $\mu = \nu$ .

For, it follows from Cor. 2 and the definition of a measure that, for every compact subset $S$ of $X$ , $\mu$ and $\nu$ take on the same values in $\mathscr{K}(X, S; \mathbf{C})$ .

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

Cited by

Theorem 5, § 4
Corollary, § 1

Cites

Cor. 2