When all Borel subsets of a Borel set B have measure 0 for signed $\mu$, prove within Bourbaki MT that B has measure 0 for $|\mu|$ (total variation)

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Suppose X is a locally compact space and $\mu$ a bounded and regular signed measure defined on the Borel sets of X, $B$ one of these sets. Prove that if $\mu(B_1)=0$ for all Borel sets $B_1$ contained in $B$, then $|\mu|(B)=0$.

Motivation: Rudin's Functional Analysis, chapter on bounded operators in Hilbert spaces. I found that I needed the result above (which I hope is true!) with complex measures to confirm the validity of the proof of the first non trivial theorem about resolutions of the identity (12.21 in my edition) ... and I was able to deduce that complex case from the real (signed) case

(Bounded regular measures are those that are associated with linear maps from the space of continuous complex-valued functions with compact support defined in X to $\mathbb C$ that are continuous with respect to the sup-norm on said space.)