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I've seen two different definitions for an outer measure $\mu$ on $\mathcal{P}(X)$, where $X$ is a set, obtained from a given probability measure on $X$.

D1 = a set of full outer measure is a subset $A\subseteq X$ such that $\mu(A)=1$.

D2 = a set of full outer measure is a subset $A$ such that $\mu(X\backslash A)=0$.

Since outer measures are not additive, I'm having trouble seeing how those two definitions are equivalent. Clearly D2 implies that a set with full outer measure is measurable (for the sigma algebra generated by $\mu$), however it's not clear to me that it's the case for D1.

So my question is : are those definitions equivalent ?

My follow up question is : when we say that $C([0,1])$ has full outer measure for the Wiener measure on $\mathbb{R}^{[0,1]}$, are we using Definition D2, or D1 ?

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  • $\begingroup$ It appears to be a probability measure, so it will. $\endgroup$ – dbx Nov 28 '18 at 0:10
  • $\begingroup$ didn't read it, thanks $\endgroup$ – Robson Nov 28 '18 at 0:10
  • $\begingroup$ I suppose that $\mu$ is constructed using the usual method (method I here en.wikipedia.org/wiki/Outer_measure) from a given probability measure. I don't know if that implies that $\mu(X)=1$ however. $\endgroup$ – Phil-W Nov 28 '18 at 0:19
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D1 and D2 are not equivalent in general.

Consider $X=[0,1]$ and $\mu=$ Lebesgue outer measure

By this thread, there is a non-measurable set $V$ with $\mu(V)=1$, so $V$ is a set of full measure in the sense of D1. However, as you pointed out, any set of full measure in the sense of D2 must be measurable, so $V$ is not such a set.

I do not have knowledge on Wiener measure so I leave it to someone who know it well.

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