# Outer measure and set with full measure

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 ?

• It appears to be a probability measure, so it will. – dbx Nov 28 '18 at 0:10
• didn't read it, thanks – Robson Nov 28 '18 at 0:10
• 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. – Phil-W Nov 28 '18 at 0:19

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.