Absolutely continuous wrt Lebesgue measure

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I am reading some notes and believe this is a typo:

'We would like an invariant measure $\mu$ on $X=[0,1]$ which is absolutely continuous wrt Lebesgue measure, so that a result which is true for $\mu$-a.e point will be true for Lebesgue-a.e point.'

I believe the typo is the second part, surely it should be a 'result which is true for Leb-a.e. point is true for $\mu$-a.e point'? Since if Leb$(P)=1$, then Leb$(X\setminus P)=0$ so that $\mu(X\setminus P)=0$, ie $\mu(P)=1$.

Unless am I missing something? Could someone please correct me if I am mistaken? Thanks!

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You're right, somebody mixed up the order.

If $\mu \ll \nu$, then $\nu$-a.e. implies $\mu$-a.e. (since the exceptional set is $\nu$-null and thus by absolute continuity $\mu$-null), but not the other way round. Whether it was intended that the Lebesgue measure should be absolutely continuous with respect to $\mu$ or something else, I can't guess.