Is $m(\cap_{n=1}^\infty (A_n\setminus E))=\lim\limits_{n\to\infty} m(A_n\setminus E)$, where the $A_i$'s are measurable, and $E$ is non-measurable?

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We know the following property for Lebesgue measure $m$

If $A_1\supset A_2\supset\dots$, then $m(\cap_{I=1}^\infty A_i)=\lim\limits_{n\to\infty}m(A_n)$

Is this also true for $m(\cap_{n=1}^\infty (A_n\setminus E))$, where $E$ is a non-measurable set contained in all the $A_i$'s (so $E\subset \cap_{n=1}^\infty A_i$)? In other words, is $m(\cap_{n=1}^\infty (A_n\setminus E))=\lim\limits_{n\to\infty} m(A_n\setminus E)$?

I feel like this should be true, going by what I understand from the proof (the breaking up of the $A_i$'s into disjoint measurable sets, etc. However, I'm not sure. If the equality is false, is the following true: $m(\cap_{I=1}^\infty (A_i\setminus E)\leq \lim\limits_{n\to\infty}m(A_n\setminus E)$?

EDIT: Here $m$ is just outer measure.

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The inequality is true. If we let $\mathfrak{M}$ be the set of all measurable sets, this follows from the fact that $$m(B)=\inf\{m(A) : A\in\mathfrak{M}, B\subseteq A\}$$ for any set $B$ contained in some measurable set (so that the set under $\inf$ above is nonempty). And this in turn follows easily from the definition of the outer measure (at least the one appearing here).