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We know that that a for measure $P$ on $(\mathbb{R},\mathcal{B}(\mathbb{R})$, for every measurable $A$ such that $P(A) < \infty$ there is a $G$ open and $F$ closed such that $F \subset A \subset G$ and $P(G-F)<\epsilon$. I m tring to see if the other way can be true ...

Is a $G$ open and $F$ closed such that $G \subset A \subset F$ and $P(F-G)<\epsilon$. ?

It is not true for the Lebesgue measure but can it be true for another measure.

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up vote 7 down vote accepted

The only measure with this property is the zero measure.

Let $\mathbb{Q}$ be the set of rationals. Notice that the only open set contained in $\mathbb{Q}$ is the empty set (i.e. $\mathbb{Q}$ has empty interior). Hence we must have $P(\mathbb{Q}) = 0$. However, the set of irrationals $\mathbb{Q}^c$ also has empty interior, so $P(\mathbb{Q}^c) = 0$ as well. By additivity we have $P(\mathbb{R}) = P(\mathbb{Q}) + P(\mathbb{Q}^c) = 0$.

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