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The exercise is the following:

Let $E\subset\mathbb{R}^n$ such that there exists an $F_\sigma$-set $F\subset E$ with $\lambda(F)=\lambda^*(E)$. How can I prove that $E$ is Lebesgue measurable (even if $\lambda^*(E)=\infty$)?

obs.: Here, $\lambda^*$ is the Lebesgue outer measure on $\mathbb{R}^n$ and $\lambda$ is the Lebesgue measure (obtained by the restriction of $\lambda^*$ to the $\lambda^*$-measurable sets).

If $\lambda^* (E)<+\infty$ it is easy. I am completely stuck on the case where $\lambda^* (E) = +\infty$.

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    Every set has a value assigned to it by $\lambda^*$, measurable or otherwise. Any bounded set, for example, has a finite $\lambda^*$ but clearly they are not all measurable. – 2017-01-26
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    This is a false statement. For example, pick $(- \infty , -1) \cup V$, where $V$ denotes the Vitali set (a non-measurable set contained in $[0,1]$). – 2017-01-26
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    @Crostul nop. If $\lambda(F)=\lambda^*(E)<\infty$, since $F\subset E$, $\lambda(F)<\infty$ and $F$ is measurable, we have $\lambda^*(E\backslash F)=\lambda^*(E) - \lambda^*(F) = 0$. So $E\backslash F$ is measurable. Then $E=F\cup E\backslash F$ is measurable, as union of two measurables sets. So for $\lambda^*(E)<\infty$ it is true. – 2017-01-26
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    @Crostul So for $\lambda^*(E)=\infty$ it is false, huh? I never seen this set before, but I'll take a look. – 2017-01-26
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    Perhaps the exercise should have stated $\lambda^* (E \setminus F) = 0$? The problem above, as Crostul pointed out, is that we know very little about $E \setminus F$ if we are just give the information above. – 2017-01-26
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    @copper.hat nothing more than I wrote. There is some hypothesis missing. Thank you all. =D – 2017-01-26

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