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This is an exercise from the book Measure,Integration & Real Analysis by Sheldon Axler.
For A ⊂ R, the quantity sup {| F | : F is a closed bounded subset of R and F ⊂ A } is called the inner measure of A.
here |F| is the outer measure of F. Show that if A is a Lebesgue measurable subset of R, then the inner measure of A equals the outer measure of A.

I could prove this if F given in the definition above is just a closed subset of A. Inner measure is less than or equal to outer measure by definition. If A is lebesgue measurable, then for each $\epsilon>0$ there exists a closed subset F of such that |A| -$\epsilon$ < |F| .If F is bounded, the proof is done. But why is F bounded?

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$|A|-\epsilon <|F\cap [-N,N]|$ if $N$ is large enough. $F\cap [-N,N]$ is a closed and bounded subset of $A$.

(Note: This argument works if $A$ has finite measure. When $m(A)=\infty$ you have to argue separately).

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