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Let f be a function in $L^1(a, b)$, with $(a, b)$ a real interval, and :

$E+ := \{ x \in (a, b): f(x) > 0 \}$ a non-null set,

$E := \{ x \in (a, b): f(x) = 0 \}$ a null set,

$E- := \{ x \in (a, b): f(x) < 0 \}$ a non-null set.

Is it possible for $E+$ and $E-$ to have both empty interior robustly ?

(Namely: so that interior emptiness is not lost by changing f on a null set.)

Thanks

balddraz
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mra
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1 Answers1

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There is a Borel set $B$ such that, for every interval $(u, v)$ (with $u<v$), we have $$0<m(B\cap (u, v))< v-u,$$ that is, $B$ is neither null nor full measure on $(u, v)$. (See e.g. Construct a Borel set on R such that it intersect every open interval with non-zero non-"full" measure.)

Now let $f(x)=1$ if $x\in B$ and $f(x)=-1$ if $x\not\in B$. If you like (as your question is phrased), restrict $f$ to some interval $(a, b)$.

Community
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Noah Schweber
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    Note that this is really the only way to solve the question - if $f$ is a function as desired, then $E-$ (or E+) has non-zero, non-full measure on every subinterval of $(a, b)$. – Noah Schweber Oct 11 '15 at 09:41