Construct a set of intersections using a Lebesgue measurable function

Question

Let $G_\delta$ denote the collection of all intersections of countably many sets in $G$.

Show that if $f$ is Lebesgue measurable, then there is a $G_\delta$ set $G$ with $m(G)=0$ so that $f \chi_{G^c}$ is Borel measurable.

Hint: Write $f$ as a linear combination of positive functions. If $f\geq0$, let $\{ r_n : n \geq 1\}$ be dense in $[0, \infty)$ and consider $f^{-1}([0,r_n))$. Remove a set of measure $0$ from each to get an $F_\sigma$ set, then find. a $G_{\delta}$ set G with $m(G)=0$ containing all of them.

I'm confused about the linear combination part of the hint. I know that every simple function can be written as a linear combination of characteristic functions. But not sure why a Lebesgue measurable function can also do so. From my understanding, we know that for each $n\in \mathbb{N}$, the set $f^{-1}([0, r_n ))$ is Lebesgue measurable. So we're taking the union of each $F_\sigma$ to get the $G_\delta$ set?

Thank you in advance

$G_\delta$ sets are countable intersections of open sets, not sure what you were trying to say in the first line, but I don't think it's right. — Physical Mathematics, Jan 31 '21 at 20:24
Sorry yes! I meant intersection. Thank, just edited the question now. — Toasted_Brain, Jan 31 '21 at 20:52

score 1 · Accepted Answer · answered Jan 31 '21 at 20:29

Note that Lebesgue measure is just the completion of the associated Borel measure. We have the following lemma.

Lemma: Suppose $(X, \mathcal M, \mu)$ is a measure space and $(X, \overline{\mathcal M}, \overline \mu)$ is it completion, then $f: X \to Y$ for some measurable space $(Y, \mathcal N)$ is $\overline{\mathcal M}$ measurable iff $f = g$ $\overline{\mu}$ a.e. for some $\mathcal M$ measurable function $g$. See here for a proof in the special case we care about here.

Thus if $f$ is Lebesgue measurable, $f = g$ a.e. where $g$ is Borel measurable. Then let $E$ a Lebesgue null set s.t. $f=g$ on $E^C$. Then by outer regularity of Lebesgue measure, $m(E) = \inf \{m(U) : U \text{ open}, E \subseteq U\}$, so let $U_n$ a sequence of open sets containing $E$ s.t. $m(U_n) \to m(E) = 0$. Let $G = \bigcap_n U_n$, then $G$ is $G_\delta$ set with $0$ measure containing $E$. Thus $f =g $ on $G^C$ also, and so $f \circ \chi_{G^C} = g \circ \chi_{G^C}$, but $\chi_{G^C}$ is Borel measurable and $G$ is Borel, so $g \circ \chi_{G^C}$ is Borel. Thus $f \circ \chi_{G^C}$ is also. Thus $G$ is the set we were looking for.

Construct a set of intersections using a Lebesgue measurable function

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