Htykuut

This question is generalization of Lusin's theorem, though it appears long.

Let $(X, rho)$ and $(Y, sigma)$ be metric spaces with $(Y, sigma)$ separable, and let $mu$ be a finite Borel measure on $X$. Let $f : X to Y$ be $mathcal B_X, mathcal B_Y)$-measurable and let $epsilon > 0$. Prove that there is a closed set $F subset X$ such that $mu(F^c) < epsilon$ and $f|_F$ is continuous.

Here, $mathcal B_X$ denote the Borel $sigma$ algwbra on $X$, and similarly for $mathcal B_Y$. I have proven that $mathcal B_Y$ is generated by the collection of all open balls in $Y$.

I am to model after a version of Lusin's theorem that uses regularity of finite measures, as the following:

"The real line has only countably many open intervals whose end-points are both rational. Let ${ I_n }$ be an enumeration of them.

For each $n$, the regularity of $µ$ gives closed subsets $A_n ⊂ f^{-1}[I_n]$ and $B_n ⊂ f^{-1}[mathbb R - I_n]$ such that $$µ((A_n ∪ B_n)^c) = µ(f^{-1}[I_n] - A_n) + µ(f^{-1}[mathbb R - I_n] - B_n) < ε2^{-n}.$$

Let $F := bigcap_n(A_n ∪ B_n)$. This is closed, and it satisfies

$$mu(F^c) = µ( bigcup (A_n ∪ B_n)^c) ≤ sum µ((A_n ∪ B_n)^c) < ε.$$

Let us show that $f|F$ is continuous. Let $x ∈ F$ and $ε > 0.$ By the density of $mathbb Q$ in $ mathbb R$, there is a rational open interval In which satisfies $f(x) ∈ I_n ⊂ (f(x) − ε, f(x) + ε)$.Thus $ x$ lies in both $F$ and $f^{-1}[I_n]$, and so $x ∈ A_n ⊂ B_n^c.$ Since $B_n$ is closed, we can now choose $ δ > 0$ so that $B_δ(x)$ is disjoint from $B_n$. Therefore, if $y ∈ F$ satisfies that $ρ(x, y) < δ$, then it cannot lie in $B_n$, so it must also lie in $A_n$. This implies that $f(y) ∈ I_n ⊂ (f(x) − ε, f(x) + ε).$

In adapting this proof we assume that for a given borel set $I$ of $Y$ (which, by what I have proved, is an open ball- and specifically, an open ball of rational radius) there is a closed set $K$ and an open set $U$ such that $K subset I subset U$, with all sets concentric balls and $K$ and $U$ with radius close enough to $I$ such that $mu(U-K)$ small.

And then proceed similarly as above.

Is my reasoning correct?