[Math] Characteristic function approximated by continuous function

continuitymeasure-theoryreal-analysis

I am trying to do the following problem

Let $E \subset \mathbb R^d$ be measurable and let $\epsilon>0$. Show that if $A \subset E$ is measurable, then there is $f:E \to \mathbb R$ continuous such that $$|\{x \in E: f(x) \neq \mathcal X_A(x)\}|<\epsilon$$

Since $A$ is measurable, there exist $F \subset A \subset G$ with $F$ closed, $G$ open and $|G \setminus F|<\epsilon$. Then $G^c$ is a closed set and since $G^c \subset F^c$, then $G^c \cap F=\emptyset$. By Urysohn's lemma, there exists $f$ continuous with $f|_{F}=1$ and $f|_{G^c}=0$. In particular, $f|_{E}$ is continuous.

Let $S=\{x \in E: f(x) \neq \mathcal X_A(x)\}$. If $x \in S$, then if $x$ can't be in $G^c \cap E$ since here $f(x)=0=\mathcal X_A(x)$. So $x \in E \cap G \subset G$, but $x$ can't be in $F$ since $f(x)=1=\mathcal X_A(x)$. So we have $S \subset G \setminus F$.

Note that $S=f^{-1}(\{f-\mathcal X_A>0\}) \cap E$, since both of these subsets are measurable, then $S$, its intersection, is measurable.

We have $$|S| \leq |G \setminus F|<\epsilon$$

I would like to check if my solution is correct, thanks in advance.

Best Answer

In general, we can hardly expect to find $F,G$ such that $F \subseteq A \subseteq E \subseteq G$ and $|G \backslash F| < \epsilon$ (as $|E \backslash A|>\epsilon$ for $\epsilon$ sufficiently small).

Hints:

  1. Using Urysohn's lemma, prove that the claim holds true for $E = \mathbb{R}^d$ and $A$ open. (That's basically what you already did).
  2. Conclude that the claim holds true for arbitrary $E$ and $A$ open (in $E$).
  3. Define $$\mathcal{D} := \{A \in \mathcal{B}(E); \forall \epsilon>0: \text{claim holds true}\}.$$ Show that $\mathcal{D}$ defines a Dynkin system. Conclude $\mathcal{D}= \mathcal{B}(E)$.
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