$f_\epsilon=\inf\{f(y):|y-x|<\epsilon\}$ is measurable Borel [closed]

Question

I found this problem I have tried but it has been a bit complicated for me,

Let $f:\mathbb{R}\to\mathbb{R}$ a bounded function. For each $\epsilon>0$, let $f_\epsilon (x)=\inf\{f(y):|y-x|<\epsilon\}$. They ask us to prove that:

1. For each $\epsilon>0$, the function $f_\epsilon$ is measurable Borel.

I tried to show that for each $a\in\mathbb{R}$ we have $f_\epsilon^{-1}((-\infty,a))$ is open in $\mathbb{R}$. Let $x\in f_\epsilon^{-1}((-\infty,a))\Leftrightarrow f_\epsilon(x)<a$. On the other hand we have that for all $R>0$ exists $y_0\in (x-\epsilon,x+\epsilon)$ such that $f(y_0)<f_\epsilon(x)+R<a+R$. Then taking $r=\epsilon-|x-y_0|>0$. Let $z\in (x-r,x+r)$, then $|z-y_0|=|z-x|+|x-y_0|<\epsilon\Rightarrow f_\epsilon(z)\leq f(y_0)<a+R$, for all $R>0$. Thus $f_\epsilon(z)\leq a$, then I need prove that $f_\epsilon(z)\neq a$ that way $(x-r,x+r)\subset f_\epsilon^{-1}((-\infty,a))$ but I couldn't get that $f_\epsilon(z)\neq a$. There are any idea for that?

2. The function $h:\mathbb{R}\to\mathbb{R}$ given by $h(x)=\sup\{f_\epsilon(x):\epsilon>0\}$ is measurable Borel too.

I don't know how to start this part, since it is a supreme over an arbitrary set that is not even countable.

Answer 1

As noted, your idea is right. Let me try to do the proof with greater clarity; to make it easier for the reader to understand.

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$f_\epsilon (x)=\inf\{f(y):|y-x|<\epsilon\}$.

Fix $\epsilon>0$. We claim that $f_\epsilon$ is Borel measurable. [In fact, even more is true: $f_\epsilon$ is upper semicontinuous.]

Fix $\lambda \in \mathbb R$. We claim: $E := \{x : f_\epsilon(x) < \lambda\}$ is open.

Take any point $z \in E$. That means $f_\epsilon(z) < \lambda$. That is, $$ \inf\{f(y) : |z-y| < \epsilon\} < \lambda $$ so there exists $y_1$ with $|z-y_1|< \epsilon$ and $f(y_1) < \lambda$. Define $\delta = \epsilon - |z-y_1| > 0$.
We claim the open ball $B_\delta(z) \subseteq E$. Indeed, let $x \in B_\delta(z)$. Then $|x-z| < \delta$ so $$ |x-y_1| \le |x-z|+|z-y_1| < \delta + |z-y_1| = \big(\epsilon - |z-y_1|\big)+|z-y_1| = \epsilon . $$ Thus $$ f_\epsilon(x) = \inf\{f(y):|x-y|<\epsilon\} \le f(y_1) < \lambda $$ so $x \in E$. This completes the proof that $B_\delta(z) \subseteq E$.

We have seen: for every $z \in E$, there is $\delta > 0$ so that $B_\epsilon(z) \subseteq E$. This shows $E$ is open.

Finally, we have seen: for every $\lambda \in \mathbb R$, the set $\{x : f_\epsilon(x) < \lambda\}$ is open (and therefore Borel measurable). It follows that $f_\epsilon$ is a Borel measurable function (in fact a lower semicontinuous function).

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$h(x)=\sup\{f_\epsilon(x):\epsilon>0\}$. Claim: $h$ is Borel measurable.

First note: if $\epsilon_1 > \epsilon_2$, then $f_{\epsilon_1}(x) \le f_{\epsilon_2}(x)$. So we have $$ h(x) = \sup\{f_\epsilon(x):\epsilon>0\} = \lim_{\epsilon \to 0^+} f_\epsilon(x) = \lim_{n \to \infty}f_{1/n}(x) , $$ a pointwise limit of a sequence of Borel measruable functions. So $h$ is a Borel measurable function.