Several definitions of approximate continuity of real functions

Question

I found the definition of approximate continuity stated as follows:

A function $f:\mathbb R \to \mathbb R$ is approximately continuous at $x_0$ iff there exists a set $A\in \mathcal L$ such that $x_0\in \Phi(A)$ and $$\lim\limits_{x\to x_0,\ x\in A}f(x)=f(x_0)$$

where, $\mathcal L$ is the set of all Lebesgue measurable subsets in $\mathbb R$, and $\Phi(A)$ is the set of all density points of $A\subset \mathbb R$.

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Question1: Can we write the above definition in $\epsilon$-$\delta$ form as follows?

A function $f:\mathbb R \to \mathbb R$ is approximately continuous at $x_0$ if and only if for each $\epsilon>0$ there exist $\delta>0$ and $A\in \mathcal L$ with $x_0\in \Phi(A)$ such that $$|f(x)-f(x_0)|<\epsilon\quad \text{whenever}\quad x\in (x_0-\delta, x_0+\delta)\cap A$$

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Question2: Also, can we write the definition of "Approximate continuity" as follow?

A function $f:\mathbb R \to \mathbb R$ is approximately continuous at $x_0$ if and only if for each $\epsilon>0$ the set $\{y\in \mathbb R: |f(y)-f(x_0)|<\epsilon\}$ has $x_0$ as a density point.

Answer 1

I will refer to the three definitions of approximate continuity given as AC0, AC1 and AC2 respectively and show they are all equivalent.

AC1 iff AC2: Given a set $A$ by AC1, $A$ has density 1, $A$ is a subset of the set in AC2, and so the set in definition 2 has density $1$ at $x_0$ Conversely, given such a set in AC2, take $A$ to be that set.

AC0 implies AC1: Take $A$ to be the set given in AC0 for all $\epsilon$.

AC2 implies AC0: Assume that $f$ satisfies AC2. Thus for every positive integer $k$, there exists a set $A_k$ of density $1$ at $x_0$ such that the set of all $x$ in $A_k$ such that $|f(x_0) - f(x)| < \frac{1}{2^k}$ has $x_0$ as a density point.

Now, since the finite intersection of sets with density $1$ at a point again has density $1$ there, given any finite natural $n$, there exists an $r_n > 0$ such that the intersection of the $A_k$ from $1$ to $n$, denoted $C_n$ satisfies $m(C_n \cap B_r (x0))/2r > 1 - 1/2^n$ for all $r \leq r_n$. Let us further choose $r_n$ to be monotonically decreasing to $0$ and $r_{n-1} < r_n/2^n$.

Denote by $D_n$ the set of $x$ such that $r_{n+1} \leq |x - x0| \leq r_n$.

Then the set $A := \bigcup_i C_i \cap D_i$ is the set required in AC0.

Indeed, given $\epsilon$, choose $n$ so large such that $1/2^{n-2} < \epsilon$. Then for all $r < r_n$, the measure of $A$ in $B_r (x)$ is at least $2r(1 - \epsilon)$, so $A$ has density $1$ at $x_0$.

It is easy to see that $A$ satisfies the limit condition, so $A$ is the required set.

Answer 2

Sorry, but I do not see how the final assertion on the measure of $A$ in $B_r(x_0)$ follows for every $r<r_n$, with your construction (it follows immediately only for SOME $r$). Moreover, the intersection of the sets $A_k$, for $k=1,...,n$ is simply $A_n$ (considering simply $A_k$ the set where $|f(x)−f(x_0)|<1/2^k$, or am I wrong?).