# Several definitions of Approximate continuity of real functions

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 $$A\subset \mathbb R$$.

## 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 $$|f(x)-f(x_0)|<\epsilon\quad \text{whenever}\quad x\in (x_0-\delta, x_0+\delta)\cap A$$

## 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.

• Jan 15 '20 at 11:16

## 1 Answer

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.

• Let me know if anything needs clearing up! Jan 19 '20 at 16:51