Can we always find a continuous a.e. probability density function?

Question

Given an absolutely continuous cumulative distribution function, we know the corresponding probability density function is not unique, but it is determined almost everywhere. My question is: For any absolutely continuous cumulative distribution function, can we always find a continuous almost everywhere probability density function?

For example, $f(x)*1\{x\in\mathbb{R} - \mathbb{Q}\}$ is nowhere continuous, where $f(x)$ is a continuous probability density function. But $f(x)=f(x)*1\{x\in\mathbb{R} - \mathbb{Q}\}$ a.e. So $f(x)$ is a desired probability density function.

Answer 1

There may be a simpler argument but here is one that shows that the answer is in the negative. Let $E$ be any set of positive finite measure. If $I_E=f$ a.e. where $f$ is continuous a.e. then $\{f<1\}$ differs from an open set only by a set of measure zero. This means that $E$ differs from an open set only by a set of measure zero. Not every measurable set of finite measure has this property.

Answer 2

The answer is negative. The following is a proof.

The following web shows that one construct a Borel set $A⊂[0,1]$ such that $0<μ(A∩I)<μ(I)$ for every interval $I ⊂ [0,1]$. Hence $0<μ(A)<1$.

Construction of a Borel set with positive but not full measure in each interval

Now we prove $f(x)=\frac{1\{x\in A\}}{μ(A)}$ satisfies that there is no a continuous a.e. function $g$ such that $f=g$ a.e..

Suppose there exists such $g$. Denote $C:=\{x:f(x)\neq g(x)\}$, and $D:=\{x: g(x) \text{ is not continuous at } x\}$. Then $\mu (C)=\mu (D)=0$, and $\mu (C\cup D)=0$. For $ x_0\in A-C\cup D$, $g(x_0)=f(x_0)=\frac{1}{μ(A)}$. Since $g$ is continuous at $x_0$, for any $\epsilon>0$, there exists a $\delta>0$ such that $|g(x)-\frac{1}{μ(A)}|<\epsilon$ for $x\in (x_0-\delta,x_0+\delta)$. Choose $\epsilon<\frac{1}{μ(A)}$, then $g(x)>0,x\in (x_0-\delta,x_0+\delta)$. On the other hand, by the definition of $A$, we know that $0<\mu((x_0-\delta,x_0+\delta)\cap A)<2\delta$. Hence $0<\mu((x_0-\delta,x_0+\delta)\cap ([0,1]-A))<2\delta$. Observe that for $x\in (x_0-\delta,x_0+\delta)\cap ([0,1]-A))$, $f(x)=0$ and $g(x)>0$, hence $\mu(C)\ge \mu((x_0-\delta,x_0+\delta)\cap ([0,1]-A))>0$. This contradicts with the assumption $\mu(C)=0$.