A function which is not equal almost everywhere (wrt Lebesgue measure) to a continuous function and not almost everywhere continuous function

Question

I know a function which is not equal a.e to a continuous function is the step function or the characteristic of any interval and I also know the Dirichlet function is not an a.e continuous function but I want an example of a function with both properties.

Answer 1

Choose any 'highly' discontinuous function, like nowhere continuous functions. Examples include Conway's base-13 function and real additive functions that are not linear.

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If you would like a more complete answer, this (freely accessible) paper fully characterizes functions $f:\mathbb R\longrightarrow \mathbb R$ which are almost everywhere equal to some continuous function.

Let $I_r(z)=\{x\in\mathbb R\,;\,|x-z|<r\}$. We say that $f$ has a $C$-limit $l$ at $x$ if

$$\forall \epsilon>0,\,\exists\delta>0,\quad \mu\Big(I_\delta(x)\cap f^{-1}\big(\mathbb R\setminus I_\epsilon(l)\big)\Big)=0.$$

Let $\mathscr C_f$ be the set of $x\in\mathbb R$ at which $f$ admits some $C$-limit. Then $f:\mathbb R\longrightarrow \mathbb R$ is almost everywhere equal to some continuous function if and only if $\mathscr C_f=\mathbb R$.

Answer 2

Proposition: If $g:\Bbb R\to \Bbb R$ is continuous and not constant then $\{x:0\ne g(x)\ne 1\}$ is not Lebesgue-null. Proof: $g^{-1}(\Bbb R \setminus \{0,1\})$ is open so if it is not empty it is not Lebesgue-null. But if it is empty then $g$ is a continuous real function that takes values only in $\{0,1\},$ which requires $g$ to be constant.

For $S\subset \Bbb R$ let $\chi_S:\Bbb R\to \Bbb R$ denote the characteristic (indicator) function of $S.$

Corollary to the Proposition: If $S\subset \Bbb R$ and $S$ is not Lebesgue-measurable and if $g:\Bbb R \to \Bbb R$ is continuous then $\neg (g=\chi_S\; a.e.).$ ....Proof of Corollary:

(i). If $g$ is not constant then $\{x:g(x)\ne \chi_S(x)\}\supset \{x:0\ne x\ne 1\},$ which, by the Proposition, is not Lebesgue-null.

(ii). If $g=1$ or if $g=0$ then $\{x:g(x)\ne \chi_S(x)\}\in \{S,\Bbb R \setminus S\}$ and neither $S$ nor $\Bbb R \setminus S$ is Lebesgue-null as neither is Lebesgue-measurable.

There is an elementary example of a non-Lebesgue $S\subset \Bbb R$ which is dense and co-dense. Then $\chi_S$ is nowhere continuous (because $S$ and $\Bbb R \setminus S$ are both dense), and by the above Corollary, $\neg (\chi_S=g \;a.e.)$ for any continuous $g:\Bbb R\to \Bbb R$.