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Since our calculus lectures, we know that there are nowhere continuous functions (like the indicator function of the rationals). However, if we change this Dirichlet function on a set of measure zero, then we get a continuous function (the zero function). So my question ist

Does there exist a function $f: \mathbb{R}\rightarrow \mathbb{R}$ such that every function which equals $f$ almost everywhere (with respect to the Lebesgue measure) is nowhere continuous? Is it possible to choose $f$ borel-measurable?

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See my answer to this question for the construction of an $F_\sigma$ set $M\subset\mathbb R$ such that $0\lt m(M\cap I)\lt m(I)$ for every finite interval $I,$ where $m$ is the Lebesgue measure.

Let $f$ be the indicator function of $M.$ Clearly $f$ is Borel-measurable. If $g(x)=f(x)$ almost everywhere, then $g^{-1}(0)$ and $g^{-1}(1)$ are everywhere dense, whence $g$ is nowhere continuous.

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  • $\begingroup$ Thank you for this elegant counterexample. $\endgroup$ – Severin Schraven Mar 3 '18 at 18:41
  • $\begingroup$ You're welcome. $\endgroup$ – bof Mar 3 '18 at 19:29
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Let $r_1,r_2, \dots$ be the rationals. Let $f(x) = x^{-1/2}, x\in (0,1),$ $f=0$ elsewhere. For $x\in \mathbb R,$ define

$$g(x) = \sum_{n=1}^{\infty}\frac{f(x-r_n)}{2^n}.$$

Then $g$ is a Borel measurable function from $\mathbb R$ to $[0,\infty].$ Note that $\lim_{x\to r_n^+} g(x)=\infty$ for all $n.$

Now $\int_{\mathbb R} g <\infty$ by the monotone convergence theorem. Thus $g<\infty$ a.e. Define $h=g$ wherever $g$ is finite, $h=0$ where $g=\infty.$ The function $h$ fits the bill.

To be sure, there are some things to check, but I'll leave it here for now

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  • $\begingroup$ I like your counterexample, I had a hard time deciding which answer I should accept. Thank you. $\endgroup$ – Severin Schraven Mar 3 '18 at 18:43
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Any discontinuous function s.t. $f(x + y) = f(x) + f(y)$.

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    $\begingroup$ Additive discontinuous functions are not Lebesgue measurable, so this example cannot be used for the second part of the question. $\endgroup$ – Kabo Murphy Mar 2 '18 at 8:02
  • $\begingroup$ @KaviRamaMurthy Is it difficult to prove that additive discontinuous functions are not Lebesgue measurable? Do you have a reference? $\endgroup$ – Severin Schraven Mar 3 '18 at 18:44
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    $\begingroup$ @SeverinSchraven, math.stackexchange.com/questions/45861/…. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 3 '18 at 19:19
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    $\begingroup$ @Severin, suppose f is additive and measurable. The $\mathbb R =\cup f^{-1} (-n,n)$ so there must be some n such that $E \equiv f^{-1} (-n,n)$ has positive measure. This implies E-E contains an interval around 0. [Ref. Halmos 's Measure Theory]. From this and additivity prove that f is bounded on some interval $(-a,a)$, say by M. Now $|f(x)|=\frac 1 n f(nx)| \leq \frac 1 n M$ if n is large enough. Conclude that f is continuous at 0 and use additivity to get continuity at all points. $\endgroup$ – Kabo Murphy Mar 5 '18 at 4:51
  • $\begingroup$ @KaviRamaMurthy and Martin: Thank you for the references. $\endgroup$ – Severin Schraven Mar 5 '18 at 8:16

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