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Recalling how nowhere continuous functions such as the Dirichlet function can sometimes be modified on a $\lambda$-null set of points (in this instance, a countable set) to become everywhere continuous, I was wondering whether a Weierstrass nowhere differentiable function can be modified on a $\lambda$-null set of points to become piecewise (or even everywhere) differentiable. If not, is there an example of such a nowhere differentiable function that remains nowhere differentiable irrespective of modifications on $\lambda$-null sets? And such nowhere continuous functions?

($\lambda$ denotes Lebesgue measure)

Sai
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1 Answers1

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You can't fix a continuous nowhere differentiable function by redefining it on a set of measure zero.

Claim. Suppose $f$ is continuous and $f'(a)$ does not exist. If $f=g$ almost everywhere, then $g'(a)$ does not exist.

Proof. Suppose $g'(a)$ exists. Then $g(a)=f(a)$; otherwise $g$ would not be even continuous at $a$. Subtracting a linear function from both $f,g$ we can make $g'(a)=0$. Since it's not true that $f'(a)=0$, there is $c>0$ such that the set $$U = \{x: |f(x)-f(a)|> c|x-a|\}$$ has $a$ as its limit point. Since $f$ is continuous, $U$ is open. Therefore, the intersection of $U$ with $(a-r,a+r)$ has positive measure for every $r$. Let $$ V=\{x : |g(x)-g(a)|> c|x-a|\}$$ Since $$\lambda(V\cap (a-r,a+r)) = \lambda(U\cap (a-r,a+r)) >0$$ the intersection $V\cap (a-r,a+r)$ is nonempty. It follows that $a$ is a limit point of $V$, which contradicts $g'(a)=0$. $\Box$

And such nowhere continuous functions?

Yes, there are nowhere continuous functions that remain nowhere continuous, no matter how they are redefined on a null set. The characteristic function of this set is an example.

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  • Very nice proof. One minor point: You only assume $f$ to be continuous at $a$, but use continuity in at least a neighbourhood of $a$ to conclude that $U$ is open (openness of $U\cap V$ for some neighbourhood $V$ of $a$ would suffice). – PhoemueX Jun 09 '14 at 20:39
  • @PhoemueX Thanks for the correction. The continuity only at $a$ indeed wouldn't be enough. To keep things simple, in the revised version I assumed $f$ to be continuous everywhere. This is enough to answer the question, and it should be clear to the reader that the matters of continuity and differentiability are local. –  Jun 09 '14 at 20:43
  • Does the fact $a$ is a limit point of $U$ also depend on the continuity of $f$? – Sai Jun 09 '14 at 22:31
  • @Sai No, that one does not. The openness of $U$ does. And the openness is used to show that if $a$ will remain a limit point even if we remove a null set from $U$. –  Jun 09 '14 at 22:39
  • Thanks for your answer and your reply! Could you be a bit more detailed? In particular, why does the intersection have positive measure, and how does it follow from openness that the second set also has $a$ as a limit point? – Sai Jun 09 '14 at 22:49
  • @Sai Two hints: (1) a nonempty open set has positive measure. (2) definition of limit point. –  Jun 09 '14 at 22:51
  • Hehehe, thanks. – Sai Jun 09 '14 at 22:54
  • I don't understand why having positive measure implies keeping $a$ as a limit point :\ – Sai Jun 09 '14 at 23:49
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    @Sai I edited the answer to clarify. –  Jun 09 '14 at 23:52
  • I can only thank you again :) – Sai Jun 09 '14 at 23:53