real analysis - Modifications of Weierstrass's continuous, nowhere differentiable functions

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)

Answer

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 $g'(a)=0$. $\Box$

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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.