Problem: Show that there exists a continuous strictly increasing function $f$ on $\mathbb{R}$ such that $f'(x) = 0$ almost everywhere.
Attempt:
Perhaps we could modify the Cantor Function which is non-decreasing, continuous, and constant on each interval in the complement of the Cantor Set in $[0,1]$.
Of course, we'd need to do the following:
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Make the Cantor function strictly increasing (since the Cantor Function is constant on certain intervals, it's certainly not strictly increasing as is).
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Extend our modified Cantor Function from $[0,1]$ to all of $\mathbb{R}$.
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Verify that $f'(x) = 0$ for all irrational $x$ (or for all but a countable subset of $\mathbb{R}$).
Best Answer
Aside. Your item 3 sets an unattainable goal. If a continuous function $f$ satisfies $f'=0$ for all but countably many points of $\mathbb R$, then $f$ is a constant function. See Set of zeroes of the derivative of a pathological function.
Here is a more explicit construction than in the post link in comments. Fix $r\in (0,1/2)$. Let $f_0(x)=x$. Inductively define $f_{n+1}$ so that
Here is a piece of this function with $r=1/4$:
and for clarity, here is the family $f_0$,$f_1$,... shown together. All the construction does is replace each line segment of previous step with two; sort of like von Koch snowflake, but less pointy. (It's a monotonic version of the blancmange curve).
To check that the properties hold, you can proceed as follows:
(It may be more enjoyable to prove part 3 using the Law of Large Numbers from probability.)