How a measurable function becomes a differentiable function almost everywhere Let $f(x): (0,\infty)\to \mathbb R$ be a measurable function on $\mathbb R^+=(0,\infty)$. Furthermore, assume that $f(x)\geq x$ for all $x>0$. I think $f$ may has not derivative almost everywhere, but I could not construct such type of this function. 
 A: Are you asking for a measurable function $f:(0,\infty)\to(0,\infty)$ with $f(x)\geq x$ which is not differentiable on a set with measure greater $0$?
$$
f(x):=\begin{cases}
x+1 & x\in\mathbb{Q}\\
x & x\in\mathbb{R}\setminus \mathbb{Q}.
\end{cases}
$$
This function is measurable and nowhere differentiable and $f(x)\geq x$.
Proof: 
First we see that $f(x)\in\mathbb{Q}\Leftrightarrow x\in\mathbb{Q}$. For a measurable set $A$ you conclude
$$
f^{-1}(A)=f^{-1}((A\cap\mathbb{Q})\dot\cup(A\cap\mathbb{R}))=f^{-1}(A\cap\mathbb{Q})\cup f^{-1}(A\cap\mathbb{R})=(A-1)\cup A
$$
 where $A-1:=\{a-1~:~a\in A\}$. Since $A$ is measurable so is $A-1$ and therefore $(A-1)\cup A=f^{-1}(A)$ and $f$ is measurable.
And $f$ is nowhere differentiable since it is nowhere continuous because $\mathbb{Q}$ is dense in $\mathbb{R}$.
Extra:
There are a lot of funny constructions. You can also make the function not differentiable on an intervall like this:
$$
f(x):=\begin{cases}x & x\in(0,1)\cup([1,2]\cap\mathbb{R})\cup(2,\infty)\\x+1 & [1,2]\cap\mathbb{Q}\end{cases},
$$
which is differentiable on $(0,1)\cup(2,\infty)$ with $f'(x)=1$ but not in $[1,2]$.
Or just differentiable in a point like this:
$$
f(x)=\begin{cases}
x & x\in\mathbb{R}\\
x^2-2x+1 & x\in\mathbb{Q}
\end{cases},
$$
which is differentiable just in $1$ with $f'(1)=1$ but nowhere else.
