Another variant on Monotonicity and Differentiability Assume $f$ is continuous on the interval $[a,b]$. Assume further that $f$ is differentiable almost everywhere and $f'> 0$ almost everywhere. Is then $f$ increasing on $[a, b]$? (Clearly, if $f$ were increasing, then $f$ is strictly increasing.) 
Please prove or provide counterexample.
 A: The answer to your question is negative. The reason is that there are monotone continuous functions $f$ that are singular, that is, for a.e. $x$ we have that $f'(x)$ exists and equals $0$. An example is the Cantor function $f:[0,1]\to[0,1]$: This function is continuous and increasing, with $f(0)=0$, $f(1)=1$, and $f'(x)=0$ for all $x$ in the complement of the Cantor set. (There are also examples that are strictly increasing.)
With $f$ the Cantor function, let $g:[0,1]\to\mathbb R$ be given by $$g(x)=\frac x2 - f(x), $$
so $g$ is continuous, differentiable a.e., and $g'(x)=1/2-0>0$ for a.e. $x$. However, $g(0)=0$ and $g(1)=-1/2$, so $g$ is not increasing.   
That said, a sort of weakening of what you were after holds: One can prove that if $f:[a,b]\to\mathbb R$ and $D$ is the set of points where $f$ is decreasing, then $ D $ is measurable and $f$ is differentiable at almost every point $x$ of $D$, with $f'(x)\le 0$. (I do not know of an elementary argument for this. It is a corollary of a nontrivial result of Denjoy and Young. You can find a proof in Chapter 21 of A second course on real functions by van Rooij and Schikhof.) This means that if $ f $ is such that $ f'(x)> 0$ at a.e. $ x $ where $ f $ is differentiable, then the set of points $ y $ such that $ f $ is decreasing at $ y $ has measure $0$.
Finally, note that your assumptions immediately give us that $f$ is increasing almost everywhere, meaning that at a.e. point $x$ we have an interval $I_x$ about $x$ such that for any $y$ in $I_x$ we have $$(f(x)-f(y))(x-y)\ge 0.$$ This is not quite the same as saying that there is a null set $N$ such that $f$ is increasing on $[a,b]\setminus N$, meaning that if $x<y$ and $x,y\notin N$, then $f(x)\le f(y)$, which would imply that $f$ is indeed increasing on $[a,b]$ and therefore strictly increasing.
