# Is it true that $f' \le 0$ for almost everywhere in (0,1) imply $f$ is monotone decreasing in (0,1)?

Let $$f:(0,1) \to \mathbb{R}$$ be a function satisfying $$f' \le 0$$ for a.e. in $$(0,1).$$

(1) Is it true that the function $$f$$ is continuous and/or monotone decreasing in $$I$$?

(2) If we give an additional condition $$f' \in L^1(0,1),$$ what properties for $$f$$ do we get?

I would be grateful if you give any comments for my questions. Thanks in advance.

• Is $f'$ a weak derivative, or an a.e. pointwise derivative? – Calvin Khor Apr 28 at 10:16
• @CalvinKhor It is pointwise derivative. – 04170706 Apr 28 at 10:18
• well, then the example of Viktor shows you don't get either of the two properties even if $f'\in L^1$ – Calvin Khor Apr 28 at 10:19
• @CalvinKhor You are right. Thanks for your comment. – 04170706 Apr 28 at 10:21
• @CalvinKhor I asked a new question. See [math.stackexchange.com/questions/3205422/… – 04170706 Apr 28 at 10:50

The first condition need not be true: Consider $$f(x) = \begin{cases} -x, & \text{if } x \in \left(0, \frac{1}{2}\right), \\ \frac{1}{2} - x, & \text{if } x \in \left[\frac{1}{2}, 1\right). \end{cases}$$ Then, $$f'(x) = -1 \le 0$$ almost everywhere in $$(0,1)$$, but $$f$$ is neither continuous (in $$\frac{1}{2}$$) nor decreasing on $$I$$.
As @CalvinKhor mentioned in the comments, because the Lebesgue integral "doesn't see null set" (i.e a single point $$x = \frac{1}{2}$$), we also have $$f' \in L^1(0,1)$$, so we don't get any other properties with that condition.
An interesting example is Cantor's singular function. It is continuous, satisfies $$f'(x) = 0$$ almost everywhere, but is not monotone decreasing. In fact it is increasing.