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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.

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

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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.

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