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Let us assume that $f:\mathbb{R}\to \mathbb{R}$ is differentiable and $f'(x)>0$ almost everywhere.

If $f'\in L^1_{loc}$, then FTC implies that for any $x,a\in \mathbb{R}$, $$ f(x)-f(a)=\int_a^x f'(t)dt. $$ Therefore, we have $f(x)\geq f(a)$, because $\int_a^x f'(t)dt\geq \frac{1}{n} m(\{x;f'(x)>\frac 1 n\})$ for any $n\in \mathbb{N}$. Then we know that $f$ has to be strictly monotone. Otherwise, there exits an interval $(c,d)\subset [a,b]$ such that $f(x)$ is constant over $(c,d)$, which is impossibe because $f'(x)=0$ is zero over this interval and hence $f'(x)>0$ a.e. fails.

How about removing the assumption that $f'\in L^1_{loc}$? Does $f'(x)>0$ almost everywhere imply that $f$ is strictly monotone

Yuhang
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  • It suffices to show that your assumption implies nonstrict monotonicity for a monotonic function is always $ L^1_{loc} $. – Bananach Dec 17 '17 at 14:31
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    what does $L^1_{loc} mean?$ – user Dec 17 '17 at 14:40
  • @Bananach I think this is the hardest bit. If we could assume $f'(x)\ge 0$, there would be quite a few ways to conclude the proof. –  Dec 17 '17 at 16:08
  • By Darboux's theorem, if $f'<0$ anywhere then there are already uncountably many points with $f'<0$. I think the road to the proof is showing that if $f'(a) = b < 0$, then $(f ')^{-1}((b,0))$ has positive measure, leading to a contradiction. – Hyperplane Dec 17 '17 at 17:23
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    If you are willing to accept the fundamental theorem of calculus for the Henstock-Kurzweil integral, then since the integral isn’t affected by sets of zero measure, it follows that $f$ is the integral of a positive function hence increasing. – Dap Dec 17 '17 at 17:51
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    @gismusi its the space of functions that satisfy $\int_K |f| <\infty$ for any compact set $K$. – Hyperplane Dec 17 '17 at 22:11
  • @Dap You should make this into an answer, so we can lay the thing to rest. (FTC holds for the H-K integral, for any differentiable function, whether one is "willing to accept it" or not...) – David C. Ullrich Dec 18 '17 at 15:47
  • @gimusi it means locally integrable function (https://en.wikipedia.org/wiki/Locally_integrable_function) – ℋolo Dec 19 '17 at 00:14
  • i am new in this area, can someone explain to me $\frac{1}{n} m({x;f'(x)>\frac 1 n})$? – ℋolo Dec 19 '17 at 00:27
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    @Holo: the $m$ means measure; the OP is appealing to countable additivity of measure to show that the integral of a strictly positive function is strictly positive; see for example https://math.stackexchange.com/questions/978692/if-the-lebesgue-integral-of-a-strictly-positive-function-is-zero – Dap Dec 19 '17 at 08:48

1 Answers1

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This is a special case of the Goldowsky-Tonelli theorem. See Saks, Theory of the integral, Chapter 6, page 206. (Thanks to https://mathoverflow.net/a/266399/112284 for the reference)

Your argument applies verbatim without the $L^1_{loc}$ restriction using the Henstock-Kurzweil integral. For this integral, the FTC holds for any differentiable function, it isn't affected by sets of zero measure, and the integral of an a.e. positive function is strictly increasing.

Here is another proof that avoids the Henstock-Kurzweil integral. First, note it suffices to show weak monotonicity, $f(b)\geq f(a)$ for all $b>a$ - we would then only have equality if the function was constant in $[a,b],$ which would give a derivative of zero.

Apply the Vitali-Carathéodory theorem to get an upper semicontinuous function $g:[a,b]\to\mathbb R$ satisfying $g(x)\leq \min(0,f'(x))$ for all $x\in[a,b]$ and $\int_a^b g(t)dt\geq -\epsilon.$ Define $h(x)=f(x)-\int_{a}^x g(t)dt+(x-a)\epsilon.$ Using upper semicontinuity, $\limsup_{y\to x}\frac{1}{y-x}\int_x^y g(t)dt\leq g(x).$ We get $\liminf_{\substack{y\to x\\ y\in[a,b]}}\frac{h(y)-h(x)}{y-x}\geq f'(x)-g(x)+\epsilon\geq \epsilon$ for all $x\in [a,b].$ It follows that $h(b)\geq h(a)$: consider the infimal $x$ such that $h(x)<h(a)$; it must satisfy $h(x)=h(a),$ but since the lower (lim inf) derivatives are positive, we must have $h(y)>h(a)$ for sufficiently small $y-x>0.$ This gives $f(b)\geq f(a)-\epsilon(1+b-a),$ and since $\epsilon$ was arbitrary, $f(b)\geq f(a).$

Dap
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