If $f$ is measurable and differentiable almost everywhere and if there is some $g \in L^1(\mathbb{R})$ such that $|f(x+y)-f(x)| \leq g(x)|y|$ for almost all $x \in \mathbb{R}$ and all $y \in \mathbb{R}$, then $f' \in L^1(\mathbb{R})$ and $$\int_a^b f'(x)dx = f(b) - f(a)$$ if $f'(a)$ and $f'(b)$ exist.

$f' \in L^1$ is obvious since $|f'| \leq g \in L^1$ almost everywhere, so I need advice on the second part.

I have not found any way to make use of the hypotheses of $f'(a)$ and $f'(b)$ existing. All of the relevant theorems I've reviewed for when $\int_a^b f'(x)dx = f(b) - f(a)$ hold for all $a$ and $b$ require stronger assumptions that do not seem to hold here. So, the assumption that $f'(a)$ and $f'(b)$ exist must be vitally important, I presume.

If we define $F(x) = \int_{-\infty}^x f'(t)dt$, then $F' = f'$ almost everywhere so $\int_a^b F' = \int_a^b f'$, but this doesn't seem to get me anywhere. Any advice is appreciated.

  • $\begingroup$ Well, what "stronger assumptions" do you think would be helpful? $\endgroup$ Apr 12, 2018 at 20:38
  • $\begingroup$ If $f$ was absolutely continuous is one such assumption. $\endgroup$ Apr 12, 2018 at 20:44

2 Answers 2


I think the equation holds whenever $f$ is continuous at $a$ and $b$.[More generally if $a$ and $b$ are Lebesgue points of $f$]. Consider $\int_a^{b} \frac {f(x+1/n)-f(x)} {1/n} dx$. We can rewrite this as $n [\int_{a+1/n}^{b+1/n} f(y)dy -\int_a^{b} f(y)dy]$ which simplifies to $n [\int_{b}^{b+1/n} f(y)dy -\int_a^{a+1/n} f(y)dy]$. Clearly this last quantity approaches $f(b)-f(a)$ as $n \to \infty$ if $f$ is continuous at $a$ and $b$. Now apply Dominated Convergence Theorem to $\int_a^{b} \frac {f(x+1/n)-f(x)} {1/n} dx$ to see that the limit is $\int_a^{b} f'(x)dx$. Hence $\int_a^{b} f'(x)dx=f(b)-f(a)$.


Here's a partial result to "tame" $f$. We'd like to have absolute continuity, but at least we have (wlog.) continuity:

Let $E\subseteq [a,b]$ be the set of points $x$ where

  • $f'(x)$ exists and
  • $\forall y\colon |f(x+y)-f(x)|\le g(x)|y|$

By assumption, $E$ is almost all of $[a,b]$.

Pick $z\in [a,b]$ and assume there are sequences $x_n\in E\cap [a,z]$, $y_n\in E\cap [z,b]$ such that $x_n,y_n\to z$ and $\lim f(x_n),\lim f(y_n)$ exist in $\Bbb R\cup\{\pm\infty\}$. Then for every $u\in E\setminus\{z\}$, we have $$|f(y_n)-f(u)|\le g(u)|y_n-u| $$ for almost all $n$ and $$|f(x_n)-f(u)|\le g(u)|x_n-u| $$ for almost all $n$. We conclude $$ g(u)\ge \max\left\{\frac{|\lim f(y_n)-f(u)|}{|z-u|},\frac{|\lim f(x_n)-f(u)|}{|z-u|}\right\}\ge \frac{|\lim f(y_n)-\lim f(x_n)|}{2|z-u|}$$ Thus $g\in L^1$ we need $\lim f(y_n)=\lim f(x_n)$ in order to avoid the pole at $z$. Also, these limits must be $\in\Bbb R$ to make $\lim f(y_n)-f(u)$ finite. As we are allowed to take constant sequences, it follows that these limits equal $f(z)$ whenever $z\in E$. We can redefine $f(z)$ as the above limits whenever $z$ is in the nullset $[a,b]\setminus E$. In other words, we may assume wlog. that $f$ is continuous to begin with. (The existence of $f'(a)$, $f'(b)$ gives us $a,b\in E$ and so prevents us from redefining the values $f(a)$, $f(b)$).

In order to make more progress, one would also like to tame $g$. E.g., if $g$ were bounded, we'd readily get absolute continuity.


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