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In a recent answer I needed an argument of the following kind:

Let $f, g: [a, b] \to \Bbb R$ be functions with the following properties:

  • $f$ is differentiable,
  • $g$ is continuous,
  • $|f'(x)| \le g(x)$ for all $x \in [a, b]$.

Then $|f(b) - f(a)| \le \int_a^b g(x) \, dx$.

Seems pretty easy: We have $$ \tag{*} |f(b) - f(a)| = \left| \int_a^b f'(t) \, dt \right| \le \int_a^b |f'(x)| \, dx \le \int_a^b g(x) \, dx \, . $$ There is just one problem: We have used the fundamental theorem of calculus, and that requires $f'$ to be Riemann integrable (or Lebesgue integrable and absolutely continuous), compare Necessity of a hypothesis in the fundamental theorem of calculus. This is for example satisfied if $f'$ is continuous.

But the above statement holds without additional assumptions on $f'$: Let $n$ be a positive integer and $x_k = a + \frac kn (b-a)$, $0 \le k \le n$, be a partition of the interval $[a, b]$. We apply the mean-value theorem to each subinterval $[x_k, x_{k+1}]$: $$ f(x_{k+1}) - f(x_k) = (x_{k+1} - x_k) f'(c_{n, k}) $$ for some $c_{n, k} \in [x_k, x_{k+1}]$. It follows that $$ |f(b) - f(a)| \le \sum_{k=0}^{n-1}| f(x_{k+1}) - f(x_k)| \le \sum_{k=0}^{n-1} (x_{k+1} - x_k) g(c_{n, k}) \, . $$ $g$ is continuous and therefore Riemann integrable. The expression on the right is a Riemann sum for $\int_a^b g(x) \, dx$ with the partition size $(b-a)/n$. It follows that $$ \lim_{n \to \infty } \sum_{k=0}^{n-1} (x_{k+1} - x_k) g(c_{n, k}) = \int_a^b g(x) \, dx $$ and therefore $|f(b) - f(a)| \le \int_a^b g(x) \, dx$.

My question: Is there a simpler proof of the statement?

It seems pretty basic, therefore I wonder if there is a simpler proof, without going into technical details such as partitions and Riemann sums. Perhaps it is a consequence of some other theorem about integration which I failed to find?

Martin R
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  • Is $g$ assumed to be integrable in $[a,b]$? – Logos Aug 01 '20 at 16:47
  • @Logos: For simplicity I assumed that $g$ is continuous, which implies that it is (Riemann + Lebesgue) integrable. – One could also ask what happens if $g$ is only Lebesgue integrable. – Martin R Aug 01 '20 at 16:49
  • Sorry, but I don’t understand where the problem is: if $g$ is integrable and dominates $f’$ such is $f’$ and you only need to integrate. What am I missing? – Logos Aug 01 '20 at 16:51
  • @Logos: The problem is that $f'$ may not be integrable and you cannot use $f(b)-f(a) = \int_a^b f'(x)dx$. Compare https://math.stackexchange.com/a/593053/42969. – Martin R Aug 01 '20 at 16:53
  • @Logos, Although not a counter-example, consider the Cantor-Lebesgue function. Then $f$ is differentiable almost everywhere with $f'=0$, but FToC does not hold. – Sangchul Lee Aug 01 '20 at 16:53
  • Okay, I get the problem now. Sorry I didn’t get it earlier! – Logos Aug 01 '20 at 16:54
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    When it comes to justifying $\text{()}$, you may use either of the following theorems: (1) If $f$ is everywhere* differentiable on $[a,b]$ and $f'$ is integrable, then the Lebesgue integral of $f'$ satisfies $$\int_{a}^{b}f'(x),\mathrm{d}x=f(b)-f(a).$$ (2) If $f$ is everywhere differentiable on $[a,b]$, then $f'$ is Henstock-Kurzweil integrable and its Henstock-Kurzweil integral satisfies $$\int_{a}^{b}f'(x),\mathrm{d}x=f(b)-f(a).$$ – Sangchul Lee Aug 01 '20 at 16:56
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    I'd try splitting into 2 cases, and look at primitives... $h(t) = G(t) - f(t)$ as one of the cases... $h'(t) \geq 0$ so it is a non-decreasing function... – user8675309 Aug 01 '20 at 17:00

1 Answers1

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For $\epsilon \in \{1, -1\}$, define

$$ k_{\epsilon}(x) = \int_{a}^{x} g(t) \, \mathrm{d}t - \epsilon(f(x) - f(a)). $$

Then

$$ k_{\epsilon}(a) = 0 \qquad \text{and} \qquad k_{\epsilon}'(x) = g(x) - \epsilon f'(x) \geq 0 , $$

and so, we have $k_{\epsilon}(x) \geq 0$ for all $x \in [a, b]$ and $\epsilon \in \{1, -1\}$. This proves the desired claim.

Sangchul Lee
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