(From Analysis 1 by Terence Tao ,Exercise 11.9.3)

Let $a<b$ be real numbers, and let $f : [a, b] → R$ be a monotone increasing function. Let $F : [a, b] → R$ be the function $F(x) :=\int_{[a,x]} f$. Let $x_0$ be an element of $[a, b]$. Show that $F$ is differentiable at $x_0$ if and only if $f$ is continuous at $x_0$. (Hint: one direction is taken care of by one of the fundamental theorems of calculus. For the other, consider left and right limits of $f$ and argue by contradiction.)

Now ,as mentioned in the hint, one way is already known from the First fundamental theorem of calculus. To prove the second one - (going by the hint ) $\lim_{x\to x_0^+}\frac{F(x)-F(x_0)}{x-x_0}=\lim_{x\to x_0^-}\frac{F(x)-F(x_0)}{x-x_0}$


$\lim_{\delta\to0}\frac{\int_{x_0}^{x_0+\delta} f(x)}{\delta}=\lim_{\delta\to0}\frac{\int_{x_0-\delta}^{x_0} f(x)}{\delta}$

I am kind of stuck at this point. Any suggestions on how to move forward ?

  • 1
    $\begingroup$ At a discontinuity, $f$ has different limits from the left, right. $\endgroup$
    – zhw.
    Jul 1, 2021 at 16:00
  • 1
    $\begingroup$ Since $f$ is monotonically increasing, we have $$\lim_{x \to x_0^-} f(x) \le \lim_{x \to x_0^+} f(x)$$ for every $x_0 \in (a,b)$. As @zhw. says, these limits must be different. $\endgroup$ Jul 1, 2021 at 16:14

1 Answer 1


I will expand on my comment slightly. The key to this is monotonicity. To avoid boundary cases, I'll consider $x_0 \in (a,b)$. Suppose $f$ is discontinuous at $x_0$. Let $L^- = \lim_{x \to x_0^-} f(x)$ and $L^+ = \lim_{x \to x_0^+} f(x)$. Since $f$ is monotonically increasing, we have that $$f(y) \le L^- \le L^+ \le f(x)$$ for all $y < x_0 < x$. Moreover, discontinuity ensures that $L^- < L^+$.

I would suggest that you take $\alpha,\beta \in (L^-,L^+)$, $\alpha < \beta$, and estimate the values $$\int_{x_0}^{x_0 + \epsilon} f \qquad \text{and} \qquad \int_{x_0-\epsilon}^{x_0} f $$ for $\epsilon > 0$ sufficiently small. Can you show that the one-sided limits $$\lim_{\epsilon \to 0^+} \frac{F(x_0-\epsilon) - F(x_0)}{-\epsilon} \qquad \text{and} \qquad \lim_{\epsilon \to 0^+} \frac{F(x_0+\epsilon) - F(x_0)}{\epsilon}$$ are different?

  • $\begingroup$ Isn't it possible that the left or right limits do not exist? (Nothing given in the question seems to prevent that possibility .) In that case the proof here wouldn't hold (According to me.) $\endgroup$
    – aryaryary
    Jul 2, 2021 at 5:37
  • 1
    $\begingroup$ @aryaryary Montonicity guarantees the existence of these limits. For the left limit, consider a sequence $(x_n)_{n \in \mathbb{N}}$ such that $x_n < x_0\ \forall n \in \mathbb{N}$ and which converges to $x_0$. Without loss of generality, we may assume that $x_n < x_{n+1}$. Then $(f(x_n))_{n \in \mathbb{N}}$ is an increasing sequence of real numbers bounded above by $f(x_0)$. Such a sequence must converge. Can you show that the limit of $(f(x_n))$ is independent of the choice of sequence $(x_n)$? From this it follows that the left limit $L^-$ exists. The right limit is handled analogously. $\endgroup$ Jul 2, 2021 at 12:13
  • $\begingroup$ @MatthewBuck Not seeing this question initially, I asked a question about the same problem here math.stackexchange.com/questions/4346761/…, but specifically about showing the two sides of the derivative are different. If you have a moment, could you offer a little insight, either here or there? $\endgroup$ Jan 2, 2022 at 4:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .