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If $f$ is integrable on $[0, 1]$, does that imply that an antiderivative of $f$ is differentiable on $[0, 1]$?

I'm sorry if this comes across as too basic of a question, but I'm struggling to accept this as obvious. The Fundamental Theorem of Calculus seems to support this statement, but perhaps I have yet to fully grasped the theorem and its nuances to easily take this statement as true/false. I hope to receive some clarification on this. Thank you!

ten_to_tenth
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    Differentiable on $(0, 1)$. Also, no, the Fundamental Theorem of Calculus is not obvious at all, hence the word "Fundamental"! – Chee Han Mar 14 '24 at 15:11
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    This is not too basic. On the contrary, this is surprisingly deep question! To truly answer it, you need some early graduate level analysis. At the level of calculus, it may be easier to just strengthen the hypothesis: if $f$ is continuous, then the antiderivative $F(x) = \int_0^x f(t)dt$ is differentiable. If $f$ is merely integrable, you need to first specify what you mean by integrable (you need integrability in the Lebesgue sense), and then you need to prove a much deeper version of the theorem which tells you that $F(x)$ is differential almost everywhere (in a technical sense). – User8128 Mar 14 '24 at 15:15
  • @CheeHan Thank you! Is it because without knowing the left side limit at $0$ and the right side limit at $1$, we could never know if the function is continuous outside of $[0, 1]$? – ten_to_tenth Mar 14 '24 at 16:48
  • @User8128 Thank you very much! Julio's comment below seems to reflect what you suggested. I just provided a list of points representing my understanding. If you don't mind, I'd greatly appreciate it if you could review them and give me some feedback! Thank you! – ten_to_tenth Mar 14 '24 at 17:05

1 Answers1

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Firstly, not all Riemann Integrable functions have primitives. Consider $f: [0,1]\to\mathbb{R}$ given by $f(x)=0, x\in[0,1/2)$ and $f(x)=1, x\in[1/2,1]$.

$f$ cannot be the derivative of a function because derivatives satisfy the Darboux condition and $f$ does not.

Now, a function $g$ is Riemann integrable on $[0,1]$ iff it is continuous except for a set of null Lebesgue Measure. Moreover, it is well known that if $g$ is integrable and continuous at some point $c$, then $F(x)=\displaystyle\int_{0}^{x} g(t)dt$ verifies that $F'(c)=g(c)$.

Thus, we will have a function whose derivative exists and equals $g$ except for a set of null Lebesgue measure.

Regarding the points of discontinuity, we cannot assure anything about $F$: take $f$ as defined above, then $F(x)=0$ for $x\in[0,1/2]$ and $F(x)=x-1/2$ for $x\in (1/2,1]$, so $F$ is not differentiable at $1/2$.

Julio Puerta
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  • Thank you very much! I'm just starting my calculus so this feels a little overwhelming to take in at once, but I'm eager to have as nuanced an understanding as possible. Here's my summary of your answer. Please let me know if it's correct (I may present them in a series of replies for better organization): – ten_to_tenth Mar 14 '24 at 16:53
  • If $f(x)$ is differentiable, $f(x)$ is continuous.
    • – ten_to_tenth Mar 14 '24 at 17:01
  • If $f(x)$ is continuous on $[a, b]$, it is Riemann integrable with antiderivative $F(x) = \int_0^x f(x)dx$ that is differentiable on $(a, b).$ The derivative of $F(x)$ is $f(x).$
    • – ten_to_tenth Mar 14 '24 at 17:02
  • If $f(x)$ is Riemann integrable on $[a, b]$, it may be discontinuous at some points in $[a, b].$ Then, if $f(x)$ does not satisfy the Darboux condition either, it does not have an antiderivative. On the other hand, if it satisfies the Darboux condition, it has antiderivative $F$ which may not be differentiable at all points in $(a, b)$.
    • – ten_to_tenth Mar 14 '24 at 17:02