Proof showing that if F is an antiderivative of f, then f must be continuous.

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1. Oct 26, 2014

werty32

1. The problem statement, all variables and given/known data
Show that if F is an antiderivative of f on [a,b] and c is in (a,b), then f cannot have a jump or removable discontinuity at c. Hint: assume that it does and show that either F'(c) does not exist or F'(c) does not equal f(c).

2. The attempt at a solution
I attempted a proof by contradiction where I said that for the sake of contradiction we should let f have a discontinuity at point c, and I would like to use this to prove that F'(c) doesn't exist, but I'm not quite sure how a discontinuity on f affects F given that f is the derivative of F.

2. Oct 26, 2014

LCKurtz

I haven't worked your problem out, but since you don't see how a discontinuity in $f$ affects $F$, I would suggest you try some examples. For example look at what happens if $f(x)=0,~0\le x\le \frac 1 2$ and $f(x) = 1,~\frac 1 2 < x \le 1$. What does $F$ look like in that case? That might give you some ideas.

3. Oct 27, 2014

gopher_p

If I'm interpreting the problem correctly, it is basically saying the following:

Assume $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$ with $f=F'$. Show that $f$ has no jump or removable discontinuities on $(a,b)$.

The "jump or removable" part of the problem is essential. Derivatives can be discontinuous; look at the function $$f(x)= \begin{cases} x^2\sin\frac{1}{x} & x\neq 0\\ 0 & x=0\end{cases}$$
The function is differentiable everywhere, but the derivative is not continuous at $0$.

I would focus on one-sided limits of the derivative if I were you; i.e. show that if $c\in (a,b)$ and $\lim\limits_{x\rightarrow c^\pm}f'(x)$ exists, then $\lim\limits_{x\rightarrow c^\pm}f'(x)=f'(c)$. That's the essence of what distinguishes the jump/removable discontinuities from the less "tame" kinds; the (one-sided) limit exists, but isn't equal to the value of the function.