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Let be $F:[a,b]\to\mathbb{R}$, where $F(x):=\int\limits_a^xf(t)dt$. We assume $f$ to be continuous at point $x_0$ and Riemann-integrable over $[a,b]$. Then we know that $F'(x_0)=f(x_0)$.


My (edited) approach (we assume $x<x_0$):

Let's define the function $M:[a,x_0]\to\mathbb{R}$, where $M(x):=\sup\{\left|f(t)-f(x_0)\right|\mid t\in[x,x_0]\}$.

If we choose an arbitrary $\frac{\epsilon}{2}>0$ then by continuity of $f$ at $x_0$ there exists a $\delta>0$ such that for all $t\in [x,x_0]$ with $|t-x_0|<\delta$ we have $|f(t)-f(x_0)|<\frac{\epsilon}{2}$. By definition of the supremum there exists a $t'\in[x,x_0]$ such that $M(x)-\frac{\epsilon}{2}<|f(t')-f(x_0)|$ and therefore $M(x)<\frac{\epsilon}{2}+|f(t')-f(x_0)|<\epsilon$. So if we only admit $x\in [a,x_0]$ such that $|x-x_0|<\delta$ we see that $M(x)$ is continuous at $x_0$.

Keeping $|x-x_0|<\delta$ in mind, allows us to conclude:

$$ \left|\frac{\int\limits_x^{x_0}f(t)dt}{x-x_0}-f(x_0)\right|=\left|\frac{\int\limits_x^{x_0}f(t)-f(x_0)dt}{x-x_0}\right|\leq \left|\frac{M(x)(x_0-x)}{x-x_0}\right| = M(x) <\epsilon. $$ The case "$x>x_0$" is basically is the same. Hence, $F'(x_0)=f(x_0)$.


Is this proof o.k.?

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  • $\begingroup$ I don't think you can assert the existence of $t'$. For instance, say $f$ is constant on $[x,x_0]$. Then $M(x)=0$, but for all $t'\in [x,x_0]$, you have $|f(t')-f(x_0)|=0=M(x)$, you never get the strict inequality you claim. The accurate claim would be that there exists $t'\in [x,x_0]$ such that $M(x)-\frac{\epsilon}{2}\lt |f(t')-f(x_0)| \leq M(x)$. $\endgroup$ – Arturo Magidin Jan 27 at 19:14
  • $\begingroup$ Also, you say "$M(x)$ is continuous"; continuous where? It is not necessarily continuous on all of $[a,x_0]$, as $f$ is not assumed continuous on the whole interval. For example, say $f(x) = 0$ for all $x$ in $[a,x_0]$, except at a point $t_0$ where $f(x)=1$. Then $M(x)=1$ for $a\leq x\leq t_0$, and $M(x)=0$ for $t_0\lt x\leq x_0$, so $M$ is not continuous on the entire interval. I also wonder if you have given all of the hypothesis: surely we must put some integrability conditions on $f(x)$, for starters! $\endgroup$ – Arturo Magidin Jan 27 at 19:21
  • $\begingroup$ @ArturoMagidin, I have put an edit because when I presented my solution to the professor I mentioned that $M(x)$ is only continuous at $x_0$ and that we must assume $f$ to be integrable. $\endgroup$ – Philipp Jan 27 at 19:28
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    $\begingroup$ I've written up how to fix it up; there is also an error/unjustified step in the final chain of inequalities, since you cannot go from $f(t)\leq g(t)$ to $|\int f(t)\,dt|\leq|\int g(t)\,dt|$ in general. So rather than try to justify that this can be done in this specific instance, I do a slight change. $\endgroup$ – Arturo Magidin Jan 27 at 20:39
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    $\begingroup$ Careful: That assume $a\leq b$. Say $f(t)=t$ on $[0,1]$, $a=1$, $b=0$. The right hand side is $-1$, the left hand side is $-\frac{1}{2}$. $\endgroup$ – Arturo Magidin Jan 29 at 20:23
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You are fairly close, but as I note in the comments, there are a few problems with what you write. There is an additional issue at the end. I had already written up a very similar approach, which is how I usually do it, but let me address how to fix your argument:

Let us assume $f(x)$ is integrable on $[a,b]$ (with $x_0\in (a,b)$) so that $F(x)$ makes sense.

To show continuity of $M(x)$ from the left at $x_0$ (equivalently, show its limit as $x\to x_0^-$ is $0$), proceeding as you do until you pick $t'$ (which was a problem prior to the edits as I noted in comments), let us do this instead: we know there exists $t'\in [x,x_0]$ such that $$M(x) - \frac{\epsilon}{2} \lt |f(t')-f(x_0)| \leq M(x).$$ Hence $$0\leq M(x) \lt |f(t')-f(x_0)|+\frac{\epsilon}{2} \lt \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon.$$ This holds for all $\epsilon\gt 0$, so we conclude the limit is indeed $0$.

Finally, there is a small correction I think you need to make to the final chain of inequalities, namely $r\leq s$ does not imply that $|\int_x^{x_0} r\,dt| \leq |\int_x^{x_0}s\,dt|$. (For example, integrade $-3$ on $[0,1]$, compraed to the integral of $-1$ on $[0,1]$).

Instead, you should remember that in general $$\left|\int_a^b f(t)\,dt\right| \leq \int_a^b |f(t)|\,dt,$$ assuming both integrals exist, so that you can say $$\begin{align*} \left|\frac{\int_x^{x_0}f(t)\,dt}{x-x_0} - f(x_0)\right| &= \left|\frac{\int_x^{x_0}(f(t)-f(x_0))\,dt}{x-x_0}\right|\\ &\leq \frac{\int_x^{x_0}|f(t)-f(x_0)|\,dt}{|x-x_0|}\\ &\leq \frac{\int_x^{x_0}M(x)\,dt}{|x-x_0|}\\ &= \frac{M(x)(x_0-x)}{x_0-x} = M(x), \end{align*}$$ and then taking the limit as $x\to x_0^-$ you get the desired $$\lim_{x\to x_0^-}\frac{\int_x^{x_0} f(t)\,dt}{x-x_0} = f(x_0).$$ And then a similar correction and fix is needed for the continuity from the right and the derivative from the right.


As I had already written up the stuff below, I keep it here. This is how I usually do this.

Rather than work with the supremum of the absolute value of the difference on the interval, I'm just going to work with the supremum and infimum of the values on that interval. It should be straightforward (though wordy) to convert this to your function.

I will restrict to $x\leq x_0$, as you do. Let me define functions that are close to your $M(x)$: for $x\lt x_0$, $$\begin{align*} m(x) &= \inf\{ f(t)\mid x\leq t\leq x_0\}\\ M(x) &= \sup\{ f(t)\mid x\leq t\leq x_0\}. \end{align*}$$ Because $f(x)$ is integrable on $[x,x_0]$, we know it is bounded, so $m(x)$ and $M(x)$ both make sense.

I claim that $$\lim_{x\to x_0^-}(M(x)-m(x)) = 0.$$ Indeed: let $\epsilon\gt 0$. Since $f$ is continuous at $x_0$, we know that there exists $\delta\gt 0$ such that if $t\in [a,x_0]$ and $|t-x_0|\lt\delta$, then $|f(t)-f(x_0)|\lt \frac{\epsilon}{4}$. Shrinking $\delta$ if necessary, we may assume that $a\leq x_0-\delta$, so that if $0\leq x_0-t\lt\delta$, then $t$ is in $[a,x_0]$.

Let $x$ be such that $0\leq x_0-x\lt\delta$. Since $M(x)$ is a supremum, there exists $t_M\in [x,x_0]$ such that $M(x)-\frac{\epsilon}{4}\lt f(t)\leq M(x)$; and there exists $t_m\in [x,x_0]$ such that $m(x)\leq f(t_m)\lt m(x)+\frac{\epsilon}{4}$. Then $$\begin{align*} |M(x)-m(x)| &= |M(x) - f(t_M) + f(t_M) -f(x_0)+f(x_0)- f(t_m) + f(t_m)-m(x)|\\ &\leq |M(x)-f(t_M)| + |f(t_M)-f(x_0)| \\ &\qquad\mathop{+} |f(x_0)-f(t_m)| + |f(t_m)-m(x)|\\ &\lt \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon}{4}\\ &= \epsilon, \end{align*}$$ where the first $\frac{\epsilon}{4}$ follows from $M(x)-\frac{\epsilon}{4}\lt f(t_M)$; the second from the continuity of $f$ at $x_0$ and the fact that $|t_M-x_0|\lt\delta$; the third from continuity of $f$ at $x_0$ and the fact that $|t_m-x_0|\lt\delta$; and the fourth from the fact that $f(t_m)\lt m(x)+\frac{\epsilon}{4}$.

Thus, the limit of $M(x)-m(x)$ is $0$ as $x\to x_0^-$.

(This argument can be simplified if you assume $f$ is continuous on the entire interval and you take the Extreme Value Theorem for granted, since then we can actually achieve $M(x)$ and $m(x)$, so we don't have to approximate them.)

Likewise, since $m(x)\leq f(x_0)\leq M(x)$, we have $$\begin{align*} 0&\leq \lim_{x\to x_0^-}(f(x_0)-m(x))\\ &\leq \lim_{x\to x_0^-}(M(x)-m(x))\\ &= 0, \end{align*}$$ so $\lim_{x\to x_0^-}m(x) = f(x_0)$; similarly, $\lim_{x\to x_0^-}M(x)=f(x_0)$.

So now, $$\lim_{x\to x_0^-}\frac{F(x)-F(x_0)}{x-x_0} = \lim_{x\to x_0^-}\frac{1}{x-x_0}\int_{x_0}^x f(t)\,dt.$$ Since $m(x)\leq f(t)\leq M(x)$ for all $t\in [x,x_0]$, $$m(x) \leq \frac{1}{x-x_0} \int_{x_0}^xf(t)\,dt \leq M(x).$$ Therefore, $$0 \leq \left(\frac{1}{x-x_0}\int_{x_0}^x f(t)\,dt\right) - m(x) \leq M(x)-m(x).$$ Letting $x\to x_0^{-}$, the Squeeze Theorem tells us that $$\lim_{x\to x_0^-}\left(\frac{1}{x-x_0}\int_{x_0}^x f(t)\,dt\right) - m(x) = 0.$$ Since $\lim_{x\to x_0^-}m(x) = f(x_0)$, then $$\lim_{x\to x_0^-}\frac{1}{x-x_0}\int_{x_0}^x f(t)\,dt$$ exists, and equals $\lim_{x\to x_0^-}m(x) = f(x_0)$, as required.

A similar argument works if $x_0\leq x\leq b$.


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  • $\begingroup$ Is the last part of your proof necessary? I am referring to everything that comes after "Therefore, [...] and equals$\lim\limits_{x\to x_0^-}m(x)=f(x_0)$, as required". I thought that $$m(x) \leq \frac{1}{x-x_0} \int_{x_0}^xf(t)\,dt \leq M(x)$$ and then letting $\lim\limits_{x\to x_0^-}$ will already deliver the desired result by using the squeeze theorem? Or was it just an explanatory issue to add the last part? $\endgroup$ – Philipp Jan 30 at 20:35
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    $\begingroup$ @Philipp: I was trying to bring it closer to your development where you ahd the difference between the integral and $f(x_0)$, instead of just the integral; as I mentioned at the top, I had written it before giving the direct modifications of your argument to fix its small issues. $\endgroup$ – Arturo Magidin Jan 30 at 20:41
  • $\begingroup$ really great answer, learnt a lot from it! $\endgroup$ – Philipp Jan 31 at 0:03

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