past Thursday I had my Calculus exam on integrals. I found a question of the test really challenging and also very interesting. The problem asked the following:
Let f be an integrable (Riemann) function in the interval [0,2] whose minimum is f(1)=0. We define $F:[0,2] \leftarrow \mathbb R$ by $F(x) = \int_1^x f(t)dt$. Do the following statements always hold? (i) $F'(1)=0$, (ii) $F$ is a increasing function, (iii)$F'(x)=f(x)$ except for a finite number of points.
For (i) I proposed $ |\sin(\frac{1}{x-1})|$ if $x \not = 1$ and $0$ if $x=1$ as a counterexample. However, I'm not sure this function would work, I just tried to find a function continuous in every point of $[0,2]$ except for $x=1$ and that $lim_{x \to 1} f(x)$ didn't exist.
For (ii) I argued it was true as, given $x,y \in [0,2]: x < y$, then note that $f(x)\geq0, \forall x \in [0,2]$, thus $F(y)-F(x) = \int_1^y f- \int_1^x f= \int_x^y f \geq (y-x) \cdot 0 = 0$.
For (iii) I tried to look for a function discontinuous in an infinite number of points but integrable as a counterexample. I inspired in Spivak's problem which proved that $g(x)=0$ if x is irrational, $g(\frac{p}{q}) = \frac{1}{q}$, where $\frac{p}{q}$ is a irreducible fraction, is integrable in $[0,1]$ and its integral is exactly 0. Thus, I thought of defining f as g in $[0,1)$, and $f(x)=0$ for $x\in [1,2]$. I think that F(x) would then be a constant 0 function hence $F'(x) \not = f(x)$ for every rational point in $[0,1)$, which is an infinite number of points.
I did this question in less than 10 minutes as I left it as the last one, so I'm quite sure I'm wrong in some of the answers (specially (i) and (iii)). Thus, I would like some feedback and specially other approaches to the activity. I find quite difficult to find good counterexamples in analysis, so I would also appreciate advice/good references to practice this sort of questions.
Your reasoning for ii) is correct.
For a counterexample to both i) and iii), define $f=0$ at the points $1,1/2,1/3,\dots$ Define $f=1$ elsewhere else. Then $f$ is Riemann integrable on $[0,2]^*,$ $f\ge 0$ everywhere, and $f(1)=0$ is the minimum value of $f.$
Since the values of $f$ at $1,1/2,\dots$ do not affect the integral of $f,$ we have
$$F(x)=\int_1^x f(t)\,dt = \int_1^x 1\,dt =(x-1),\,x\in [0,2].$$
It follows that $F'(x)=1$ everywhere in $[0,2].$ But $f=0$ at the points $1,1/2,1/3, \dots$ Thus $F'(x)\ne f(x)$ for $x=1,1/2,1/3, \dots$ It follows that i) and iii) both fail for this $f.$
Proof of *: Let $\epsilon>0.$ We know $f$ is RI on $[\frac{\epsilon}{2},1].$ Thus there is a partition $P$ of $[\epsilon/2,1]$ such that $U(P,f)-L(P,f)<\frac{\epsilon}{2}.$ Let $Q=\{0\}\cup P.$ Then $Q$ is a partition of $[0,1]$ and
$$U(Q,f)-L(Q,f)= (1-0)\cdot \frac{\epsilon}{2}+ U(P,f)-L(P,f)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$
It follows that $f$ is RI on $[0,1].$