Some kind of a converse of Leibniz-Newton

Question

To prove: Let $f,g:[a,b] \to \mathbb{R}$ be two functions such that $f$ is Riemann integrable and $g$ has an antiderivative $G$ on $[a,b]$. If $$\int_a^b f(x)dx=G(b)-G(a) \text { and } f \le g,$$ then $g$ is also Riemann integrable.

I thought about using Darboux's criterion. I have some trouble proving that $g$ is bounded, but nevertheless, if we consider a sequence of partitions $\Delta_n=(x_0=a,x_1,...,x_n=b)$ such that $||\Delta_n ||\to 0$, we have that $$G(b)-G(a)=\sum_{k=1}^n\left[G(x_k)-G(x_{k-1})\right]\stackrel{\text{MVT}}{=}\sum_{k=1}^n (x_k-x_{k-1})g(c_k)$$ and then I tried to somehow bound the upper and lower Darboux sums, but I wasn't succesful.

Answer 1

If we were lucky enough to have $f$ continuous, we could do this: Define

$$h(x) = G(x)-G(a) -\int_a^x f(t)\,dt.$$

Now $G'=g$ is given. And by the FTC, the derivative of the integral function is $f(x).$ Thus

$$h'(x) =g(x)-f(x)\ge 0.$$

Thus $h$ is nondecreasing on $[a,b].$ But $h(a)=h(b)=0.$ A nondecreasing function that is $0$ at the endpoints must be identically $0.$ Therefore $g\equiv f$ and $g$ is Riemann integrable.

Answer 2

In general, it is false that $g$ is Riemann integrable. In fact, $g$ may fail to be Lebesgue integrable. Reason: The LHS $\int_a^b f(x) dx$ and RHS $G(b)-G(a)$ are just two real numbers and do not impose any restriction on $g$.

You may search: A differentiable function whose derivative fails to be integrable. Then, the derivative is your $g$.

Answer 3

We will prove that $F'(x)=g(x)$. Until then, we need the following:

Lemma 1: $\forall c,d\in[a,b]$ we have $\int_c^df(x)dx\leq G(d)-G(c)$.

Proof: Let $\Delta=(c=x_1<x_2<...<x_n=d)$ a division of $[c,d]$. Then as you noticed $G(d)-G(c)=\sum\limits_{k=1}^{n-1}G(x_{k+1})-G(x_k)\overset{MVT}{=}\sum_\limits{k=1}^{n-1}(x_{k+1}-x_k)g(c_k)\geq \sum\limits_{k=1}^{n-1}(x_{k+1}-x_k)\inf\limits_{x\in[x_{k},x_{k+1}]}f(x)=s_{\Delta}$

where $s_{\Delta}$ is the lower Darboux sum correponding to the division $\Delta$.

Hence $G(d)-G(c)\geq\sup\limits_{\Delta}s_{\Delta}=\int_c^df(x)dx$. $\square$

Now, let $F(x)=\int_a^xf(t)dt$.

Then $G(b)-G(a)=G(b)-G(c)+G(c)-G(a)\leq F(b)-F(c)+F(c)-F(a)=F(b)-F(a)=G(b)-G(a)$, therefore $F(c)=G(c)-G(a)\Rightarrow F'(x)=g(x),\forall x\in[a,b]$, as stated.

Let's prove one more lemma before finishing the proof

Lemma 2: $\sup\limits_{z\in[c,d]}g(z)\leq\sup\limits_{z\in[c,d]}f(z)$.

Proof: This lemma is based on the following observation:

$\frac{F(x)-F(y)}{x-y}=\frac{1}{x-y}\int_y^xf(t)dt\leq\sup\limits_{z\in[c,d]}f(z)$, for any $c\leq x<y\leq d$. Hence $g(x)=\lim\limits_{y\to x}\frac{F(x)-F(y)}{x-y}\leq\sup\limits_{z\in[c,d]}f(z)\Rightarrow\sup\limits_{z\in[c,d]}g(z)\leq\sup\limits_{z\in[c,d]}f(z)$, as claimed. $\square$

Let $\varepsilon>0$ and $\Delta=(a=x_0<x_1<...<x_n=b)$ a division for which $S_{\Delta}(f)-s_{\Delta}(f)<\varepsilon$.Then $S_{\Delta}(g)-s_{\Delta}(g)=\sum\limits_{i=0}^{n-1}(\sup\limits_{x_i\leq z\leq x_{i+1}}g(z)-\inf\limits_{x_i\leq z\leq x_{i+1}}g(z))\leq\sum\limits_{i=0}^{n-1}(\sup\limits_{x_i\leq z\leq x_{i+1}}f(z)-\inf\limits_{x_i\leq z\leq x_{i+1}}f(z))=S_{\Delta}(f)-s_{\Delta}(f)<\varepsilon$

This entails that $g$ is Riemann integrable. $\square$