Some kind of a converse of Leibniz-Newton

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.

• I guess there can be a counterexample, for instance, when $a=0$, $b=1$, $f\equiv -1$, and $g(x)$ in unbounded in each neighborhood of the point $a$. Feb 28, 2020 at 17:21
• @Alex Ravsky You are most likely right. If you can exhibit a counterexample, I would be glad to award you the bounty Mar 3, 2020 at 8:25
• It holds if $f$ is continuous.
– zhw.
Mar 3, 2020 at 16:15
• @zhw and how do you prove it in this case? Mar 5, 2020 at 13:38
• I posted an answer on this below.
– zhw.
Mar 5, 2020 at 17:13

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.

• thanks, I awarded you the bounty because you managed to fix this statement, which is, by the way, taken from an old olympiad. Mar 5, 2020 at 17:44
• Well thank you. I didn't even know a bounty was possible at this late hour.
– zhw.
Mar 5, 2020 at 20:20
• +1 Can't we make this argument work when $h'$ is non-negative almost everywhere? Mar 6, 2020 at 1:49
• It appears from this answer that non-negative derivative almost everywhere does not imply non-decreasing nature of the function. Mar 6, 2020 at 2:45

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 $$g$$.

• Ah, thank you, I forgot that $f \le g$. Now the statement should be ok. I am sorry for my forgetfulness. Feb 25, 2020 at 6:17

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 $$\Delta=(c=x_1 a division of $$[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 $$c\leq x. Hence $$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 $$\Delta=(a=x_0 a division for which $$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. $$\square$$