# Proving the Second Mean Value Theorem for Integrals using Abel's Lemma.

This is a problem and solution from Spivak's Calculus. I'm trying to fill in the details of the final paragraph in the solution, to prove the conclusion in problem (c).

Problem:

$\rm(b)$ Suppose that $\{b_n\}$ is nonincreasing, with $b_n\geq0$ for each $n$, and that $$m\leq a_1+\cdots+a_n\leq M$$ for all $n$. Prove Abel's Lemma: $$b_1m\leq a_1b_1+\cdots+a_nb_n\leq b_1M.$$ (And, moreover, $$b_km\leq a_kb_k+\cdots+a_nb_n\leq b_kM,$$ a formula which only looks more general, but really isn't.)
$\rm(c)$ Let $f$ be integrable on $[a,b]$ and let $\phi$ be nonincreasing on $[a,b]$ with $\phi(b)=0$. Let $P=\{t_0,\ldots,t_n\}$ be a partition of $[a,b]$. Show that the sum $$\sum_{i=1}^n f(t_{i-1})\phi(t_{i-1})(t_i-t_{i-1})$$ lies between the smallest and the largest of the sums $$\phi(a)\sum_{i=1}^k f(t_{i-1})(t_i-t_{i-1}).$$ Conclude that $$\int_a^b f(x)\phi(x)\,dx$$ lies between the minimum and the maximum of $$\phi(a)\int_a ^xf(t)\,dt,$$ and that it therefore equals $\phi(a)\displaystyle\int_a^\xi f(t)\,dt$ for some $\xi$ in $[a,b]$.

Solution:

$(\rm c)$ If we set $$a_i=f(x_i)(t_i-t_{i-1})$$ and let \begin{align}m&=\text{smallest of the }\sum_{i=1}^k f(x_i)(t_i-t_{i-1})\\M&=\text{largest of the }\sum_{i=1}^k f(x_i)(t_i-t_{i-1})\end{align} then $m\leq a_1+\cdots+a_k\leq M$ for all $k$. Letting $b_k=\phi(x_k)$ in part $\rm(b)$, we find that $$\sum_{i=1} ^n f(x_i)\phi(x_i)(t_i-t_{i-1})$$ lies between the smallest and the largest of the sums $$\phi(x_1)\sum_{i=1}^k f(x_i)(t_i-t_{i-1}).$$ $\,\,\,$ Since we can approximate $\int_a^b f(x_i)\phi(x)\,dx$ by sums $\sum\limits_{i=1}^n f(x_i)\phi(x_i)(t_i-t_{i-1}),$ and $\int_a^x f(t)\,dt$ by sums like $\sum\limits_{i=1}^k f(x_i)(t_i-t_{i-1})$, the final result should follow from the above. However, some care is required for the argument: $\,\,\,$ Given $\varepsilon>0$ we can choose $\delta>0$ so that whenever all $t_i-t_{i-1}<\delta$ we have $$\tag{1}\left|\int_a^b f(x)\,dx-\sum_{i=1}^n f(t_{i-1})(t_i-t_{i-1})\right|<\varepsilon$$ We claim that for any $\varepsilon'>\epsilon$ it also follow that for each $k$ $$\left|\int_a^{t_k}f(x)\,dx-\sum_{i=1}^k f(t_{i-1})(t_i-t_{i-1})\right|<\varepsilon'.$$ The idea is that if we had $$\tag{2}\left|\int_a^{t_k}f(x)\,dx-\sum_{i=1}^k f(t_{i-1})(t_i-t_{i-1})\right|\geq\varepsilon',$$ then by choosing some $p>n$ and new $t_{k+1}<t_{k+2}<\cdots<t_p=b$ we could make the sums on $[t_k,b]$ so close to $\int_{t_k}^b f(x)\, dx$ that inequality $(2)$ would contradict $(1)$. More precisely, choose $t_{k+1},\ldots, t_p$, still with $t_i-t_{i-1}<\delta$, so that $$\tag{3}\left|\int_{t_k}^bf(x)\,dx-\sum_{i=k+1}^p f(t_{i-1})(t_i-t_{i-1})\right|<\varepsilon'-\varepsilon>0.$$

My work:

Let $F(x)=\int_a^x f(t)dt$ on $[a,b]$, then since $F$ is continuous on $[a,b]$, it attains its minimum and maximum on some $t_k$ and $t_l$.

Now given any $\epsilon \gt 0$, there is some $\delta \gt 0$ such that for any partition with mesh less than $\delta$, we have

$|\int_a^b f(x)dx-\sum_{i=1}^n f(t_{i-1})(t_i-t_{i-1})|\lt \epsilon.$

Then by the claim after (1) in the solution, we have

$|\int_a^{t_k} f(x)dx- \sum_{i=1}^k f(t_{i-1})(t_i-t_{i-1})|\le \epsilon.$ for any $k\in \{1,\cdots, n\}$.

Hence for a given partition with mesh less than $\delta$ having the specified $t_k$ and $t_l$ as endpoints, we get

$|\phi(a)\int_a^{t_k} f(x)dx- \phi(a)\sum_{i=1}^k f(t_{i-1})(t_i-t_{i-1})|\le \phi(a)\epsilon.$

and

$|\phi(a)\int_a^{t_l} f(x)dx- \phi(a)\sum_{i=1}^l f(t_{i-1})(t_i-t_{i-1})|\le \phi(a)\epsilon.$

Clearly, $\phi(a)\int_a^{t_k} f(x)dx$ and $\phi(a)\int_a^{t_l} f(x)dx$ are the minimum and maximum of $\phi(a)\int_a^x f(t)dt$ for $x\in [a,b]$.

Now this is where my solution has hit a problem. Assume that the smallest and largest of the sums $\phi(a)\sum_{i=1}^k f(t_{i-1})(t_i-t_{i-1})$ for $k \in \{1, \dots, n\}$, are $\phi(a)\sum_{i=1}^k f(t_{i-1})(t_i-t_{i-1})$, and $\phi(a)\sum_{i=1}^l f(t_{i-1})(t_i-t_{i-1})$, respectively.

Then we have, combining the inequalities (1) and (2) in the solution and the first thing proven in problem (c),

$\phi(a)\int_a^{t_k} f(x)dx-(\phi(a)+1)\epsilon \le \phi(a)\sum_{i=1}^k f(t_{i-1})(t_i-t_{i-1})-\epsilon \le \sum_{i=1}^n f(t_{i-1})\phi(t_{i-1})(t_i-t_{i-1})-\epsilon \lt \int_a^b f(x)\phi(x)dx \lt \sum_{i=1}^n f(t_{i-1})\phi(t_{i-1})(t_i-t_{i-1})+\epsilon \le \phi(a)\sum_{i=1}^l f(t_{i-1})(t_i-t_{i-1})+\epsilon \le \phi(a)\int_a^{t_l} f(x)dx+(\phi(a)+1)\epsilon.$

Since $\epsilon$ is arbitrary, we get

$\phi(a) \int_a^{t_k} f(x)dx \le \int_a^b f(x)\phi(x)dx \le \phi(a) \int_a^{t_l} f(x)dx$, which is the desired conclusion.

However, this answer depends on the "assumption" I made in the bold sentence above. I've tried to prove this part, but have been unsuccessful. I think I followed the intentions of the written sketch in the last paragraph of the solution, but I can't get this part straight by any means. I would immensely appreciate it if anyone could show me how to solve this part, thanks for reading this long writing.

Your assumption is not valid. The smallest and largest sums do not necessarily coincide with the smallest and largest values of $F(x)$, but you can salvage your proof.

Denote as follows

$$S(P) = \sum_{i=1}^n f(t_{i-1})\phi(t_{i-1})(t_i - t_{i-1}), \\ S(P_j) = \sum_{i=1}^j f(t_{i-1})(t_i - t_{i-1}).$$

We have that

$$m \phi(a) \leqslant S(P) \leqslant M \phi(a),$$

where for $1 \leqslant j \leqslant n$

$$m \leqslant S(P_j) \leqslant M.$$

For any $\epsilon >0$ if the mesh of the partition is sufficiently small, then we have both

$$S(P)-\epsilon < \int_a^b f(t) \phi(t) \, dt < S(P) + \epsilon ,$$

and for $1 \leqslant j \leqslant n$

$$\int_a^{x_j} f(t) \, dt -\epsilon < S(P_j) < \int_a^{x_j} f(t) \, dt + \epsilon .$$

Using the first inequality it follows that

$$m\phi(a) - \epsilon < \int_a^b f(t)\phi(t) \, dt < M\phi(a) + \epsilon.$$

The second inequality implies

$$\inf_{x \in [a,b]}\int_a^{x} f(t) \, dt - \epsilon < m \leqslant M < \sup_{x \in [a,b]}\int_a^{x} f(t) \, dt + \epsilon.$$

Combining we find that for any $\epsilon > 0$ we have

$$\phi(a)\inf_{x \in [a,b]}\int_a^{x} f(t) \, dt - \epsilon(1 + \phi(a)) \leqslant \int_a^b f(t)\phi(t) \, dt \leqslant \phi(a)\sup_{x \in [a,b]}\int_a^{x} f(t) \, dt + \epsilon(1 + \phi(a)),$$

and the conclusion follows since $\epsilon$ can be made arbitrarily small.

• I think I almost got it from your answer but just a few questions. First, shouldn't $S(P_j)$ be $\sum_{i=1}^j f(t_{i-1})(t_i-t_{i-1})$? I don't see how the second inequality makes sense if $\phi(t_{i-1})$ is includes in $S(P_j)$. Second, I'm guessing that the inequality involving $S(P_j)$ and $\int_a^{x_j}f$ follows from the claim proven in the solution? And in that case shouldn't it be $\le$ instead of the strict inequality? Finally, how does the second inequality imply the infimum of $\int_a^x f -\epsilon \le m$ and the right side as well? Commented Sep 21, 2015 at 1:44
• @takecare: On your first comment - sorry, that is a typo on my part. Will correct.
– RRL
Commented Sep 21, 2015 at 2:00
• Regarding the second question on the inequality involving $S(P_j)$ and the integral -- choosing a sufficiently fine partition we can ensure that a finite number of Riemann sums are all within $\epsilon$ of the respective integrals.
– RRL
Commented Sep 21, 2015 at 2:09
• If $\int_a^{x_j} f(t) \, dt -\epsilon < S(P_j)$ and $m = \min_{1 \leqslant j \leqslant n} S(P_j)$, then $\inf_{x} \int_a^{x} f(t) \, dt -\epsilon < m.$
– RRL
Commented Sep 21, 2015 at 3:15