# Theorem 6.12 (b) in Baby Rudin: If $f_1 \leq f_2$ on $[a, b]$, then $\int_a^b f_1 d\alpha \leq \int_a^b f_2 d\alpha$

Suppose $f_1$ and $f_2$ are Riemann-integrable with respect to $\alpha$ over $[a, b]$. If $f_1(x) \leq f_2(x)$ on $[a, b]$, then $$\int_a^b f_1 d \alpha \leq \int_a^b f_2 d \alpha.$$

This is (essentially) Theorem 6.12 (b) in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition.

Here is my proof:

As $f_1 \leq f_2$ on $[a, b]$, so, for any partition $P = \left\{ \ x_0, \ldots, x_n \right\}$ of $[a, b]$, we have $$\inf_{x_{i-1} \leq x \leq x_i } f_1(x) \leq \inf_{x_{i-1} \leq x \leq x_i } f_2(x), \ \mbox{ and } \ \sup_{x_{i-1} \leq x \leq x_i } f_1(x) \leq \sup_{x_{i-1} \leq x \leq x_i } f_2(x)$$ for each $i = 1, \ldots, n$, and therefore $$L \left( P, f_1, \alpha \right) \leq L \left( P, f_2, \alpha \right), \ \mbox{ and } \ U \left( P, f_1, \alpha \right) \leq U \left( P, f_2, \alpha \right) \tag{0}$$ for every partition $P$ of $[a, b]$.

Now as $f_1 \in \mathscr{R}(\alpha)$ and $f_2 \in \mathscr{R}(\alpha)$, so, for $j = 1, 2$, we have
$$\sup \left\{ \ L \left( P, f_j, \alpha \right) \ \colon \ P \mbox{ is a partition of } [a, b] \ \right\} = \int_a^b f_j d \alpha = \inf \left\{ \ U \left( P, f_j, \alpha \right) \ \colon \ P \mbox{ is a partition of } [a, b] \ \right\}.$$ Therefore, for $j = 1, 2$, we have $$L \left( P, f_j, \alpha \right) \leq \int_a^b f_j d \alpha \leq U \left( P, f_j, \alpha \right) \tag{1}$$ for every partition $P$ of $[a, b]$; moreover, for every real number $\delta > 0$, there exist partitions $P_j$, $Q_j$ of $[a, b]$ such that $$\int_a^b f_j d \alpha - \delta < L \left( P_j, f_j, \alpha \right), \mbox{ and } U \left( Q_j, f_j, \alpha \right) < \int_a^b f_j d \alpha + \delta, \tag{2}$$ and, hence if $S_j$ is any partition of $[a, b]$ such that $S_j \supset P_j$ and $S_j \supset Q_j$, then (by Theorem 6.4 in Baby Rudin, 3rd edition) we must have $$L \left( P_j, f_j, \alpha \right) \leq L \left( S_j, f_j, \alpha \right) \leq U \left( S_j, f_j, \alpha \right) \leq U \left( Q_j, f_j, \alpha \right). \tag{3}$$ From (2) and (3) we can conclude that, for each $j = 1, 2$, there exists a partition $S_j$ of $[a, b]$ such that $$\int_a^b f_j d \alpha - \delta < L \left( S_j, f_j, \alpha \right) \leq U \left( S_j, f_j, \alpha \right) < \int_a^b f_j d \alpha + \delta. \tag{4}$$ Now let $P$ be any partition of $[a, b]$ such that $P \supset S_1$ and $P \supset S_2$. Then (again by Theorem 6.4 in Baby Rudin, 3rd edition) we have for each $j = 1, 2$,
$$L \left( S_j, f_j, \alpha \right) \leq L \left( P, f_j, \alpha \right) \leq U \left( P, f_j, \alpha \right) \leq U \left( S_j, f_j, \alpha \right). \tag{5}$$

Thus, for every real number $\delta > 0$, we see that \begin{align} \int_a^b f_1 d\alpha &\leq U \left( P, f_1, \alpha \right) \qquad \mbox{ [ by (1) above ] } \\ &\leq U \left( P, f_2, \alpha \right) \qquad \mbox{ [ by (0) above ] } \\ & \leq U \left( S_2, f_2, \alpha \right) \qquad \mbox{ [ by (5) ] } \\ & < \int_a^b f_2 d \alpha + \delta \qquad \mbox{ [ by (4) ] }, \end{align} which implies that $$\int_a^b f_1 d \alpha \leq \int_a^b f_2 d \alpha,$$ as required.

Is this proof correct? If so, then is my presentation clear and optimal enough? If not, then where lie the pitfalls in my reasoning? Have I superfluously used any of the partitions $P_j$, $Q_j$, $S_j$ for $j = 1, 2$, or the partition $P$ at the end?

• The proof may be simplified by using the linearity (proved in text by Rudin), and $\int_a^b g d \alpha \geq 0$ if $g$ is nonegative, for which the proof can be made shorter, at least to save some notations. – Zhanxiong Jul 10 '17 at 3:52
• @Zhanxiong but the second one has not been stated or proved by Rudin until this point in the book, but it can of course be derived as a consequence of this result. – Saaqib Mahmood Jul 10 '17 at 3:54
• I mean, instead of proving this result directly, you can prove my second statement instead. The result you want to prove then follows from the linearity. The possible gain of doing so (though no essential difference from attacking the original statement directly) is to borrow an existed result and circumvent defining notations for two functions. Nevertheless, your proof is good. – Zhanxiong Jul 10 '17 at 4:09

Your proof is correct, but it can be shortened. Since, for each partition $$P$$, $$L(f_1,P,\alpha)\leqslant L(f_2,P,\alpha)$$,$$\sup\bigl\{L(f_1,P,\alpha)\,|\,P\text{ is a partition of }[a,b]\bigr\}\leqslant\sup\bigl\{L(f_2,P,\alpha)\,|\,P\text{ is a partition of }[a,b]\bigr\}.$$Therefore, $$\displaystyle\int_a^bf_1\,\mathrm d\alpha\leqslant\int_a^bf_2\,\mathrm d\alpha$$.

Your proof is correct. I think, however, you can also shorten the proof by simply using the definition of the (simple) Riemann integral that is not formulated in terms of lower and upper sums. I'm using the following definition:

$f \in \mathcal{R}[a,b]$ if and only if for eveyr $\varepsilon > 0$, there exists a $\delta > 0$ such that for every partition, $P$ such that $|| {P} || < \delta,$ we have that $| \; S(f, P) - \int_{a}^{b} f \;| < \varepsilon$. Note that here, $S(P, f)$ denotes the Riemann Sum and the integrator is asssumed to be $x$. So this is just the standard Riemman integral.

We simply apply this definition twice, and use the triangle inequality. For $\varepsilon > 0,$ there exists partitions, $P_1$ and $P_2$ and $\delta_1$ and $\delta_2$ such that,

\begin{align} ||P_1|| < \delta_1 \quad & \text{implies} \quad \Bigg| \; S(f_1, P_1) - \int_{a}^{b} f_1 \; \Bigg| < \varepsilon/2 \quad \; \text{implies} \; - \varepsilon/2 + \int_{a}^{b} f_1 < S(f_1, P_1), \tag{1} \\ ||P_2|| < \delta_2 \quad & \text{implies} \quad \Bigg| \; S(f_2, P_2) - \int_{a}^{b} f_2 \; \Bigg| < \varepsilon/2 \quad \; \text{implies} \; \; \; \; S(f_2, P_2) < \varepsilon/2 + \int_{a}^{b} f_2 \tag{2} \end{align}

Now, since $f_1 \leq f_2$, we have that the $S(f_1, P_1) \leq S(f_2, P_2)$. Therefore, using $(1)$ and $(2)$, we have that,

\begin{align} \int_{a}^{b} f_1 < \int_{a}^{b} f_1 + \varepsilon \quad \text{for every} \; \varepsilon > 0. \end{align}

Hence, it is clear that the stated conclusion holds.

• thank you for your answer. However, since I'm doing a problem in Baby Rudin, it is more appropriate to use the definitions and theorems that he has placed at our disposal. Are you from Pakistan? If so, then where? – Saaqib Mahmood Aug 9 '17 at 14:23
• Yes, absolutely. Apostol's textbook takes the same approach to the Riemann integral, and I like that book more. Lahore. – Junaid Aftab Aug 9 '17 at 15:07
• @SaaqibMahmuud Also, this integral is a special case of the more general integral, so I think it's good to know multiple approaches. Of course, one forgets the details, but I think it's good to see all of them. – Junaid Aftab Aug 9 '17 at 15:08
• I have yet to read the chapter on integration in Apostol. I've so far only managed to cover that wonderful text upto Chap. 5, and part of Chap. 6. Are you a student? If so, at what level and at which university? – Saaqib Mahmood Aug 9 '17 at 15:26
• @SaaqibMahmuud Oh, I have read it, but I forget details over time so I need to revise it as well. I just graduated. I should have been a mathematics major. – Junaid Aftab Aug 9 '17 at 17:05