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?
Best Answer
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, $\displaystyle\int_a^bf_1\,\mathrm d\alpha\leqslant\int_a^bf_2\,\mathrm d\alpha$.