Proof. Let $F$ be a function, and $[a, b]$ an interval over which $F$ is integrable. We shall first show that, for every partition $\sigma$ of $[a, b]$, the upper and lower sums, $U_\sigma$ and $L_\sigma$, satisfy the inequalities
$$
L_\sigma \leq W(F, a, b) \leq U_\sigma
$$
To do this, we let $\sigma=\left{x_0, \ldots, x_n\right}$ and assume the usual ordering:
$$
a=x_0 \leq x_1 \leq \cdots \leq x_n=b .
$$
As we have done in the past, we denote by $M_i$ the least upper bound of the values of $F$ on the ith subinterval $\left[x_{i-1}, x_i\right]$, and by $m_i$ the greatest lower bound. Then
$$
m_i \leq F(x) \leq M_i, \quad \text { whenever } x_{i-1} \leq x \leq x_i
$$
The two constant functions with values Mi and mi, respectively, are certainly integrable over the subinterval $\left[x_{i-1}, x_i\right]$. Following the common practice of denoting a constant function by its value, we know, as a result of (5.2), that
$$
W\left(m_i, x_{i-1}, x_i\right) \leq W\left(F, x_{i-1}, x_i\right) \leq W\left(M_i, x_{i-1}, x_i\right)
$$
Using (5.3), we obtain
$$
\begin{aligned}
& W\left(m_i, x_{i-1}, x_i\right)=m_i\left(x_i-x_{i-1}\right) \
& W\left(M_i, x_{i-1}, x_i\right)=M_i\left(x_i-x_{i-1}\right)
\end{aligned}
$$
Hence
$$
m_i\left(x_i-x_{i-1}\right) \leq W\left(F, x_{i-1}, x_i\right) \leq M_i\left(x_i-x_{i-1}\right) .
$$
Adding these inequalities for $i=1, \ldots, n$, we get
$$
\sum_{i=1}^n m_i\left(x_i-x_{i-1}\right) \leq \sum_{i=1}^n W\left(F, x_{i-1}, x_i\right) \leq \sum_{i=1}^n M_i\left(x_i-x_{i-1}\right) .
$$
The left and right sides of the inequalities in the preceding equation are precisely $L_\sigma$ and $U_\sigma$, respectively. It follows from repeated use that
$$
\sum_{i=1}^n W\left(F, x_{i-1}, x_i\right)=W(F, a, b)
$$
and we have therefore proved that the inequalities do hold.
now essentially complete. Let $\sigma$ and $\tau$ be two arbitrary partitions of $[a, b]$. The union $U \cup T$ is the partition which is the common refinement of both. It is shown in the last line of the proof of Proposition (1.1), page 168, that
$$
L_\sigma \leq L_{\sigma \cup \tau} \leq U_{\sigma \cup \tau} \leq U_\tau
$$

Proof. If $a=b$, the conclusion follows at once since $\int_a^b f=\int_a^a f=0$. Hence we shall assume that $a<b$. To be specific, we shall furthermore assume that $f$ is continuous at every point of $[a, b]$ except at $a$. The necessary mod)fication in the argument if a discontinuity occurs at $b$ (or at both $a$ and $b$ ) should be obvious. According to the definition of integrability (page 168), it is sufficient to prove that there exist partitions $\sigma$ and $\tau$ of $[a, b]$ such that $U_\sigma-L_\tau$, the difference between the corresponding upper and lower sums, is arbitrarily small. For this purpose, we choose an arbitrary positive number $\epsilon$. Since $f$ is bounded on $[a, b]$, there exists a positive number $k$ such that $|f(x)| \leq k$, for every $x$ in $[a, b]$. We next pick a point $a^{\prime}$ which is in $[a, b]$ and sufficiently close to a that
$$
0<a^{\prime}-a<\frac{\epsilon}{3 k}
$$
(see Figure 26). Since $f$ is continuous on the smaller interval $\left[a^{\prime}, b\right]$, we know that $f$ is integrable over it. Hence there exist partitions $\sigma^{\prime}$ and $\tau^{\prime}$ of $\left[a^{\prime}, b\right]$ such that the upper sum $U_{\sigma^{\prime}}$, and lower sum $L_{\tau^{\prime}}$ for $f$ satisfy
$$
\left|U_{\sigma^{\prime}}-L_{\tau^{\prime}}\right|<\frac{\epsilon}{3}
$$
Let $\sigma$ and $\tau$ be the partitions of $[a, b]$ obtained from $\sigma^{\prime}$ and $\tau^{\prime}$ respectively, by adjoining the point $a$; i.e., $\sigma=\sigma^{\prime} \cup{a}$ and $\tau=\tau^{\prime} \cup{a}$. Since the maximum value of $|f(x)|$ on the subinterval $\left[a, a^{\prime}\right]$ is less than or equal to $k$, it follows that
$$
\left|U_\sigma-U_{\sigma^{\prime}}\right| \leq k\left(a^{\prime}-a\right)<k \cdot \frac{\epsilon}{3 k}=\frac{\epsilon}{3} .
$$
By the same argument, we have
$$
\left|L_{\tau^{\prime}}-L_\tau\right| \leq k\left(a^{\prime}-a\right)<k \cdot \frac{\epsilon}{3 k}=\frac{\epsilon}{3} .
$$
Next, consider the algebraic identity
$$
U_\sigma-L_\tau=\left(U_\sigma-U_{\sigma^{\prime}}\right)+\left(L_{\tau^{\prime}}-L_\tau\right)+\left(U_{\sigma^{\prime}}-L_{\tau^{\prime}}\right) .
$$

Proof. It is sufficient to prove this theorem under the assumption that the values of $f$ and $g$ differ at only a single point $c$ in the interval $[a, b]$ (because the result can then be iterated). To be specific, we shall assume that $f(c)<g(c)$. The proof is completed if we can show that there exist upper and lower sums for $g$ which differ from the integral $\int_a^b f$ by an arbitrarily small amount. For this purpose, we choose an arbitrary positive number $\epsilon$. Since $f$ is, by hypothesis, integrable over $[a, b]$, there exists a partition $\tau$ of $[a, b]$ such that the corresponding lower sum for $f$, which we denote by $L_\tau(f)$, satisfies
$$
\int_a^b f L_\tau(f)<\epsilon .
$$
However, every lower sum for $f$ is also a lower sum for $g$. Hence we may substitute $L_\tau(g)$ for $L_\tau(f)$ in the preceding inequality and obtain
$$
\int_a^b f-L_\tau(g)<\epsilon .
$$
We next derive a similar inequality involving an upper sum for $g$. The integrability of $f$ also implies the existence of a partition $\sigma^{\prime}$ of $[a, b]$ such that the corresponding upper sum for $f$ satisfies
$$
U_{\sigma^{\prime}}(f)-\int_a^b f<\frac{\epsilon}{2} .
$$
By possibly adjoining to $\sigma^{\prime}$ a point on either side of $c$, we can assure ourselves of getting a partition $\sigma=\left{x_0, \ldots, x_n\right}$ of $[a, b]$ with the property that if $c$ lies in the ith subinterval $\left[x_{i-1}, x_i\right]$, then
$$
x_i-x_{i-1}<\frac{\epsilon}{2[g(c)-f(c)]}
$$