Exercise 7.2.5

Since both $U(f,P_n)$ and $U(f)$ are defined in terms of supremums (and likewise for $L$ and infimums), it’s useful to have the following lemma: For all $n\in N$, let $A_n$ be a set of real numbers with supremum $s_n$ and infimum $i_n$, and let $B$ be a set with supremum $s$ and infimum $i$. If $\forall <!--\; FUNCTION\; APPLICATION\; -->𝜖>0,\exists <!--\; FUNCTION\; APPLICATION\; -->N$ so that $n>N$ implies that $\forall <!--\; FUNCTION\; APPLICATION\; -->a\in A_n,\exists <!--\; FUNCTION\; APPLICATION\; -->b\in B$ such that $|a-b|<𝜖$, then $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{s}_n=s$ and $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{i}_n=i$.

Proof: Let $𝜖>0$, and choose $N$ large enough so that for $n>N$, $\forall <!--\; FUNCTION\; APPLICATION\; -->a\in A_n,\exists <!--\; FUNCTION\; APPLICATION\; -->b\in B$ with $|a-b|<𝜖∕2$. Note this implies $a<b+𝜖∕2$. Since $s_n$ is the supremum of $A_n$ we have that for some $a\in A_n$,

A similar argument in reverse shows that $s<s_n+𝜖$, or $|s_n-s|<𝜖$, as desired. The proof is identical for $i_n\to i$.

Back to the main proof — consider a particular interval $[c,d]$ which is part of a partition $P_m$. We can consider

to be the contribution of the interval $[c,d]$ to $U(f_n,P_m)$, and similarly define

. Because $(f_n)\to f$ uniformly, we can apply our lemma to claim that as $n\to \infty$, $u_{n,c,d}\to u_{c,d}$. Since the interval $[c,d]$ is arbitrary, we can apply this to each interval in $P_m$ to get $U(f_n,P_m)\to U(f,P_m)$.

This, plus using our lemma again, implies that $U(f_n)\to U(f)$. A similar argument shows $L(f_n)\to L(f)$. But since $U(f_n)=L(f_n)$, we must have $U(f)=L(f)$ and therefore $f$ is integrable on $[a,b]$.