I'm reading Abbot's Understanding Analysis, and have stumbled across this problem, but there is a step that I don't fully understand.
The problem is:
Assume that for each $n$, $f_n$ is an (Riemann) integrable function on $[a, b]$. If $(f_n) \rightarrow f$ uniformly on $[a, b]$, prove that $f$ is also integrable on this set.
Solution:
$$U(f, P) – L(f, P)$$$$=U(f, P) – U(f_N, P) + U(f_N, P) – L(f_N, P) + L(f_N, P) – L(f, P)$$$$\leq |U(f, P) – U(f_N, P)| + (U(f_N, P) – L(f_N, P)) + |L(f_N, P) – L(f, P)|$$
Let $\epsilon > 0$ be arbitrary. Because $f_n \rightarrow f$ uniformly, we can choose $N$ so that:
$|f_N(x) – f(x)| \leq \dfrac{\epsilon}{3(b-a)}$ for all $x \in [a, b]$
Now the function $f_N$ is integrable and so there exists a partition $P$ for which
$U(f_N, P) – L(f_N, P) < \dfrac{\epsilon}{3}$
Let's consider a particular subinterval $[x_{k-1}, x_k]$ from this partition. If
$M_k = \sup\{f(x): x \in [x_{k -1}, x_k]\}$ and $N_k = \sup\{f_N(x): x \in [x_{k -1}, x_k]\}$
then our choice of f_N guarantees that
$|M_k – N_k| \leq \dfrac{\epsilon}{3 (b – a)}$
The proof goes on and it demonstrates that:
$U(f, P) – L(f, P) < \epsilon / 3 + \epsilon / 3 + \epsilon / 3$
My doubt is why is it so that
$|M_k – N_k| \leq \dfrac{\epsilon}{3 (b – a)}$
We have that
$|f_N(x) – f(x)| \leq \dfrac{\epsilon}{3(b-a)}$ for all $x \in [a, b]$
But this only work for the same $x$. The supremums can have different $x$ value, or they may even not have a $x$ value.
Best Answer
$M_k-N_k>\frac{\epsilon}{3(b-a)}$ would imply $f(x)-N_k>\frac{\epsilon}{3(b-a)}$ for some $x$, and this would imply $f_N(x)>N_k$.