Riemann integrability and uniform convergence

[lys04]

Was reading the Reimann integrals chapter of Understanding Analysis by Stephen Abbott and got stuck on exercise 7.2.5. In the solutions they went from having |f-f_n|<epsilon/3(a-b) to having |M_k-N_k|<epsilon/3(a-b), but I’m confused how did they do this. We know that fn uniformly converges to f, that means for any epsilon greater than 0 I can find natural number M st when n is greater than M then the distance from fn to f is less than epsilon for all x in the interval [a,b], but the supremums of fn and f in a sub interval [x_k-1,x_k] might occur at different values and I’m not sure how they’re meant to be at most epsilon apart.

 

[Office_Shredder]

I suspect it's a typo and the 3 in the denominator is supposed to be gone.

Edit: actually I take it back. Let's say ##M_k>N_k##. Then where ##M_k## is realized, ##f_N## is within ##\epsilon/(3(b-a))## of ##M_k## and is guaranteed to not be larger than ##N_k##, showing the inequality in your post is correct. If ##N_k\leq M_k## just do the logic in the other direction

 

[pasmith]

We have for [itex]n > N[/itex] that (dropping the constant multiple of [itex]\epsilon[/itex]) [itex]|f_n(x) - f(x)| < \epsilon[/itex]. We can write this in the following two ways: [tex]
\begin{split}
f_n(x) - \epsilon &< f(x) < f_n(x) + \epsilon \\
f(x) - \epsilon &< f_n(x) < f(x) + \epsilon. \end{split}[/tex] Ignoring the lower bounds and making the right hand sides as large as possible, we find that on the particular subinterval [tex]
\begin{split}
f(x) &< f_n(x) +\epsilon \leq N_k + \epsilon \\
f_n(x) &< f(x) + \epsilon \leq M_k + \epsilon.\end{split}[/tex] Now the first tells us that [itex]N_k + \epsilon[/itex] is an upper bound for [itex]f(x)[/itex], so that [tex]M_k \leq N_k + \epsilon.[/tex] Simiilarly the second tells us that [itex]M_k + \epsilon[/itex] is an upper bound for [itex]f_n(x)[/itex] so that [tex]
N_k \leq M_k + \epsilon.[/tex] Putting these two inequlities together gives [tex]
|N_k - M_k| \leq \epsilon[/tex] as required.