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