Another nontrivial (trick) question in Chapter 7

[rudinreader]

Problem 13 in Chapter 7 of baby Rudin states:

Assume $$\{f_n\}$$ is a sequence of monotonically increasing functions on R with $$0 \leq f_n(x) \leq 1$$ for all x and all n.
(a) Prove that there is a function f and a sequence $$\{n_k\}$$ such that $$f(x) = \lim_{k \rightarrow \infty}f_{n_k}(x)$$ for every x in R.
(b) If, moreover, f is continuous, prove that $$f_{n_k} \rightarrow f$$ uniformly on R.

Anyone else notice a problem with (b)?

 

[rudinreader]

I have boldy decided to post what I think is a misleading error in baby Rudin. And it's slightly less trivial than the error in Definition 1.5 (ii). I urge anyone to correct me if I'm wrong.

Nonetheless, I can even strengthen the hypothesis with $$f_1 \leq f_2 \leq ...$$ and get a counterexample to (b). Set $$f_n(x) = 0, x \leq -n, f_n(x) = 1, x > -n$$. I.e. $$f_n(x) = I(x+n)$$ is a sequence of step functions. In particular, step functions are monotonic increasing, so they satisfy the hypothesis. This is a pointwise convergent sequence, every subsequence converges to the same thing, $$f(x) = 1$$. f is continuous, but the convergence is not uniform.

However, it converges uniformly on compact subsets. In summary, I think that's a typo. (b) should read $$f_{n_k} \rightarrow f$$ uniformly on compact subsets of R.

P.S. You can also construct a sequence of strictly increasing functions that will yield the same counterexample. I think you can modify $$f_n(x) = tan^{-1}(x+n)$$ appropriately.