$\{f_n\}$ converges uniformly but $\lim_{n\rightarrow\infty}f_n$ doesn't exist

Question

I'm asked to give an example of a sequence $\left\{f_{n}\right\}$ of functions and a function $f$ such that

(a) $f_{\mathrm{n}} \in \mathscr{R}[a, b]$ for every positive integer $n$

(b) $f \in \mathscr{R}[a, b]$

(c) $\lim _{n \rightarrow \infty} \int_{a}^{b} f_{n}=\int_{a}^{b} f$

(d) $\lim _{n \rightarrow \infty} f_{n}(x)$ does not exist for any $x \in[\mathrm{a}, \mathrm{b}]$

I think this is equivalent to giving a sequence of function $\{g_n\}$ such that $g_n$ converges to $g$ uniformly but $g_n'$ doesn't exist. But I can't find such an example

Answer 1

As another poster pointed out, there is a classic example that fulfills your request. But since this seems like a homework problem, it might be nice to see what the thought process might go like to get this yourself.

Without loss of generality, we can instead ask for a sequence $\{g_n\}$ of functions on $[a,b]$ such that each $g_n$ is integrable, $\lim_{n \to \infty} \int_a^b g_n =0$, but $\{g_n\}$ converges pointwise nowhere. There are basically two ways to make the integral go to zero:

(1) each $g_n$ is, say, positive but has 'mass' trending to zero, or

(2) each $g_n$ has positive and negative parts which cancel out in the integral.

Think about the ways you can get these conditions to happen but in a somehow incoherent way, so that for each $x \in [a,b]$ the sequence of values $g_n(x)$ does just bounces around without limit.

Answer 2

The typewriter sequence works. Take $f=0$ to get what you want.

Answer 3

For $x \in [0,1]$ define $f_n(x) = (-1)^n([2x]-\frac{1}{2})$ (where $[\ ]$ is integer part), and $f(x)=0$