Good evening/afternoon/morning!
I am in a situation here where I think I have found a solution but it all feels too simple and good to be true. Here is our problem:
Let $f_n(x):[0,1] \to [0,1]$ for $n\in\mathbb{N}$ by :
$f_n(x)= \begin{cases} nx & x\in[0,\frac{1}{n}] \\ 1 & x \in(\frac{1}{n},1] \end{cases}$
Let $f$ be the Pointwise Limit of $f_n$ Show that there is no set $B ⊂ [0, 1]$ of Lebesgue measure $0$ so that $f_n$ to $f$ converges uniformly on the set $[0, 1] \backslash B$ (= $[0,1]\cap B^c$)
Here is a sketch of the "fantastic" solution I think I have found:
Firstly $f = 1$ everywhere but $f(0) = 0 $
Let $I:= [0,1] \backslash B$ be the domain in which $f_n$ converges uniformly. Uniform convergence (which is the same limit of pointwise) implies Riemann Integrability which inturn implies a Lebesgue Integral with the same value. Also due to the uniform convergence the limit of the integral is the integral of the limit and hence:
$ \lim_{n \to \infty}\int_If_ndx=\int_Ifdx = \int_Ifd\lambda=\int_I1\cdot d\lambda = \lambda(I)$
And hence all I would need to do in theory would be to show that $ \lim_{n \to \infty}\int_If_ndx < 1 $ And im done as $I$ and $B$ are disjoint and $\lambda(B) = 1-\lambda(I) > 0$
Any pointers or help would be greatly appreciated. Also any way to show $ \lim_{n \to \infty}\int_If_ndx < 1 $ would be amazing!
You can argue in this way.
Assume, by the way of contradiction, that there is $B\subset[0,1]$ with $m(B)=0$ and such that $f_n\to f$ uniformly on $[0,1]\setminus B$.
For any $n\in\Bbb N$, let $I_n=\left(0,\frac{1}{n^2}\right)$. Since $m(I_n)=\frac{1}{n^2}>0$ and $m(B)=0$, then $m(I_n\setminus B)=m(I_n)-m(I_n\cap B)=\frac{1}{n^2}-0>0$; and, in particular, $I_n\setminus B\ne\varnothing$ and there is $x_n\in I_n\setminus B$. So we have constructed a sequences $(x_n)_n\subset[0,1]\setminus B$ with $0<x_n<\frac{1}{n^2}$, in particular, $0<x_n<\frac{1}{n}$. Hence $$\sup_{x\in [0,1]\setminus B} |f_n(x)-f(x)| \ge |f_n(x_n)-f(x_n)|=\left|n x_n-1\right|\to1\ (n\to\infty)$$ (observe that $0<nx_n<n\frac{1}{n^2}=\frac{1}{n}$)wich is a contradiction with $f_n\to f$ uniformly on $[0,1]\setminus B$.